Sorting upon the first element of a nested list and show that as numerical values
$begingroup$
I wish to sort the below list upon the first elements:
{{1/2 (-1 - Sqrt[2]), {{1, 11}, {1, 12}, {2, 11}}},
{-1 - Sqrt[2], {{5, 11}, {5, 12}}},
{1/2 (1 - Sqrt[2]), {{1, 9}, {1, 10}}},
{1 - Sqrt[2], {{7, 9}, {7, 10}}},
{1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
The desired output can be shaped as:
{{-2.414, {{5, 11}, {5, 12}}},
{-1.207, {{1, 11}, {1, 12},{2,11}}},
{-0.414, {{7, 9}, {7, 10}}},
{-0.207, {{1, 9}, {1,10}}},
{0.207, {{1, 7}, {1, 8}}}}
where the first elements are numeric values (without Sqrt symbols) and just for 3-digit precision.
I have tried with N function. But the problem gives rise the effects on any pairs for example output be shaped as {-2.41421, {{5., 11.}, {5., 12.}}} (with a dot after 5 or 11.
Another problem is related to sorting of list just based on the first elements and NOT other elements.
list-manipulation numerics sorting
edited May 16 at 20:01
Unbelievable
2,240931
asked May 16 at 9:47
Inzo BabariaInzo Babaria
51629
$endgroup$
add a comment |
$begingroup$
I wish to sort the below list upon the first elements:
{{1/2 (-1 - Sqrt[2]), {{1, 11}, {1, 12}, {2, 11}}},
{-1 - Sqrt[2], {{5, 11}, {5, 12}}},
{1/2 (1 - Sqrt[2]), {{1, 9}, {1, 10}}},
{1 - Sqrt[2], {{7, 9}, {7, 10}}},
{1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
The desired output can be shaped as:
{{-2.414, {{5, 11}, {5, 12}}},
{-1.207, {{1, 11}, {1, 12},{2,11}}},
{-0.414, {{7, 9}, {7, 10}}},
{-0.207, {{1, 9}, {1,10}}},
{0.207, {{1, 7}, {1, 8}}}}
where the first elements are numeric values (without Sqrt symbols) and just for 3-digit precision.
I have tried with N function. But the problem gives rise the effects on any pairs for example output be shaped as {-2.41421, {{5., 11.}, {5., 12.}}} (with a dot after 5 or 11.
Another problem is related to sorting of list just based on the first elements and NOT other elements.
list-manipulation numerics sorting
edited May 16 at 20:01
Unbelievable
2,240931
asked May 16 at 9:47
Inzo BabariaInzo Babaria
51629
$endgroup$
$begingroup$
If you want to keep the exact forms of the first element you can useSortBy[lst, First[N[#]]&]
$endgroup$
– N.J.Evans
May 16 at 17:43
add a comment |
$begingroup$
I wish to sort the below list upon the first elements:
{{1/2 (-1 - Sqrt[2]), {{1, 11}, {1, 12}, {2, 11}}},
{-1 - Sqrt[2], {{5, 11}, {5, 12}}},
{1/2 (1 - Sqrt[2]), {{1, 9}, {1, 10}}},
{1 - Sqrt[2], {{7, 9}, {7, 10}}},
{1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
The desired output can be shaped as:
{{-2.414, {{5, 11}, {5, 12}}},
{-1.207, {{1, 11}, {1, 12},{2,11}}},
{-0.414, {{7, 9}, {7, 10}}},
{-0.207, {{1, 9}, {1,10}}},
{0.207, {{1, 7}, {1, 8}}}}
where the first elements are numeric values (without Sqrt symbols) and just for 3-digit precision.
I have tried with N function. But the problem gives rise the effects on any pairs for example output be shaped as {-2.41421, {{5., 11.}, {5., 12.}}} (with a dot after 5 or 11.
