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How to find the largest number(s) in a list of elements, possibly non-unique?
How do I check if a list is empty?Finding the index of an item given a list containing it in PythonWhat's the simplest way to print a Java array?Convert bytes to a string?Getting the last element of a list in PythonHow to make a flat list out of list of lists?How do I get the number of elements in a list in Python?How do I list all files of a directory?Use of *args and **kwargsHow do I remove a particular element from an array in JavaScript?
Here is my program,
item_no = []
max = 0
for i in range(5):
input_no = int(input("Enter an item number: "))
item_no.append(input_no)
for i in item_no:
if i > max:
max = i
high = item_no.index(max)
print (item_no[high])
Example input: [5, 6, 7, 8, 8]
Example output: 8
How can I change my program to output the same highest numbers in an array?
Expected output: [8, 8]
python arrays python-3.x
New contributor
|
show 2 more comments
Here is my program,
item_no = []
max = 0
for i in range(5):
input_no = int(input("Enter an item number: "))
item_no.append(input_no)
for i in item_no:
if i > max:
max = i
high = item_no.index(max)
print (item_no[high])
Example input: [5, 6, 7, 8, 8]
Example output: 8
How can I change my program to output the same highest numbers in an array?
Expected output: [8, 8]
python arrays python-3.x
New contributor
3
Band indent at line 8
– DirtyBit
Mar 18 at 7:21
1
You can use Dictionary to store the maximum number and its count as item_no = , if you max is different then original in item_no, reinitialize it and add that item and add count =1
– dkb
Mar 18 at 7:21
3
("Band" probably is a misspelling for "Bad")
– tripleee
Mar 18 at 7:26
1
@tripleee Indeed. bulls-eye!
– DirtyBit
Mar 18 at 7:39
1
As answers show, it's more Pythonic to do it with list comprehensions, rather than loops. But if you use loops to iterate over the list, a style tip: don't use the same variable namei
both for indicesi in range(5)
then also for items/values:for i in item_no
. Better to dofor no in item_no
– smci
Mar 19 at 0:35
|
show 2 more comments
Here is my program,
item_no = []
max = 0
for i in range(5):
input_no = int(input("Enter an item number: "))
item_no.append(input_no)
for i in item_no:
if i > max:
max = i
high = item_no.index(max)
print (item_no[high])
Example input: [5, 6, 7, 8, 8]
Example output: 8
How can I change my program to output the same highest numbers in an array?
Expected output: [8, 8]
python arrays python-3.x
New contributor
Here is my program,
item_no = []
max = 0
for i in range(5):
input_no = int(input("Enter an item number: "))
item_no.append(input_no)
for i in item_no:
if i > max:
max = i
high = item_no.index(max)
print (item_no[high])
Example input: [5, 6, 7, 8, 8]
Example output: 8
How can I change my program to output the same highest numbers in an array?
Expected output: [8, 8]
python arrays python-3.x
python arrays python-3.x
New contributor
New contributor
edited Mar 19 at 0:38
smci
15.4k678109
15.4k678109
New contributor
asked Mar 18 at 7:19
user11206537user11206537
91112
91112
New contributor
New contributor
3
Band indent at line 8
– DirtyBit
Mar 18 at 7:21
1
You can use Dictionary to store the maximum number and its count as item_no = , if you max is different then original in item_no, reinitialize it and add that item and add count =1
– dkb
Mar 18 at 7:21
3
("Band" probably is a misspelling for "Bad")
– tripleee
Mar 18 at 7:26
1
@tripleee Indeed. bulls-eye!
– DirtyBit
Mar 18 at 7:39
1
As answers show, it's more Pythonic to do it with list comprehensions, rather than loops. But if you use loops to iterate over the list, a style tip: don't use the same variable namei
both for indicesi in range(5)
then also for items/values:for i in item_no
. Better to dofor no in item_no
– smci
Mar 19 at 0:35
|
show 2 more comments
3
Band indent at line 8
– DirtyBit
Mar 18 at 7:21
1
You can use Dictionary to store the maximum number and its count as item_no = , if you max is different then original in item_no, reinitialize it and add that item and add count =1
– dkb
Mar 18 at 7:21
3
("Band" probably is a misspelling for "Bad")
– tripleee
Mar 18 at 7:26
1
@tripleee Indeed. bulls-eye!
