How to check is there any negative term in a large list?Issue with very large lists in MathematicaHow to check if all the members of list lies in specific rangeQuery Dataset to check if row contains any from a set of valuesDeleting any list that contains a negative numberDefining a function that detects square matricesHow to use Contains functions on matrices?Ordering real numeric quantitiescount the pairs in a set of DataSpeed up Flatten[] of a large nested listHow to check expression depends on symbol in a particular way

How to set the font color of quantity objects (Version 11.3 vs version 12)?

What is a Recurrent Neural Network?

Why is current rating for multicore cable lower than single core with the same cross section?

Toggle Overlays shortcut?

Why do computer-science majors learn calculus?

Is creating your own "experiment" considered cheating during a physics exam?

Why are the 2nd/3rd singular forms of present of « potere » irregular?

How can Republicans who favour free markets, consistently express anger when they don't like the outcome of that choice?

Was there a Viking Exchange as well as a Columbian one?

If Earth is tilted, why is Polaris always above the same spot?

Will tsunami waves travel forever if there was no land?

Pressure to defend the relevance of one's area of mathematics

Where did the extra Pym particles come from in Endgame?

Why the difference in metal between 銀行 and お金?

Is GOCE a satellite or aircraft?

How does a Swashbuckler rogue "fight with two weapons while safely darting away"?

What are the spoon bit of a spoon and fork bit of a fork called?

Reverse the word in a string with the same order in javascript

Subtleties of choosing the sequence of tenses in Russian

Can I get candy for a Pokemon I haven't caught yet?

Python "triplet" dictionary?

When India mathematicians did know Euclid's Elements?

Packing rectangles: Does rotation ever help?

Can someone publish a story that happened to you?



How to check is there any negative term in a large list?


Issue with very large lists in MathematicaHow to check if all the members of list lies in specific rangeQuery Dataset to check if row contains any from a set of valuesDeleting any list that contains a negative numberDefining a function that detects square matricesHow to use Contains functions on matrices?Ordering real numeric quantitiescount the pairs in a set of DataSpeed up Flatten[] of a large nested listHow to check expression depends on symbol in a particular way













8












$begingroup$


I want to check if a data set of size $10^10$ contains any non-positive elements. Positive[Name of dataset] returns a list of True and False of length $10^10$. I want only a single True if all terms of that dataset are positive and False otherwise.










share|improve this question











$endgroup$







  • 3




    $begingroup$
    VectorQ[list, Positive]?
    $endgroup$
    – J. M. is away
    Mar 27 at 14:26






  • 1




    $begingroup$
    Use Apply as in And @@ Positive[list]
    $endgroup$
    – Bob Hanlon
    Mar 27 at 14:52










  • $begingroup$
    How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (use Positive), or do you need all terms to be zero or positive (use NonNegative)?
    $endgroup$
    – Roman
    Mar 27 at 15:01















8












$begingroup$


I want to check if a data set of size $10^10$ contains any non-positive elements. Positive[Name of dataset] returns a list of True and False of length $10^10$. I want only a single True if all terms of that dataset are positive and False otherwise.










share|improve this question











$endgroup$







  • 3




    $begingroup$
    VectorQ[list, Positive]?
    $endgroup$
    – J. M. is away
    Mar 27 at 14:26






  • 1




    $begingroup$
    Use Apply as in And @@ Positive[list]
    $endgroup$
    – Bob Hanlon
    Mar 27 at 14:52










  • $begingroup$
    How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (use Positive), or do you need all terms to be zero or positive (use NonNegative)?
    $endgroup$
    – Roman
    Mar 27 at 15:01













8












8








8


2



$begingroup$


I want to check if a data set of size $10^10$ contains any non-positive elements. Positive[Name of dataset] returns a list of True and False of length $10^10$. I want only a single True if all terms of that dataset are positive and False otherwise.










share|improve this question











$endgroup$




I want to check if a data set of size $10^10$ contains any non-positive elements. Positive[Name of dataset] returns a list of True and False of length $10^10$. I want only a single True if all terms of that dataset are positive and False otherwise.







list-manipulation expression-test






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 27 at 19:01









mjw

1,33910




1,33910










asked Mar 27 at 13:57









a ba b

712




712







  • 3




    $begingroup$
    VectorQ[list, Positive]?
    $endgroup$
    – J. M. is away
    Mar 27 at 14:26