Another problem is related to sorting of list just based on the first elements and NOT other elements.
list-manipulation numerics sorting
edited May 16 at 20:01
Unbelievable
2,240931
asked May 16 at 9:47
Inzo BabariaInzo Babaria
51629
$endgroup$
I wish to sort the below list upon the first elements:
{{1/2 (-1 - Sqrt[2]), {{1, 11}, {1, 12}, {2, 11}}},
{-1 - Sqrt[2], {{5, 11}, {5, 12}}},
{1/2 (1 - Sqrt[2]), {{1, 9}, {1, 10}}},
{1 - Sqrt[2], {{7, 9}, {7, 10}}},
{1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
The desired output can be shaped as:
{{-2.414, {{5, 11}, {5, 12}}},
{-1.207, {{1, 11}, {1, 12},{2,11}}},
{-0.414, {{7, 9}, {7, 10}}},
{-0.207, {{1, 9}, {1,10}}},
{0.207, {{1, 7}, {1, 8}}}}
where the first elements are numeric values (without Sqrt symbols) and just for 3-digit precision.
I have tried with N function. But the problem gives rise the effects on any pairs for example output be shaped as {-2.41421, {{5., 11.}, {5., 12.}}} (with a dot after 5 or 11.
Another problem is related to sorting of list just based on the first elements and NOT other elements.
list-manipulation numerics sorting
list-manipulation numerics sorting
edited May 16 at 20:01
Unbelievable
2,240931
asked May 16 at 9:47
Inzo BabariaInzo Babaria
51629
edited May 16 at 20:01
Unbelievable
2,240931
asked May 16 at 9:47
Inzo BabariaInzo Babaria
51629
edited May 16 at 20:01
Unbelievable
2,240931
edited May 16 at 20:01
Unbelievable
2,240931
edited May 16 at 20:01
Unbelievable
2,240931
2,240931
asked May 16 at 9:47
Inzo BabariaInzo Babaria
51629
asked May 16 at 9:47
Inzo BabariaInzo Babaria
51629
asked May 16 at 9:47
Inzo BabariaInzo Babaria
51629
51629
$begingroup$
If you want to keep the exact forms of the first element you can useSortBy[lst, First[N[#]]&]
$endgroup$
– N.J.Evans
May 16 at 17:43
add a comment |
$begingroup$
If you want to keep the exact forms of the first element you can useSortBy[lst, First[N[#]]&]
$endgroup$
– N.J.Evans
May 16 at 17:43
$begingroup$
If you want to keep the exact forms of the first element you can use
SortBy[lst, First[N[#]]&]$endgroup$
– N.J.Evans
May 16 at 17:43
$begingroup$
If you want to keep the exact forms of the first element you can use
SortBy[lst, First[N[#]]&]$endgroup$
– N.J.Evans
May 16 at 17:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
SortBy[MapAt[N, lst, {All, 1}], First]
{{-2.41421, {{5, 11}, {5, 12}}}, {-1.20711, {{1, 11}, {1, 12}, {2,
11}}}, {-0.414214, {{7, 9}, {7, 10}}}, {-0.207107, {{1, 9}, {1,
10}}}, {0.207107, {{1, 7}, {1, 8}}}}
SortBy[N @* First] @ lst
{{-1 - Sqrt[2], {{5, 11}, {5, 12}}}, {1/2 (-1 - Sqrt[2]), {{1,
11}, {1, 12}, {2, 11}}}, {1 -
Sqrt[2], {{7, 9}, {7, 10}}}, {1/2 (1 - Sqrt[2]), {{1, 9}, {1,
10}}}, {1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
answered May 16 at 9:57
kglrkglr
198k10223449
$endgroup$
$begingroup$
Thankxxx. It is very professional solution. Can you write theSortBy[N @* First] @ lstin a simpler way to understand how it does work.
$endgroup$
– Inzo Babaria
May 16 at 10:04
$begingroup$
I could not understand the mean of@*
$endgroup$
– Inzo Babaria
May 16 at 10:05
1
$begingroup$
@Inzo, it is short form for compositionComposition[N, First],N@*Firstis the same function asN[First@#]&.