– DirtyBit
Mar 18 at 7:39
1
As answers show, it's more Pythonic to do it with list comprehensions, rather than loops. But if you use loops to iterate over the list, a style tip: don't use the same variable namei
both for indicesi in range(5)
then also for items/values:for i in item_no
. Better to dofor no in item_no
– smci
Mar 19 at 0:35
3
3
Band indent at line 8
– DirtyBit
Mar 18 at 7:21
Band indent at line 8
– DirtyBit
Mar 18 at 7:21
1
1
You can use Dictionary to store the maximum number and its count as item_no = , if you max is different then original in item_no, reinitialize it and add that item and add count =1
– dkb
Mar 18 at 7:21
You can use Dictionary to store the maximum number and its count as item_no = , if you max is different then original in item_no, reinitialize it and add that item and add count =1
– dkb
Mar 18 at 7:21
3
3
("Band" probably is a misspelling for "Bad")
– tripleee
Mar 18 at 7:26
("Band" probably is a misspelling for "Bad")
– tripleee
Mar 18 at 7:26
1
1
@tripleee Indeed. bulls-eye!
– DirtyBit
Mar 18 at 7:39
@tripleee Indeed. bulls-eye!
– DirtyBit
Mar 18 at 7:39
1
1
As answers show, it's more Pythonic to do it with list comprehensions, rather than loops. But if you use loops to iterate over the list, a style tip: don't use the same variable name
i
both for indices i in range(5)
then also for items/values: for i in item_no
. Better to do for no in item_no
– smci
Mar 19 at 0:35
As answers show, it's more Pythonic to do it with list comprehensions, rather than loops. But if you use loops to iterate over the list, a style tip: don't use the same variable name
i
both for indices i in range(5)
then also for items/values: for i in item_no
. Better to do for no in item_no
– smci
Mar 19 at 0:35
|
show 2 more comments
6 Answers
6
active
oldest
votes
Just get the maximum using max
and then its count
and combine the two in a list-comprehension.
item_no = [5, 6, 7, 8, 8]
max_no = max(item_no)
highest = [max_no for _ in range(item_no.count(max_no))]
print(highest) # -> [8, 8]
Note that this will return a list of a single item in case your maximum value appears only once.
A solution closer to your current programming style would be the following:
item_no = [5, 6, 7, 8, 8]
max_no = 0 # Note 1
for i in item_no:
if i > max_no:
max_no = i
high = [i]
elif i == max_no:
high.append(i)
with the same results as above of course.
Notes
- I am assuming that you are dealing with N* (1, 2, ...) numbers only. If that is not the case, initializing with
-math.inf
should be used instead.
Note that the second code snippet is less efficient than the first by quite a margin. Python allows you to be more efficient than these explicit, fortran-like loops and it is more efficient itself when you use it properly.
Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.
– user11206537
Mar 18 at 7:42
2
@user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!
– Ev. Kounis
Mar 18 at 7:44
One last question, how do you find the index of the result (high) in item_no?
– user11206537
Mar 18 at 8:32
3
@user11206537 To get the index as well, you can modify the code slightly and useenumerate
onitem_no
. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this
– Ev. Kounis
Mar 18 at 9:32
1
@Drew Yes, you're right; ifitem_no[0] == 0
, the code crashes ashigh
isn't defined when reachinghigh.append(i)
.
– PiCTo
Mar 19 at 0:03
|
show 4 more comments
You can do it even shorter:
item_no = [5, 6, 7, 8, 8]
#compute once - use many
max_item = max(item_no)
print(item_no.count(max_item) * [max_item])
output:
[8, 8]
One last question, how do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:48
add a comment |
You could use list
comprehension for that task following way:
numbers = [5, 6, 7, 8, 8]
maxnumbers = [i for i in numbers if i==max(numbers)]
print(*maxnumbers,sep=',')
output:
8,8
*
operator in print
is used to unpack values, sep
is used to inform print
what seperator to use: ,
in this case.
EDIT: If you want to get indices of biggest value and call max
only once then do:
numbers = [5, 6, 7, 8, 8]
biggest = max(numbers)
positions = [inx for inx,i in enumerate(numbers) if i==biggest]
print(*positions,sep=',')
Output:
3,4
As you might check numbers[3]
is equal to biggest
and numbers[4]
is equal to biggest
.
4
note that your solution callsmax
a total oflen(numbers)
times. Storing it outside the list and then using that would be better.
– Ev. Kounis
Mar 18 at 7:31
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
@Ev.Kounis I assume the Python interpreter optimizes this, no?
– user1717828
Mar 18 at 16:38
@user1717828 No it does not.