  • 1




    $begingroup$
    Use Apply as in And @@ Positive[list]
    $endgroup$
    – Bob Hanlon
    Mar 27 at 14:52










  • $begingroup$
    How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (use Positive), or do you need all terms to be zero or positive (use NonNegative)?
    $endgroup$
    – Roman
    Mar 27 at 15:01












  • 3




    $begingroup$
    VectorQ[list, Positive]?
    $endgroup$
    – J. M. is away
    Mar 27 at 14:26






  • 1




    $begingroup$
    Use Apply as in And @@ Positive[list]
    $endgroup$
    – Bob Hanlon
    Mar 27 at 14:52










  • $begingroup$
    How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (use Positive), or do you need all terms to be zero or positive (use NonNegative)?
    $endgroup$
    – Roman
    Mar 27 at 15:01







3




3




$begingroup$
VectorQ[list, Positive]?
$endgroup$
– J. M. is away
Mar 27 at 14:26




$begingroup$
VectorQ[list, Positive]?
$endgroup$
– J. M. is away
Mar 27 at 14:26




1




1




$begingroup$
Use Apply as in And @@ Positive[list]
$endgroup$
– Bob Hanlon
Mar 27 at 14:52




$begingroup$
Use Apply as in And @@ Positive[list]
$endgroup$
– Bob Hanlon
Mar 27 at 14:52












$begingroup$
How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (use Positive), or do you need all terms to be zero or positive (use NonNegative)?
$endgroup$
– Roman
Mar 27 at 15:01




$begingroup$
How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (use Positive), or do you need all terms to be zero or positive (use NonNegative)?
$endgroup$
– Roman
Mar 27 at 15:01










4 Answers
4






active

oldest

votes


















13












$begingroup$

Alternate solution:



list = RandomReal[1, 10^6];
Min[list] >= 0





share|improve this answer









$endgroup$








  • 3




    $begingroup$
    ...i.e. NonNegative[Min[list]].
    $endgroup$
    – J. M. is away
    Mar 27 at 16:08






  • 3




    $begingroup$
    Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
    $endgroup$
    – Henrik Schumacher
    Mar 27 at 18:01






  • 2




    $begingroup$
    This is about 100 times faster than any of the other solutions. Impressive!
    $endgroup$
    – Roman
    Mar 27 at 20:03










  • $begingroup$
    Thank you very much.
    $endgroup$
    – a b
    Mar 28 at 8:40


















9












$begingroup$

Since you have a very large list, you should look at the timing



list = RandomReal[1, 10^6];

(And @@ Positive[list]) // AbsoluteTiming (* Hanlon *)

(* 0.050573, True *)

VectorQ[list, Positive] // AbsoluteTiming (* J.M. *)

(* 0.261642, True *)

(AnyTrue[list, Negative] // Not) // AbsoluteTiming (* Morbo *)

(* 0.324062, True *)

And @@ (list /. x_?Negative -> False,
x_?Positive -> True) // AbsoluteTiming (* Alrubaie *)

(* 1.00664, True *)


EDIT: As suggested by mjw, encountering a nonpositive value early in the list significantly alters the results.



list2 = ReplacePart[list, 1000 -> -1];

(And @@ Positive[list2]) // AbsoluteTiming (*Hanlon*)

(* 0.277642, False *)

VectorQ[list2, Positive] // AbsoluteTiming (*J.M.*)

(* 0.000223, False *)

(AnyTrue[list2, Negative] // Not) // AbsoluteTiming (*Morbo*)

(* 0.000262, False *)

And @@ (list2 /. x_?Negative -> False,
x_?Positive -> True) // AbsoluteTiming (*Alrubaie*)

(* 1.43026, False *)





share|improve this answer











$endgroup$












  • $begingroup$
    Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
    $endgroup$
    – mjw
    Mar 27 at 15:34






  • 1




    $begingroup$
    Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
    $endgroup$
    – mjw
    Mar 27 at 15:37






  • 1




    $begingroup$
    @mjw, And[] does short-circuit evaluation.
    $endgroup$
    – J. M. is away
    Mar 27 at 15:41






  • 1




    $begingroup$
    @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
    $endgroup$
    – mjw
    Mar 27 at 16:03






  • 1




    $begingroup$
    @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
    $endgroup$
    – J. M. is away
    Mar 27 at 16:06


