$endgroup$
– kglr
May 16 at 10:12
add a comment |
$begingroup$
list={{1/2 (-1 - Sqrt[2]), {{1, 11}, {1, 12}, {2, 11}}},{-1 - Sqrt[2], {{5, 11}, {5, 12}}},{1/2 (1 - Sqrt[2]), {{1, 9}, {1, 10}}}, {1 - Sqrt[2], {{7, 9}, {7, 10}}},{1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
SortBy[N@list,First]
{{-2.41421, {{5., 11.}, {5., 12.}}}, {-1.20711, {{1., 11.}, {1.,
12.}, {2., 11.}}}, {-0.414214, {{7., 9.}, {7.,
10.}}}, {-0.207107, {{1., 9.}, {1., 10.}}}, {0.207107, {{1.,
7.}, {1., 8.}}}}
answered May 16 at 9:57
J42161217J42161217
5,425525
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
SortBy[MapAt[N, lst, {All, 1}], First]
{{-2.41421, {{5, 11}, {5, 12}}}, {-1.20711, {{1, 11}, {1, 12}, {2,
11}}}, {-0.414214, {{7, 9}, {7, 10}}}, {-0.207107, {{1, 9}, {1,
10}}}, {0.207107, {{1, 7}, {1, 8}}}}
SortBy[N @* First] @ lst
{{-1 - Sqrt[2], {{5, 11}, {5, 12}}}, {1/2 (-1 - Sqrt[2]), {{1,
11}, {1, 12}, {2, 11}}}, {1 -
Sqrt[2], {{7, 9}, {7, 10}}}, {1/2 (1 - Sqrt[2]), {{1, 9}, {1,
10}}}, {1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
answered May 16 at 9:57
kglrkglr
198k10223449
$endgroup$
$begingroup$
Thankxxx. It is very professional solution. Can you write theSortBy[N @* First] @ lstin a simpler way to understand how it does work.
$endgroup$
– Inzo Babaria
May 16 at 10:04
$begingroup$
I could not understand the mean of@*
$endgroup$
– Inzo Babaria
May 16 at 10:05
1
$begingroup$
@Inzo, it is short form for compositionComposition[N, First],N@*Firstis the same function asN[First@#]&.
$endgroup$
– kglr
May 16 at 10:12
add a comment |
$begingroup$
SortBy[MapAt[N, lst, {All, 1}], First]
{{-2.41421, {{5, 11}, {5, 12}}}, {-1.20711, {{1, 11}, {1, 12}, {2,
11}}}, {-0.414214, {{7, 9}, {7, 10}}}, {-0.207107, {{1, 9}, {1,
10}}}, {0.207107, {{1, 7}, {1, 8}}}}
SortBy[N @* First] @ lst
{{-1 - Sqrt[2], {{5, 11}, {5, 12}}}, {1/2 (-1 - Sqrt[2]), {{1,
11}, {1, 12}, {2, 11}}}, {1 -
Sqrt[2], {{7, 9}, {7, 10}}}, {1/2 (1 - Sqrt[2]), {{1, 9}, {1,
10}}}, {1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
answered May 16 at 9:57
kglrkglr
198k10223449
$endgroup$
$begingroup$
Thankxxx. It is very professional solution. Can you write theSortBy[N @* First] @ lstin a simpler way to understand how it does work.
$endgroup$
– Inzo Babaria
May 16 at 10:04
$begingroup$
I could not understand the mean of@*
$endgroup$
– Inzo Babaria
May 16 at 10:05
1
$begingroup$
@Inzo, it is short form for compositionComposition[N, First],N@*Firstis the same function asN[First@#]&.