– Ev. Kounis
Mar 19 at 7:04
@Ev.Kounis, Oh man, I didn't believe this so I tested it out.python -m timeit "numbers = [5, 6, 7, 8, 8]; max_number = max(numbers); maxnumbers = [i for i in numbers if i==max_number]"
is twice as fast as when the max calculation is in the comprehension. I've gotten very spoiled from using languages with compilers.
– user1717828
Mar 19 at 9:53
|
show 1 more comment
Count the occurrence of max number
iterate over the list to print the max number for the range of the count (1)
Hence:
item_no = [5, 6, 7, 8, 8]
counter = item_no.count(max(item_no)) # 2
print([max(item_no) for x in range(counter)])
OUTPUT:
[8, 8]
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:56
add a comment |
This issue can be solved in one line, by finding an item which is equal to the maximum value:
To improve performance store max in var
Mvalue=max(item_no)[i for i in item_no if i==Mvalue]
1
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]
– Pradeep Pandey
Mar 18 at 11:56
But isn'tmax
called n times? Not efficient.
– Sachin Dangol
2 days ago
add a comment |
I think it would be better if we evaluate the max
in the array and its count
in one iteration
def maxs(iterable):
max = None
count = 0
for index, value in enumerate(iterable):
if index == 0 or value >= max:
if value != max:
count = 0
max = value
count += 1
return count * [max]
print (maxs([5, 6, 7, 8, 8])) # [8, 8]
print (maxs([3, 2, 4, 5, 1, 2, 4, 5, 2, 5, 0])) # [5, 5, 5]
print (maxs([])) # []
Give it a Try!!
I did! And it worked! That's a very efficient solution.
– user11206537
2 days ago
@user11206537 Did you measure and compared the efficiency among the solutions. Can you share with us?
– Grijesh Chauhan
2 days ago
This kind of solution was the one that was actually needed as a part of my school project and my teacher thought of it as a very efficient solution. This code is also written along the lines of my current programming style, which is exactly what I wanted.
– user11206537
yesterday
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
Just get the maximum using max
and then its count
and combine the two in a list-comprehension.
item_no = [5, 6, 7, 8, 8]
max_no = max(item_no)
highest = [max_no for _ in range(item_no.count(max_no))]
print(highest) # -> [8, 8]
Note that this will return a list of a single item in case your maximum value appears only once.
A solution closer to your current programming style would be the following:
item_no = [5, 6, 7, 8, 8]
max_no = 0 # Note 1
for i in item_no:
if i > max_no:
max_no = i
high = [i]
elif i == max_no:
high.append(i)
with the same results as above of course.
Notes
- I am assuming that you are dealing with N* (1, 2, ...) numbers only. If that is not the case, initializing with
-math.inf
should be used instead.
Note that the second code snippet is less efficient than the first by quite a margin. Python allows you to be more efficient than these explicit, fortran-like loops and it is more efficient itself when you use it properly.
Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.
– user11206537
Mar 18 at 7:42
2
@user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!
– Ev. Kounis
Mar 18 at 7:44
One last question, how do you find the index of the result (high) in item_no?
– user11206537
Mar 18 at 8:32
3
@user11206537 To get the index as well, you can modify the code slightly and useenumerate
onitem_no
. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this
– Ev. Kounis
Mar 18 at 9:32
1
@Drew Yes, you're right; ifitem_no[0] == 0
, the code crashes ashigh
isn't defined when reachinghigh.append(i)
.
– PiCTo
Mar 19 at 0:03
|
show 4 more comments
Just get the maximum using max
and then its count
and combine the two in a list-comprehension.
item_no = [5, 6, 7, 8, 8]
max_no = max(item_no)
highest = [max_no for _ in range(item_no.count(max_no))]
print(highest) # -> [8, 8]
Note that this will return a list of a single item in case your maximum value appears only once.
A solution closer to your current programming style would be the following:
item_no = [5, 6, 7, 8, 8]
max_no = 0 # Note 1
for i in item_no:
if i > max_no:
max_no = i
high = [i]
elif i == max_no:
high.append(i)
with the same results as above of course.
Notes
- I am assuming that you are dealing with N* (1, 2, ...) numbers only. If that is not the case, initializing with
-math.inf
should be used instead.
Note that the second code snippet is less efficient than the first by quite a margin. Python allows you to be more efficient than these explicit, fortran-like loops and it is more efficient itself when you use it properly.
Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.