4












$begingroup$

Ah, maybe this is too simple, but works for exactly what you're doing:



data = Table[RandomReal[-1,1],i,1,1000];
AnyTrue[data,Negative] // Not
(*False*)

data2 = Table[RandomReal[], i, 1, 10^2];
AnyTrue[data2, Negative] // Not
(*True*)





share|improve this answer











$endgroup$












  • $begingroup$
    AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
    $endgroup$
    – Roman
    Mar 27 at 14:55










  • $begingroup$
    @Roman - the poster is using AnyTrue not AllTrue
    $endgroup$
    – Bob Hanlon
    Mar 27 at 15:00






  • 1




    $begingroup$
    Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
    $endgroup$
    – Roman
    Mar 27 at 15:04











  • $begingroup$
    ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
    $endgroup$
    – morbo
    Mar 27 at 15:04


















3












$begingroup$

list = 1, 2, 3, 4, -5, -6, -7;

list /. x_?Negative -> True, x_?Positive -> False





share|improve this answer









$endgroup$













    Your Answer








    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "387"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194043%2fhow-to-check-is-there-any-negative-term-in-a-large-list%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    13












    $begingroup$

    Alternate solution:



    list = RandomReal[1, 10^6];
    Min[list] >= 0





    share|improve this answer









    $endgroup$








    • 3




      $begingroup$
      ...i.e. NonNegative[Min[list]].
      $endgroup$
      – J. M. is away
      Mar 27 at 16:08






    • 3




      $begingroup$
      Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
      $endgroup$
      – Henrik Schumacher
      Mar 27 at 18:01






    • 2




      $begingroup$
      This is about 100 times faster than any of the other solutions. Impressive!
      $endgroup$
      – Roman
      Mar 27 at 20:03










    • $begingroup$
      Thank you very much.
      $endgroup$
      – a b
      Mar 28 at 8:40















    13












    $begingroup$

    Alternate solution:



    list = RandomReal[1, 10^6];
    Min[list] >= 0





    share|improve this answer









    $endgroup$








    • 3




      $begingroup$
      ...i.e. NonNegative[Min[list]].
      $endgroup$
      – J. M. is away
      Mar 27 at 16:08






    • 3




      $begingroup$
      Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
      $endgroup$
      – Henrik Schumacher
      Mar 27 at 18:01






    • 2




      $begingroup$
      This is about 100 times faster than any of the other solutions. Impressive!
      $endgroup$
      – Roman
      Mar 27 at 20:03










    • $begingroup$
      Thank you very much.
      $endgroup$
      – a b
      Mar 28 at 8:40













    13












    13








    13





    $begingroup$

    Alternate solution:



    list = RandomReal[1, 10^6];
    Min[list] >= 0





    share|improve this answer









    $endgroup$



    Alternate solution:



    list = RandomReal[1, 10^6];
    Min[list] >= 0






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Mar 27 at 16:06









    sakrasakra

    2,8881429




    2,8881429







    • 3




      $begingroup$
      ...i.e. NonNegative[Min[list]].
      $endgroup$
      – J. M. is away
      Mar 27 at 16:08






    • 3




      $begingroup$
      Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
      $endgroup$
      – Henrik Schumacher
      Mar 27 at 18:01






    • 2




      $begingroup$
      This is about 100 times faster than any of the other solutions. Impressive!
      $endgroup$
      – Roman
      Mar 27 at 20:03










    • $begingroup$
      Thank you very much.
      $endgroup$
      – a b
      Mar 28 at 8:40












    • 3




      $begingroup$
      ...i.e. NonNegative[Min[list]].
      $endgroup$
      – J. M. is away
      Mar 27 at 16:08






    • 3




      $begingroup$
      Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
      $endgroup$
      – Henrik Schumacher
      Mar 27 at 18:01






    • 2




      $begingroup$
      This is about 100 times faster than any of the other solutions. Impressive!
      $endgroup$
      – Roman
      Mar 27 at 20:03










    • $begingroup$
      Thank you very much.
      $endgroup$
      – a b
      Mar 28 at 8:40







    3




    3




    $begingroup$
    ...i.e. NonNegative[Min[list]].
    $endgroup$
    – J. M. is away
    Mar 27 at 16:08




    $begingroup$
    ...i.e. NonNegative[Min[list]].
    $endgroup$
    – J. M. is away
    Mar 27 at 16:08




