How to reverse every other sublist of a list? The 2019 Stack Overflow Developer Survey Results Are InHow can the lowest non-zero value in a nested list be found and its position in the sublist returned?Comparing elements of the $n^textth$ sublist in a ragged list with the $n^textth$ member of a sequenceSublist inside simple listHow to split list into disjoint lists?Selecting every last element of a nested listAdd elements of one list to sublists of another listEliminating elements from sublists under a global conditionRemove the right sublistList manipulation - adding last element of sublist to each sublistHow to partition a 1D list according to specified lengths?
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How to reverse every other sublist of a list?
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How to reverse every other sublist of a list?
The 2019 Stack Overflow Developer Survey Results Are InHow can the lowest non-zero value in a nested list be found and its position in the sublist returned?Comparing elements of the $n^textth$ sublist in a ragged list with the $n^textth$ member of a sequenceSublist inside simple listHow to split list into disjoint lists?Selecting every last element of a nested listAdd elements of one list to sublists of another listEliminating elements from sublists under a global conditionRemove the right sublistList manipulation - adding last element of sublist to each sublistHow to partition a 1D list according to specified lengths?
$begingroup$
I have a list with sublists list = 1, 2, 3, 4, 5, 6, 7, 8. I would like to reverse every other sublist, starting from the second one. The result would look like result = 1, 2, 4, 3, 5, 6, 8, 7. What would be a simple way to do this?
list-manipulation
asked 8 hours ago
nanjunnanjun
43729
$endgroup$
add a comment |
$begingroup$
I have a list with sublists list = 1, 2, 3, 4, 5, 6, 7, 8. I would like to reverse every other sublist, starting from the second one. The result would look like result = 1, 2, 4, 3, 5, 6, 8, 7. What would be a simple way to do this?
list-manipulation
asked 8 hours ago
nanjunnanjun
43729
$endgroup$
add a comment |
$begingroup$
I have a list with sublists list = 1, 2, 3, 4, 5, 6, 7, 8. I would like to reverse every other sublist, starting from the second one. The result would look like result = 1, 2, 4, 3, 5, 6, 8, 7. What would be a simple way to do this?
list-manipulation
asked 8 hours ago
nanjunnanjun
43729
$endgroup$
I have a list with sublists list = 1, 2, 3, 4, 5, 6, 7, 8. I would like to reverse every other sublist, starting from the second one. The result would look like result = 1, 2, 4, 3, 5, 6, 8, 7. What would be a simple way to do this?
list-manipulation
list-manipulation
asked 8 hours ago
nanjunnanjun
43729
asked 8 hours ago
nanjunnanjun
43729
asked 8 hours ago
nanjunnanjun
43729
asked 8 hours ago
nanjunnanjun
43729
asked 8 hours ago
nanjunnanjun
43729
43729
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, 2 ;; ;; 2]
(* 1, 2, 4, 3, 5, 6, 8, 7 *)
answered 8 hours ago
marchmarch
17.5k22769
$endgroup$
add a comment |
$begingroup$
Using Part and Span might not seem overly elegant but it is fast:
list = RandomReal[-1, 1, 100000, 10];
a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
c = Module[result = list,
result[[2 ;; ;; 2]] =
list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
result
]; // RepeatedTiming // First
a == b == c
0.11
0.317
0.0036
True
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
59.6k582166
$endgroup$
add a comment |
$begingroup$
Another method is to use ReplacePart:
ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
i_ ? EvenQ :> Reverse@list[[i]]
]
(* 1,2,4,3,5,6,8,7 *)
answered 7 hours ago
Jason B.Jason B.