– user11206537
Mar 18 at 7:42
2
@user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!
– Ev. Kounis
Mar 18 at 7:44
One last question, how do you find the index of the result (high) in item_no?
– user11206537
Mar 18 at 8:32
3
@user11206537 To get the index as well, you can modify the code slightly and useenumerate
onitem_no
. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this
– Ev. Kounis
Mar 18 at 9:32
1
@Drew Yes, you're right; ifitem_no[0] == 0
, the code crashes ashigh
isn't defined when reachinghigh.append(i)
.
– PiCTo
Mar 19 at 0:03
|
show 4 more comments
Just get the maximum using max
and then its count
and combine the two in a list-comprehension.
item_no = [5, 6, 7, 8, 8]
max_no = max(item_no)
highest = [max_no for _ in range(item_no.count(max_no))]
print(highest) # -> [8, 8]
Note that this will return a list of a single item in case your maximum value appears only once.
A solution closer to your current programming style would be the following:
item_no = [5, 6, 7, 8, 8]
max_no = 0 # Note 1
for i in item_no:
if i > max_no:
max_no = i
high = [i]
elif i == max_no:
high.append(i)
with the same results as above of course.
Notes
- I am assuming that you are dealing with N* (1, 2, ...) numbers only. If that is not the case, initializing with
-math.inf
should be used instead.
Note that the second code snippet is less efficient than the first by quite a margin. Python allows you to be more efficient than these explicit, fortran-like loops and it is more efficient itself when you use it properly.
Just get the maximum using max
and then its count
and combine the two in a list-comprehension.
item_no = [5, 6, 7, 8, 8]
max_no = max(item_no)
highest = [max_no for _ in range(item_no.count(max_no))]
print(highest) # -> [8, 8]
Note that this will return a list of a single item in case your maximum value appears only once.
A solution closer to your current programming style would be the following:
item_no = [5, 6, 7, 8, 8]
max_no = 0 # Note 1
for i in item_no:
if i > max_no:
max_no = i
high = [i]
elif i == max_no:
high.append(i)
with the same results as above of course.
Notes
- I am assuming that you are dealing with N* (1, 2, ...) numbers only. If that is not the case, initializing with
-math.inf
should be used instead.
Note that the second code snippet is less efficient than the first by quite a margin. Python allows you to be more efficient than these explicit, fortran-like loops and it is more efficient itself when you use it properly.
edited Mar 19 at 7:10
answered Mar 18 at 7:22
Ev. KounisEv. Kounis
11.3k21751
11.3k21751
Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.
– user11206537
Mar 18 at 7:42
2
@user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!
– Ev. Kounis
Mar 18 at 7:44
One last question, how do you find the index of the result (high) in item_no?
– user11206537
Mar 18 at 8:32
3
@user11206537 To get the index as well, you can modify the code slightly and useenumerate
onitem_no
. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this
– Ev. Kounis
Mar 18 at 9:32
1
@Drew Yes, you're right; ifitem_no[0] == 0
, the code crashes ashigh
isn't defined when reachinghigh.append(i)
.
– PiCTo
Mar 19 at 0:03
|
show 4 more comments
Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.
– user11206537
Mar 18 at 7:42
2
@user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!
– Ev. Kounis
Mar 18 at 7:44
One last question, how do you find the index of the result (high) in item_no?
– user11206537
Mar 18 at 8:32
3
@user11206537 To get the index as well, you can modify the code slightly and useenumerate
onitem_no
. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this
– Ev. Kounis
Mar 18 at 9:32
1
@Drew Yes, you're right; ifitem_no[0] == 0
, the code crashes ashigh
isn't defined when reachinghigh.append(i)
.
– PiCTo
Mar 19 at 0:03
Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.
– user11206537
Mar 18 at 7:42
Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.
– user11206537
Mar 18 at 7:42
2
2
@user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!
– Ev. Kounis
Mar 18 at 7:44
@user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!
– Ev. Kounis
Mar 18 at 7:44
One last question, how do you find the index of the result (high) in item_no?
– user11206537
Mar 18 at 8:32
One last question, how do you find the index of the result (high) in item_no?