    3




    3




    $begingroup$
    Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
    $endgroup$
    – Henrik Schumacher
    Mar 27 at 18:01




    $begingroup$
    Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
    $endgroup$
    – Henrik Schumacher
    Mar 27 at 18:01




    2




    2




    $begingroup$
    This is about 100 times faster than any of the other solutions. Impressive!
    $endgroup$
    – Roman
    Mar 27 at 20:03




    $begingroup$
    This is about 100 times faster than any of the other solutions. Impressive!
    $endgroup$
    – Roman
    Mar 27 at 20:03












    $begingroup$
    Thank you very much.
    $endgroup$
    – a b
    Mar 28 at 8:40




    $begingroup$
    Thank you very much.
    $endgroup$
    – a b
    Mar 28 at 8:40











    9












    $begingroup$

    Since you have a very large list, you should look at the timing



    list = RandomReal[1, 10^6];

    (And @@ Positive[list]) // AbsoluteTiming (* Hanlon *)

    (* 0.050573, True *)

    VectorQ[list, Positive] // AbsoluteTiming (* J.M. *)

    (* 0.261642, True *)

    (AnyTrue[list, Negative] // Not) // AbsoluteTiming (* Morbo *)

    (* 0.324062, True *)

    And @@ (list /. x_?Negative -> False,
    x_?Positive -> True) // AbsoluteTiming (* Alrubaie *)

    (* 1.00664, True *)


    EDIT: As suggested by mjw, encountering a nonpositive value early in the list significantly alters the results.



    list2 = ReplacePart[list, 1000 -> -1];

    (And @@ Positive[list2]) // AbsoluteTiming (*Hanlon*)

    (* 0.277642, False *)

    VectorQ[list2, Positive] // AbsoluteTiming (*J.M.*)

    (* 0.000223, False *)

    (AnyTrue[list2, Negative] // Not) // AbsoluteTiming (*Morbo*)

    (* 0.000262, False *)

    And @@ (list2 /. x_?Negative -> False,
    x_?Positive -> True) // AbsoluteTiming (*Alrubaie*)

    (* 1.43026, False *)





    share|improve this answer











    $endgroup$












    • $begingroup$
      Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
      $endgroup$
      – mjw
      Mar 27 at 15:34






    • 1




      $begingroup$
      Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
      $endgroup$
      – mjw
      Mar 27 at 15:37






    • 1




      $begingroup$
      @mjw, And[] does short-circuit evaluation.
      $endgroup$
      – J. M. is away
      Mar 27 at 15:41






    • 1




      $begingroup$
      @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
      $endgroup$
      – mjw
      Mar 27 at 16:03






    • 1




      $begingroup$
      @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
      $endgroup$
      – J. M. is away
      Mar 27 at 16:06















    9












    $begingroup$

    Since you have a very large list, you should look at the timing



    list = RandomReal[1, 10^6];

    (And @@ Positive[list]) // AbsoluteTiming (* Hanlon *)

    (* 0.050573, True *)

    VectorQ[list, Positive] // AbsoluteTiming (* J.M. *)

    (* 0.261642, True *)

    (AnyTrue[list, Negative] // Not) // AbsoluteTiming (* Morbo *)

    (* 0.324062, True *)

    And @@ (list /. x_?Negative -> False,
    x_?Positive -> True) // AbsoluteTiming (* Alrubaie *)

    (* 1.00664, True *)


    EDIT: As suggested by mjw, encountering a nonpositive value early in the list significantly alters the results.



    list2 = ReplacePart[list, 1000 -> -1];

    (And @@ Positive[list2]) // AbsoluteTiming (*Hanlon*)

    (* 0.277642, False *)

    VectorQ[list2, Positive] // AbsoluteTiming (*J.M.*)

    (* 0.000223, False *)

    (AnyTrue[list2, Negative] // Not) // AbsoluteTiming (*Morbo*)

    (* 0.000262, False *)

    And @@ (list2 /. x_?Negative -> False,
    x_?Positive -> True) // AbsoluteTiming (*Alrubaie*)

    (* 1.43026, False *)





    share|improve this answer











    $endgroup$












    • $begingroup$
      Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
      $endgroup$
      – mjw
      Mar 27 at 15:34






    • 1




      $begingroup$
      Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
      $endgroup$
      – mjw
      Mar 27 at 15:37