$endgroup$
– kglr
May 16 at 10:12
add a comment |
$begingroup$
SortBy[MapAt[N, lst, {All, 1}], First]
{{-2.41421, {{5, 11}, {5, 12}}}, {-1.20711, {{1, 11}, {1, 12}, {2,
11}}}, {-0.414214, {{7, 9}, {7, 10}}}, {-0.207107, {{1, 9}, {1,
10}}}, {0.207107, {{1, 7}, {1, 8}}}}
SortBy[N @* First] @ lst
{{-1 - Sqrt[2], {{5, 11}, {5, 12}}}, {1/2 (-1 - Sqrt[2]), {{1,
11}, {1, 12}, {2, 11}}}, {1 -
Sqrt[2], {{7, 9}, {7, 10}}}, {1/2 (1 - Sqrt[2]), {{1, 9}, {1,
10}}}, {1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
answered May 16 at 9:57
kglrkglr
198k10223449
$endgroup$
SortBy[MapAt[N, lst, {All, 1}], First]
{{-2.41421, {{5, 11}, {5, 12}}}, {-1.20711, {{1, 11}, {1, 12}, {2,
11}}}, {-0.414214, {{7, 9}, {7, 10}}}, {-0.207107, {{1, 9}, {1,
10}}}, {0.207107, {{1, 7}, {1, 8}}}}
SortBy[N @* First] @ lst
{{-1 - Sqrt[2], {{5, 11}, {5, 12}}}, {1/2 (-1 - Sqrt[2]), {{1,
11}, {1, 12}, {2, 11}}}, {1 -
Sqrt[2], {{7, 9}, {7, 10}}}, {1/2 (1 - Sqrt[2]), {{1, 9}, {1,
10}}}, {1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
answered May 16 at 9:57
kglrkglr
198k10223449
edited May 16 at 10:02
answered May 16 at 9:57
kglrkglr
198k10223449
answered May 16 at 9:57
kglrkglr
198k10223449
answered May 16 at 9:57
kglrkglr
198k10223449
198k10223449
$begingroup$
Thankxxx. It is very professional solution. Can you write theSortBy[N @* First] @ lstin a simpler way to understand how it does work.
$endgroup$
– Inzo Babaria
May 16 at 10:04
$begingroup$
I could not understand the mean of@*
$endgroup$
– Inzo Babaria
May 16 at 10:05
1
$begingroup$
@Inzo, it is short form for compositionComposition[N, First],N@*Firstis the same function asN[First@#]&.
$endgroup$
– kglr
May 16 at 10:12
add a comment |
$begingroup$
Thankxxx. It is very professional solution. Can you write theSortBy[N @* First] @ lstin a simpler way to understand how it does work.
$endgroup$
– Inzo Babaria
May 16 at 10:04
$begingroup$
I could not understand the mean of@*
$endgroup$
– Inzo Babaria
May 16 at 10:05
1
$begingroup$
@Inzo, it is short form for compositionComposition[N, First],N@*Firstis the same function asN[First@#]&.