– user11206537
Mar 18 at 8:32
3
3
@user11206537 To get the index as well, you can modify the code slightly and use
enumerate
on item_no
. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this– Ev. Kounis
Mar 18 at 9:32
@user11206537 To get the index as well, you can modify the code slightly and use
enumerate
on item_no
. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this– Ev. Kounis
Mar 18 at 9:32
1
1
@Drew Yes, you're right; if
item_no[0] == 0
, the code crashes as high
isn't defined when reaching high.append(i)
.– PiCTo
Mar 19 at 0:03
@Drew Yes, you're right; if
item_no[0] == 0
, the code crashes as high
isn't defined when reaching high.append(i)
.– PiCTo
Mar 19 at 0:03
|
show 4 more comments
You can do it even shorter:
item_no = [5, 6, 7, 8, 8]
#compute once - use many
max_item = max(item_no)
print(item_no.count(max_item) * [max_item])
output:
[8, 8]
One last question, how do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:48
add a comment |
You can do it even shorter:
item_no = [5, 6, 7, 8, 8]
#compute once - use many
max_item = max(item_no)
print(item_no.count(max_item) * [max_item])
output:
[8, 8]
One last question, how do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:48
add a comment |
You can do it even shorter:
item_no = [5, 6, 7, 8, 8]
#compute once - use many
max_item = max(item_no)
print(item_no.count(max_item) * [max_item])
output:
[8, 8]
You can do it even shorter:
item_no = [5, 6, 7, 8, 8]
#compute once - use many
max_item = max(item_no)
print(item_no.count(max_item) * [max_item])
output:
[8, 8]
edited Mar 18 at 11:40
answered Mar 18 at 7:30
AllanAllan
8,04231535
8,04231535
One last question, how do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:48
add a comment |
One last question, how do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:48
One last question, how do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:48
One last question, how do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:48
add a comment |
You could use list
comprehension for that task following way:
numbers = [5, 6, 7, 8, 8]
maxnumbers = [i for i in numbers if i==max(numbers)]
print(*maxnumbers,sep=',')
output:
8,8
*
operator in print
is used to unpack values, sep
is used to inform print
what seperator to use: ,
in this case.
EDIT: If you want to get indices of biggest value and call max
only once then do:
numbers = [5, 6, 7, 8, 8]
biggest = max(numbers)
positions = [inx for inx,i in enumerate(numbers) if i==biggest]
print(*positions,sep=',')
Output:
3,4
As you might check numbers[3]
is equal to biggest
and numbers[4]
is equal to biggest
.
4
note that your solution callsmax
a total oflen(numbers)
times. Storing it outside the list and then using that would be better.
– Ev. Kounis
Mar 18 at 7:31
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
@Ev.Kounis I assume the Python interpreter optimizes this, no?
– user1717828
Mar 18 at 16:38
@user1717828 No it does not.
– Ev. Kounis
Mar 19 at 7:04
@Ev.Kounis, Oh man, I didn't believe this so I tested it out.python -m timeit "numbers = [5, 6, 7, 8, 8]; max_number = max(numbers); maxnumbers = [i for i in numbers if i==max_number]"
is twice as fast as when the max calculation is in the comprehension. I've gotten very spoiled from using languages with compilers.
– user1717828
Mar 19 at 9:53
|
show 1 more comment
You could use list
comprehension for that task following way:
numbers = [5, 6, 7, 8, 8]
maxnumbers = [i for i in numbers if i==max(numbers)]
print(*maxnumbers,sep=',')
output:
8,8
*
operator in print
is used to unpack values, sep
is used to inform print
what seperator to use: ,
in this case.
EDIT: If you want to get indices of biggest value and call max
only once then do:
numbers = [5, 6, 7, 8, 8]
biggest = max(numbers)
positions = [inx for inx,i in enumerate(numbers) if i==biggest]
print(*positions,sep=',')
Output:
3,4
As you might check numbers[3]
is equal to biggest
and numbers[4]
is equal to biggest
.
4
note that your solution callsmax
a total oflen(numbers)
times. Storing it outside the list and then using that would be better.
– Ev. Kounis
Mar 18 at 7:31
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
@Ev.Kounis I assume the Python interpreter optimizes this, no?
– user1717828
Mar 18 at 16:38
@user1717828 No it does not.
– Ev. Kounis
Mar 19 at 7:04
@Ev.Kounis, Oh man, I didn't believe this so I tested it out.python -m timeit "numbers = [5, 6, 7, 8, 8]; max_number = max(numbers); maxnumbers = [i for i in numbers if i==max_number]"
is twice as fast as when the max calculation is in the comprehension. I've gotten very spoiled from using languages with compilers.