    • 1




      $begingroup$
      @mjw, And[] does short-circuit evaluation.
      $endgroup$
      – J. M. is away
      Mar 27 at 15:41






    • 1




      $begingroup$
      @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
      $endgroup$
      – mjw
      Mar 27 at 16:03






    • 1




      $begingroup$
      @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
      $endgroup$
      – J. M. is away
      Mar 27 at 16:06













    9












    9








    9





    $begingroup$

    Since you have a very large list, you should look at the timing



    list = RandomReal[1, 10^6];

    (And @@ Positive[list]) // AbsoluteTiming (* Hanlon *)

    (* 0.050573, True *)

    VectorQ[list, Positive] // AbsoluteTiming (* J.M. *)

    (* 0.261642, True *)

    (AnyTrue[list, Negative] // Not) // AbsoluteTiming (* Morbo *)

    (* 0.324062, True *)

    And @@ (list /. x_?Negative -> False,
    x_?Positive -> True) // AbsoluteTiming (* Alrubaie *)

    (* 1.00664, True *)


    EDIT: As suggested by mjw, encountering a nonpositive value early in the list significantly alters the results.



    list2 = ReplacePart[list, 1000 -> -1];

    (And @@ Positive[list2]) // AbsoluteTiming (*Hanlon*)

    (* 0.277642, False *)

    VectorQ[list2, Positive] // AbsoluteTiming (*J.M.*)

    (* 0.000223, False *)

    (AnyTrue[list2, Negative] // Not) // AbsoluteTiming (*Morbo*)

    (* 0.000262, False *)

    And @@ (list2 /. x_?Negative -> False,
    x_?Positive -> True) // AbsoluteTiming (*Alrubaie*)

    (* 1.43026, False *)





    share|improve this answer











    $endgroup$



    Since you have a very large list, you should look at the timing



    list = RandomReal[1, 10^6];

    (And @@ Positive[list]) // AbsoluteTiming (* Hanlon *)

    (* 0.050573, True *)

    VectorQ[list, Positive] // AbsoluteTiming (* J.M. *)

    (* 0.261642, True *)

    (AnyTrue[list, Negative] // Not) // AbsoluteTiming (* Morbo *)

    (* 0.324062, True *)

    And @@ (list /. x_?Negative -> False,
    x_?Positive -> True) // AbsoluteTiming (* Alrubaie *)

    (* 1.00664, True *)


    EDIT: As suggested by mjw, encountering a nonpositive value early in the list significantly alters the results.



    list2 = ReplacePart[list, 1000 -> -1];

    (And @@ Positive[list2]) // AbsoluteTiming (*Hanlon*)

    (* 0.277642, False *)

    VectorQ[list2, Positive] // AbsoluteTiming (*J.M.*)

    (* 0.000223, False *)

    (AnyTrue[list2, Negative] // Not) // AbsoluteTiming (*Morbo*)

    (* 0.000262, False *)

    And @@ (list2 /. x_?Negative -> False,
    x_?Positive -> True) // AbsoluteTiming (*Alrubaie*)

    (* 1.43026, False *)






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Mar 27 at 15:53

























    answered Mar 27 at 15:19









    Bob HanlonBob Hanlon

    62k33598




    62k33598











    • $begingroup$
      Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
      $endgroup$
      – mjw
      Mar 27 at 15:34






    • 1




      $begingroup$
      Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
      $endgroup$
      – mjw
      Mar 27 at 15:37






    • 1




      $begingroup$
      @mjw, And[] does short-circuit evaluation.
      $endgroup$
      – J. M. is away
      Mar 27 at 15:41






    • 1




      $begingroup$
      @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
      $endgroup$
      – mjw
      Mar 27 at 16:03






    • 1




      $begingroup$
      @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
      $endgroup$
      – J. M. is away
      Mar 27 at 16:06
















    • $begingroup$
      Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
      $endgroup$
      – mjw
      Mar 27 at 15:34






    • 1




      $begingroup$
      Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
      $endgroup$
      – mjw
      Mar 27 at 15:37






    • 1




      $begingroup$
      @mjw, And[] does short-circuit evaluation.
      $endgroup$
      – J. M. is away
      Mar 27 at 15:41






    • 1




      $begingroup$
      @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
      $endgroup$
      – mjw
      Mar 27 at 16:03