$endgroup$
– kglr
May 16 at 10:12
$begingroup$
Thankxxx. It is very professional solution. Can you write the
SortBy[N @* First] @ lst in a simpler way to understand how it does work.$endgroup$
– Inzo Babaria
May 16 at 10:04
$begingroup$
Thankxxx. It is very professional solution. Can you write the
SortBy[N @* First] @ lst in a simpler way to understand how it does work.$endgroup$
– Inzo Babaria
May 16 at 10:04
$begingroup$
I could not understand the mean of
@*$endgroup$
– Inzo Babaria
May 16 at 10:05
$begingroup$
I could not understand the mean of
@*$endgroup$
– Inzo Babaria
May 16 at 10:05
1
1
$begingroup$
@Inzo, it is short form for composition
Composition[N, First], N@*First is the same function as N[First@#]&.$endgroup$
– kglr
May 16 at 10:12
$begingroup$
@Inzo, it is short form for composition
Composition[N, First], N@*First is the same function as N[First@#]&.$endgroup$
– kglr
May 16 at 10:12
add a comment |
$begingroup$
list={{1/2 (-1 - Sqrt[2]), {{1, 11}, {1, 12}, {2, 11}}},{-1 - Sqrt[2], {{5, 11}, {5, 12}}},{1/2 (1 - Sqrt[2]), {{1, 9}, {1, 10}}}, {1 - Sqrt[2], {{7, 9}, {7, 10}}},{1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
SortBy[N@list,First]
{{-2.41421, {{5., 11.}, {5., 12.}}}, {-1.20711, {{1., 11.}, {1.,
12.}, {2., 11.}}}, {-0.414214, {{7., 9.}, {7.,
10.}}}, {-0.207107, {{1., 9.}, {1., 10.}}}, {0.207107, {{1.,
7.}, {1., 8.}}}}
answered May 16 at 9:57
J42161217J42161217
5,425525
$endgroup$
add a comment |
$begingroup$
list={{1/2 (-1 - Sqrt[2]), {{1, 11}, {1, 12}, {2, 11}}},{-1 - Sqrt[2], {{5, 11}, {5, 12}}},{1/2 (1 - Sqrt[2]), {{1, 9}, {1, 10}}}, {1 - Sqrt[2], {{7, 9}, {7, 10}}},{1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
SortBy[N@list,First]
{{-2.41421, {{5., 11.}, {5., 12.}}}, {-1.20711, {{1., 11.}, {1.,
12.}, {2., 11.}}}, {-0.414214, {{7., 9.}, {7.,
10.}}}, {-0.207107, {{1., 9.}, {1., 10.}}}, {0.207107, {{1.,
7.}, {1., 8.}}}}
answered May 16 at 9:57
J42161217J42161217
5,425525
$endgroup$
add a comment |
$begingroup$
list={{1/2 (-1 - Sqrt[2]), {{1, 11}, {1, 12}, {2, 11}}},{-1 - Sqrt[2], {{5, 11}, {5, 12}}},{1/2 (1 - Sqrt[2]), {{1, 9}, {1, 10}}}, {1 - Sqrt[2], {{7, 9}, {7, 10}}},{1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
SortBy[N@list,First]
{{-2.41421, {{5., 11.}, {5., 12.}}}, {-1.20711, {{1., 11.}, {1.,
12.}, {2., 11.}}}, {-0.414214, {{7., 9.}, {7.,
10.}}}, {-0.207107, {{1., 9.}, {1., 10.}}}, {0.207107, {{1.,
7.}, {1., 8.}}}}
answered May 16 at 9:57
J42161217J42161217
5,425525
$endgroup$
list={{1/2 (-1 - Sqrt[2]), {{1, 11}, {1, 12}, {2, 11}}},{-1 - Sqrt[2], {{5, 11}, {5, 12}}},{1/2 (1 - Sqrt[2]), {{1, 9}, {1, 10}}}, {1 - Sqrt[2], {{7, 9}, {7, 10}}},{1/2 (-1 + Sqrt[2]), {{1, 7}, {1, 8}}}}
SortBy[N@list,First]
{{-2.41421, {{5., 11.}, {5., 12.}}}, {-1.20711, {{1., 11.}, {1.,
12.}, {2., 11.}}}, {-0.414214, {{7., 9.}, {7.,
10.}}}, {-0.207107, {{1., 9.}, {1., 10.}}}, {0.207107, {{1.,
7.}, {1., 8.}}}}
answered May 16 at 9:57
J42161217J42161217
5,425525
edited May 16 at 10:23
answered May 16 at 9:57
J42161217J42161217
5,425525
answered May 16 at 9:57
J42161217J42161217
5,425525
answered May 16 at 9:57
J42161217J42161217
5,425525
5,425525
add a comment |
add a comment |
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$begingroup$
If you want to keep the exact forms of the first element you can use
SortBy[lst, First[N[#]]&]$endgroup$
– N.J.Evans
May 16 at 17:43