– user1717828
Mar 19 at 9:53
|
show 1 more comment
You could use list
comprehension for that task following way:
numbers = [5, 6, 7, 8, 8]
maxnumbers = [i for i in numbers if i==max(numbers)]
print(*maxnumbers,sep=',')
output:
8,8
*
operator in print
is used to unpack values, sep
is used to inform print
what seperator to use: ,
in this case.
EDIT: If you want to get indices of biggest value and call max
only once then do:
numbers = [5, 6, 7, 8, 8]
biggest = max(numbers)
positions = [inx for inx,i in enumerate(numbers) if i==biggest]
print(*positions,sep=',')
Output:
3,4
As you might check numbers[3]
is equal to biggest
and numbers[4]
is equal to biggest
.
You could use list
comprehension for that task following way:
numbers = [5, 6, 7, 8, 8]
maxnumbers = [i for i in numbers if i==max(numbers)]
print(*maxnumbers,sep=',')
output:
8,8
*
operator in print
is used to unpack values, sep
is used to inform print
what seperator to use: ,
in this case.
EDIT: If you want to get indices of biggest value and call max
only once then do:
numbers = [5, 6, 7, 8, 8]
biggest = max(numbers)
positions = [inx for inx,i in enumerate(numbers) if i==biggest]
print(*positions,sep=',')
Output:
3,4
As you might check numbers[3]
is equal to biggest
and numbers[4]
is equal to biggest
.
edited Mar 18 at 10:02
answered Mar 18 at 7:29
DaweoDaweo
1,08525
1,08525
4
note that your solution callsmax
a total oflen(numbers)
times. Storing it outside the list and then using that would be better.
– Ev. Kounis
Mar 18 at 7:31
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
@Ev.Kounis I assume the Python interpreter optimizes this, no?
– user1717828
Mar 18 at 16:38
@user1717828 No it does not.
– Ev. Kounis
Mar 19 at 7:04
@Ev.Kounis, Oh man, I didn't believe this so I tested it out.python -m timeit "numbers = [5, 6, 7, 8, 8]; max_number = max(numbers); maxnumbers = [i for i in numbers if i==max_number]"
is twice as fast as when the max calculation is in the comprehension. I've gotten very spoiled from using languages with compilers.
– user1717828
Mar 19 at 9:53
|
show 1 more comment
4
note that your solution callsmax
a total oflen(numbers)
times. Storing it outside the list and then using that would be better.
– Ev. Kounis
Mar 18 at 7:31
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
@Ev.Kounis I assume the Python interpreter optimizes this, no?
– user1717828
Mar 18 at 16:38
@user1717828 No it does not.
– Ev. Kounis
Mar 19 at 7:04
@Ev.Kounis, Oh man, I didn't believe this so I tested it out.python -m timeit "numbers = [5, 6, 7, 8, 8]; max_number = max(numbers); maxnumbers = [i for i in numbers if i==max_number]"
is twice as fast as when the max calculation is in the comprehension. I've gotten very spoiled from using languages with compilers.
– user1717828
Mar 19 at 9:53
4
4
note that your solution calls
max
a total of len(numbers)
times. Storing it outside the list and then using that would be better.– Ev. Kounis
Mar 18 at 7:31
note that your solution calls
max
a total of len(numbers)
times. Storing it outside the list and then using that would be better.– Ev. Kounis
Mar 18 at 7:31
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
@Ev.Kounis I assume the Python interpreter optimizes this, no?
– user1717828
Mar 18 at 16:38
@Ev.Kounis I assume the Python interpreter optimizes this, no?
– user1717828
Mar 18 at 16:38
@user1717828 No it does not.
– Ev. Kounis
Mar 19 at 7:04
@user1717828 No it does not.
– Ev. Kounis
Mar 19 at 7:04
@Ev.Kounis, Oh man, I didn't believe this so I tested it out.
python -m timeit "numbers = [5, 6, 7, 8, 8]; max_number = max(numbers); maxnumbers = [i for i in numbers if i==max_number]"
is twice as fast as when the max calculation is in the comprehension. I've gotten very spoiled from using languages with compilers.– user1717828
Mar 19 at 9:53
@Ev.Kounis, Oh man, I didn't believe this so I tested it out.