    • 1




      $begingroup$
      @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
      $endgroup$
      – J. M. is away
      Mar 27 at 16:06















    $begingroup$
    Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
    $endgroup$
    – mjw
    Mar 27 at 15:34




    $begingroup$
    Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
    $endgroup$
    – mjw
    Mar 27 at 15:34




    1




    1




    $begingroup$
    Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
    $endgroup$
    – mjw
    Mar 27 at 15:37




    $begingroup$
    Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
    $endgroup$
    – mjw
    Mar 27 at 15:37




    1




    1




    $begingroup$
    @mjw, And[] does short-circuit evaluation.
    $endgroup$
    – J. M. is away
    Mar 27 at 15:41




    $begingroup$
    @mjw, And[] does short-circuit evaluation.
    $endgroup$
    – J. M. is away
    Mar 27 at 15:41




    1




    1




    $begingroup$
    @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
    $endgroup$
    – mjw
    Mar 27 at 16:03




    $begingroup$
    @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
    $endgroup$
    – mjw
    Mar 27 at 16:03




    1




    1




    $begingroup$
    @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
    $endgroup$
    – J. M. is away
    Mar 27 at 16:06




    $begingroup$
    @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
    $endgroup$
    – J. M. is away
    Mar 27 at 16:06











    4












    $begingroup$

    Ah, maybe this is too simple, but works for exactly what you're doing:



    data = Table[RandomReal[-1,1],i,1,1000];
    AnyTrue[data,Negative] // Not
    (*False*)

    data2 = Table[RandomReal[], i, 1, 10^2];
    AnyTrue[data2, Negative] // Not
    (*True*)





    share|improve this answer











    $endgroup$












    • $begingroup$
      AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
      $endgroup$
      – Roman
      Mar 27 at 14:55










    • $begingroup$
      @Roman - the poster is using AnyTrue not AllTrue
      $endgroup$
      – Bob Hanlon
      Mar 27 at 15:00






    • 1




      $begingroup$
      Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
      $endgroup$
      – Roman
      Mar 27 at 15:04











    • $begingroup$
      ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
      $endgroup$
      – morbo
      Mar 27 at 15:04















    4












    $begingroup$

    Ah, maybe this is too simple, but works for exactly what you're doing:



    data = Table[RandomReal[-1,1],i,1,1000];
    AnyTrue[data,Negative] // Not
    (*False*)

    data2 = Table[RandomReal[], i, 1, 10^2];
    AnyTrue[data2, Negative] // Not
    (*True*)





    share|improve this answer











    $endgroup$












    • $begingroup$
      AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
      $endgroup$
      – Roman
      Mar 27 at 14:55










    • $begingroup$
      @Roman - the poster is using AnyTrue not AllTrue
      $endgroup$
      – Bob Hanlon
      Mar 27 at 15:00






    • 1




      $begingroup$
      Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
      $endgroup$
      – Roman
      Mar 27 at 15:04











    • $begingroup$
      ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
      $endgroup$
      – morbo
      Mar 27 at 15:04













    4












    4








    4





    $begingroup$

    Ah, maybe this is too simple, but works for exactly what you're doing:



    data = Table[RandomReal[-1,1],i,1,1000];
    AnyTrue[data,Negative] // Not
    (*False*)

    data2 = Table[RandomReal[], i, 1, 10^2];
    AnyTrue[data2, Negative] // Not
    (*True*)





    share|improve this answer











    $endgroup$



    Ah, maybe this is too simple, but works for exactly what you're doing:



    data = Table[RandomReal[-1,1],i,1,1000];
    AnyTrue[data,Negative] // Not
    (*False*)

    data2 = Table[RandomReal[], i, 1, 10^2];
    AnyTrue[data2, Negative] // Not
    (*True*)






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Mar 27 at 15:03

























    answered Mar 27 at 14:29









    morbomorbo

    48428




    48428











    • $begingroup$
      AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
      $endgroup$
      – Roman
      Mar 27 at 14:55










    • $begingroup$
      @Roman - the poster is using AnyTrue not AllTrue
      $endgroup$
      – Bob Hanlon
      Mar 27 at 15:00






    • 1




      $begingroup$
      Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
      $endgroup$
      – Roman
      Mar 27 at 15:04











    • $begingroup$
      ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
      $endgroup$
      – morbo
      Mar 27 at 15:04
