48.9k389196
$endgroup$
add a comment |
$begingroup$
This will work
list = 1, 2, 3, 4, 5, 6, 7, 8;
Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], i, 1, Length[list]]
(*1, 2, 4, 3, 5, 6, 8, 7*)
```
answered 5 hours ago
cphyscphys
615
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, 2 ;; ;; 2]
(* 1, 2, 4, 3, 5, 6, 8, 7 *)
answered 8 hours ago
marchmarch
17.5k22769
$endgroup$
add a comment |
$begingroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, 2 ;; ;; 2]
(* 1, 2, 4, 3, 5, 6, 8, 7 *)
answered 8 hours ago
marchmarch
17.5k22769
$endgroup$
add a comment |
$begingroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, 2 ;; ;; 2]
(* 1, 2, 4, 3, 5, 6, 8, 7 *)
answered 8 hours ago
marchmarch
17.5k22769
$endgroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, 2 ;; ;; 2]
(* 1, 2, 4, 3, 5, 6, 8, 7 *)
answered 8 hours ago
marchmarch
17.5k22769
answered 8 hours ago
marchmarch
17.5k22769
answered 8 hours ago
marchmarch
17.5k22769
answered 8 hours ago
marchmarch
17.5k22769
17.5k22769
add a comment |
add a comment |
$begingroup$
Using Part and Span might not seem overly elegant but it is fast:
list = RandomReal[-1, 1, 100000, 10];
a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
c = Module[result = list,
result[[2 ;; ;; 2]] =
list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
result
]; // RepeatedTiming // First
a == b == c
0.11
0.317
0.0036
True
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
59.6k582166
$endgroup$
add a comment |
$begingroup$
Using Part and Span might not seem overly elegant but it is fast:
list = RandomReal[-1, 1, 100000, 10];
a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
c = Module[result = list,
result[[2 ;; ;; 2]] =
list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
result
]; // RepeatedTiming // First
a == b == c
0.11
0.317
0.0036
True
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
59.6k582166
$endgroup$
add a comment |
$begingroup$
Using Part and Span might not seem overly elegant but it is fast:
list = RandomReal[-1, 1, 100000, 10];
a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
c = Module[result = list,
result[[2 ;; ;; 2]] =
list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
result
]; // RepeatedTiming // First
a == b == c
0.11
0.317
0.0036
True
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
59.6k582166
$endgroup$
Using Part and Span might not seem overly elegant but it is fast:
list = RandomReal[-1, 1, 100000, 10];
a = MapAt[Reverse, list, 2 ;; ;; 2]; // RepeatedTiming // First
b = ReplacePart[list, i_?EvenQ :> Reverse@list[[i]]]; // RepeatedTiming // First
c = Module[result = list,
result[[2 ;; ;; 2]] =
list[[2 ;; ;; 2, -1 ;; 1 ;; -1]];
result
]; // RepeatedTiming // First
a == b == c
0.11
0.317
0.0036
True
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
59.6k582166
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
59.6k582166
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
59.6k582166
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
59.6k582166
59.6k582166
add a comment |
add a comment |
$begingroup$
Another method is to use ReplacePart:
ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
i_ ? EvenQ :> Reverse@list[[i]]
]
(* 1,2,4,3,5,6,8,7 *)
answered 7 hours ago
Jason B.Jason B.
48.9k389196
$endgroup$
add a comment |
$begingroup$
Another method is to use ReplacePart:
ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
i_ ? EvenQ :> Reverse@list[[i]]
]
(* 1,2,4,3,5,6,8,7 *)
answered 7 hours ago
Jason B.Jason B.
48.9k389196
$endgroup$
add a comment |
$begingroup$
Another method is to use ReplacePart:
ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
i_ ? EvenQ :> Reverse@list[[i]]
]
(* 1,2,4,3,5,6,8,7 *)
answered 7 hours ago
Jason B.Jason B.
48.9k389196
$endgroup$
Another method is to use ReplacePart:
ReplacePart[1, 2, 3, 4, 5, 6, 7, 8,
i_ ? EvenQ :> Reverse@list[[i]]
]
(* 1,2,4,3,5,6,8,7 *)
answered 7 hours ago
Jason B.Jason B.
48.9k389196
answered 7 hours ago
Jason B.Jason B.
48.9k389196
answered 7 hours ago
Jason B.Jason B.
48.9k389196
answered 7 hours ago
Jason B.Jason B.
48.9k389196
48.9k389196
add a comment |
add a comment |
$begingroup$
This will work
list = 1, 2, 3, 4, 5, 6, 7, 8;
Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], i, 1, Length[list]]
(*1, 2, 4, 3, 5, 6, 8, 7*)
```
answered 5 hours ago
cphyscphys
615
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This will work
list = 1, 2, 3, 4, 5, 6, 7, 8;
Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], i, 1, Length[list]]
(*1, 2, 4, 3, 5, 6, 8, 7*)
```
answered 5 hours ago
cphyscphys
615
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This will work
list = 1, 2, 3, 4, 5, 6, 7, 8;
Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], i, 1, Length[list]]
(*1, 2, 4, 3, 5, 6, 8, 7*)
```
answered 5 hours ago
cphyscphys
615
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This will work
list = 1, 2, 3, 4, 5, 6, 7, 8;
Table[If[EvenQ[i], Reverse[list[[i]]], list[[i]]], i, 1, Length[list]]
(*1, 2, 4, 3, 5, 6, 8, 7*)
```
answered 5 hours ago
cphyscphys
615
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 5 hours ago
cphyscphys
615
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 5 hours ago
cphyscphys
615
answered 5 hours ago
cphyscphys
615
615
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
cphys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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