python -m timeit "numbers = [5, 6, 7, 8, 8]; max_number = max(numbers); maxnumbers = [i for i in numbers if i==max_number]"
is twice as fast as when the max calculation is in the comprehension. I've gotten very spoiled from using languages with compilers.– user1717828
Mar 19 at 9:53
|
show 1 more comment
Count the occurrence of max number
iterate over the list to print the max number for the range of the count (1)
Hence:
item_no = [5, 6, 7, 8, 8]
counter = item_no.count(max(item_no)) # 2
print([max(item_no) for x in range(counter)])
OUTPUT:
[8, 8]
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:56
add a comment |
Count the occurrence of max number
iterate over the list to print the max number for the range of the count (1)
Hence:
item_no = [5, 6, 7, 8, 8]
counter = item_no.count(max(item_no)) # 2
print([max(item_no) for x in range(counter)])
OUTPUT:
[8, 8]
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:56
add a comment |
Count the occurrence of max number
iterate over the list to print the max number for the range of the count (1)
Hence:
item_no = [5, 6, 7, 8, 8]
counter = item_no.count(max(item_no)) # 2
print([max(item_no) for x in range(counter)])
OUTPUT:
[8, 8]
Count the occurrence of max number
iterate over the list to print the max number for the range of the count (1)
Hence:
item_no = [5, 6, 7, 8, 8]
counter = item_no.count(max(item_no)) # 2
print([max(item_no) for x in range(counter)])
OUTPUT:
[8, 8]
answered Mar 18 at 7:35
DirtyBitDirtyBit
10.1k21540
10.1k21540
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:56
add a comment |
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:56
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:56
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 8:56
add a comment |
This issue can be solved in one line, by finding an item which is equal to the maximum value:
To improve performance store max in var
Mvalue=max(item_no)[i for i in item_no if i==Mvalue]
1
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]
– Pradeep Pandey
Mar 18 at 11:56
But isn'tmax
called n times? Not efficient.
– Sachin Dangol
2 days ago
add a comment |
This issue can be solved in one line, by finding an item which is equal to the maximum value:
To improve performance store max in var
Mvalue=max(item_no)[i for i in item_no if i==Mvalue]
1
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]
– Pradeep Pandey
Mar 18 at 11:56
But isn'tmax
called n times? Not efficient.
– Sachin Dangol
2 days ago
add a comment |
This issue can be solved in one line, by finding an item which is equal to the maximum value:
To improve performance store max in var
Mvalue=max(item_no)[i for i in item_no if i==Mvalue]
This issue can be solved in one line, by finding an item which is equal to the maximum value:
To improve performance store max in var
Mvalue=max(item_no)[i for i in item_no if i==Mvalue]
edited yesterday
answered Mar 18 at 8:06
Pradeep PandeyPradeep Pandey
17917
17917
1
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]
– Pradeep Pandey
Mar 18 at 11:56
But isn'tmax
called n times? Not efficient.
– Sachin Dangol
2 days ago
add a comment |
1
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]
– Pradeep Pandey
Mar 18 at 11:56
But isn'tmax
called n times? Not efficient.
– Sachin Dangol
2 days ago
1
1
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
How do you find the index of the result in item_no?
– user11206537
Mar 18 at 9:01
By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]
– Pradeep Pandey
Mar 18 at 11:56
By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]
– Pradeep Pandey
Mar 18 at 11:56
But isn't
max
called n times? Not efficient.– Sachin Dangol
2 days ago
But isn't
max
called n times? Not efficient.– Sachin Dangol
2 days ago
add a comment |
I think it would be better if we evaluate the max
in the array and its count
in one iteration
def maxs(iterable):
max = None
count = 0
for index, value in enumerate(iterable):
if index == 0 or value >= max:
if value != max:
count = 0
max = value
count += 1
return count * [max]
print (maxs([5, 6, 7, 8, 8])) # [8, 8]
print (maxs([3, 2, 4, 5, 1, 2, 4, 5, 2, 5, 0])) # [5, 5, 5]
print (maxs([])) # []
Give it a Try!!
I did! And it worked! That's a very efficient solution.
– user11206537
2 days ago
@user11206537 Did you measure and compared the efficiency among the solutions. Can you share with us?
– Grijesh Chauhan
2 days ago
This kind of solution was the one that was actually needed as a part of my school project and my teacher thought of it as a very efficient solution. This code is also written along the lines of my current programming style, which is exactly what I wanted.
– user11206537
yesterday
add a comment |
I think it would be better if we evaluate the max
in the array and its count
in one iteration
def maxs(iterable):
max = None
count = 0
for index, value in enumerate(iterable):
if index == 0 or value >= max:
if value != max:
count = 0
max = value
count += 1
return count * [max]
print (maxs([5, 6, 7, 8, 8])) # [8, 8]
print (maxs([3, 2, 4, 5, 1, 2, 4, 5, 2, 5, 0])) # [5, 5, 5]
print (maxs([])) # []
Give it a Try!!