    • $begingroup$
      AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
      $endgroup$
      – Roman
      Mar 27 at 14:55










    • $begingroup$
      @Roman - the poster is using AnyTrue not AllTrue
      $endgroup$
      – Bob Hanlon
      Mar 27 at 15:00






    • 1




      $begingroup$
      Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
      $endgroup$
      – Roman
      Mar 27 at 15:04











    • $begingroup$
      ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
      $endgroup$
      – morbo
      Mar 27 at 15:04















    $begingroup$
    AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
    $endgroup$
    – Roman
    Mar 27 at 14:55




    $begingroup$
    AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
    $endgroup$
    – Roman
    Mar 27 at 14:55












    $begingroup$
    @Roman - the poster is using AnyTrue not AllTrue
    $endgroup$
    – Bob Hanlon
    Mar 27 at 15:00




    $begingroup$
    @Roman - the poster is using AnyTrue not AllTrue
    $endgroup$
    – Bob Hanlon
    Mar 27 at 15:00




    1




    1




    $begingroup$
    Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
    $endgroup$
    – Roman
    Mar 27 at 15:04





    $begingroup$
    Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
    $endgroup$
    – Roman
    Mar 27 at 15:04













    $begingroup$
    ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
    $endgroup$
    – morbo
    Mar 27 at 15:04




    $begingroup$
    ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
    $endgroup$
    – morbo
    Mar 27 at 15:04











    3












    $begingroup$

    list = 1, 2, 3, 4, -5, -6, -7;

    list /. x_?Negative -> True, x_?Positive -> False





    share|improve this answer









    $endgroup$

















      3












      $begingroup$

      list = 1, 2, 3, 4, -5, -6, -7;

      list /. x_?Negative -> True, x_?Positive -> False





      share|improve this answer









      $endgroup$















        3












        3








        3





        $begingroup$

        list = 1, 2, 3, 4, -5, -6, -7;

        list /. x_?Negative -> True, x_?Positive -> False





        share|improve this answer









        $endgroup$



        list = 1, 2, 3, 4, -5, -6, -7;

        list /. x_?Negative -> True, x_?Positive -> False






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 27 at 14:31









        AlrubaieAlrubaie

        692111




        692111



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Mathematica Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194043%2fhow-to-check-is-there-any-negative-term-in-a-large-list%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

            He _____ here since 1970 . Answer needed [closed]What does “since he was so high” mean?Meaning of “catch birds for”?How do I ensure “since” takes the meaning I want?“Who cares here” meaningWhat does “right round toward” mean?the time tense (had now been detected)What does the phrase “ring around the roses” mean here?Correct usage of “visited upon”Meaning of “foiled rail sabotage bid”It was the third time I had gone to Rome or It is the third time I had been to Rome

            Slayer Innehåll Historia | Stil, komposition och lyrik | Bandets betydelse och framgångar | Sidoprojekt och samarbeten | Kontroverser | Medlemmar | Utmärkelser och nomineringar | Turnéer och festivaler | Diskografi | Referenser | Externa länkar | Navigeringsmenywww.slayer.net”Metal Massacre vol. 1””Metal Massacre vol. 3””Metal Massacre Volume III””Show No Mercy””Haunting the Chapel””Live Undead””Hell Awaits””Reign in Blood””Reign in Blood””Gold & Platinum – Reign in Blood””Golden Gods Awards Winners”originalet”Kerrang! Hall Of Fame””Slayer Looks Back On 37-Year Career In New Video Series: Part Two””South of Heaven””Gold & Platinum – South of Heaven””Seasons in the Abyss””Gold & Platinum - Seasons in the Abyss””Divine Intervention””Divine Intervention - Release group by Slayer””Gold & Platinum - Divine Intervention””Live Intrusion””Undisputed Attitude””Abolish Government/Superficial Love””Release “Slatanic Slaughter: A Tribute to Slayer” by Various Artists””Diabolus in Musica””Soundtrack to the Apocalypse””God Hates Us All””Systematic - Relationships””War at the Warfield””Gold & Platinum - War at the Warfield””Soundtrack to the Apocalypse””Gold & Platinum - Still Reigning””Metallica, Slayer, Iron Mauden Among Winners At Metal Hammer Awards””Eternal Pyre””Eternal Pyre - Slayer release group””Eternal Pyre””Metal Storm Awards 2006””Kerrang! Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029