I did! And it worked! That's a very efficient solution.
– user11206537
2 days ago
@user11206537 Did you measure and compared the efficiency among the solutions. Can you share with us?
– Grijesh Chauhan
2 days ago
This kind of solution was the one that was actually needed as a part of my school project and my teacher thought of it as a very efficient solution. This code is also written along the lines of my current programming style, which is exactly what I wanted.
– user11206537
yesterday
add a comment |
I think it would be better if we evaluate the max
in the array and its count
in one iteration
def maxs(iterable):
max = None
count = 0
for index, value in enumerate(iterable):
if index == 0 or value >= max:
if value != max:
count = 0
max = value
count += 1
return count * [max]
print (maxs([5, 6, 7, 8, 8])) # [8, 8]
print (maxs([3, 2, 4, 5, 1, 2, 4, 5, 2, 5, 0])) # [5, 5, 5]
print (maxs([])) # []
Give it a Try!!
I think it would be better if we evaluate the max
in the array and its count
in one iteration
def maxs(iterable):
max = None
count = 0
for index, value in enumerate(iterable):
if index == 0 or value >= max:
if value != max:
count = 0
max = value
count += 1
return count * [max]
print (maxs([5, 6, 7, 8, 8])) # [8, 8]
print (maxs([3, 2, 4, 5, 1, 2, 4, 5, 2, 5, 0])) # [5, 5, 5]
print (maxs([])) # []
Give it a Try!!
answered 2 days ago
Grijesh ChauhanGrijesh Chauhan
45.8k1498163
45.8k1498163
I did! And it worked! That's a very efficient solution.
– user11206537
2 days ago
@user11206537 Did you measure and compared the efficiency among the solutions. Can you share with us?
– Grijesh Chauhan
2 days ago
This kind of solution was the one that was actually needed as a part of my school project and my teacher thought of it as a very efficient solution. This code is also written along the lines of my current programming style, which is exactly what I wanted.
– user11206537
yesterday
add a comment |
I did! And it worked! That's a very efficient solution.
– user11206537
2 days ago
@user11206537 Did you measure and compared the efficiency among the solutions. Can you share with us?
– Grijesh Chauhan
2 days ago
This kind of solution was the one that was actually needed as a part of my school project and my teacher thought of it as a very efficient solution. This code is also written along the lines of my current programming style, which is exactly what I wanted.
– user11206537
yesterday
I did! And it worked! That's a very efficient solution.
– user11206537
2 days ago
I did! And it worked! That's a very efficient solution.
– user11206537
2 days ago
@user11206537 Did you measure and compared the efficiency among the solutions. Can you share with us?
– Grijesh Chauhan
2 days ago
@user11206537 Did you measure and compared the efficiency among the solutions. Can you share with us?
– Grijesh Chauhan
2 days ago
This kind of solution was the one that was actually needed as a part of my school project and my teacher thought of it as a very efficient solution. This code is also written along the lines of my current programming style, which is exactly what I wanted.
– user11206537
yesterday
This kind of solution was the one that was actually needed as a part of my school project and my teacher thought of it as a very efficient solution. This code is also written along the lines of my current programming style, which is exactly what I wanted.
– user11206537
yesterday
add a comment |
user11206537 is a new contributor. Be nice, and check out our Code of Conduct.
user11206537 is a new contributor. Be nice, and check out our Code of Conduct.
user11206537 is a new contributor. Be nice, and check out our Code of Conduct.
user11206537 is a new contributor. Be nice, and check out our Code of Conduct.
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3
Band indent at line 8
– DirtyBit
Mar 18 at 7:21
1
You can use Dictionary to store the maximum number and its count as item_no = , if you max is different then original in item_no, reinitialize it and add that item and add count =1
– dkb
Mar 18 at 7:21
3
("Band" probably is a misspelling for "Bad")
– tripleee
Mar 18 at 7:26
1
@tripleee Indeed. bulls-eye!
– DirtyBit
Mar 18 at 7:39
1
As answers show, it's more Pythonic to do it with list comprehensions, rather than loops. But if you use loops to iterate over the list, a style tip: don't use the same variable name
i
both for indicesi in range(5)
then also for items/values:for i in item_no
. Better to dofor no in item_no
– smci
Mar 19 at 0:35