How to reverse every other sublist of a list?
5
$begingroup$
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 Apr 9 at 20:32
nanjunnanjun
44629
$endgroup$
add a comment |
5
$begingroup$
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 Apr 9 at 20:32
nanjunnanjun
44629
$endgroup$
add a comment |
5
5
5
2
$begingroup$
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 Apr 9 at 20:32
nanjunnanjun
44629
$endgroup$
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 Apr 9 at 20:32
nanjunnanjun
44629
asked Apr 9 at 20:32
nanjunnanjun
44629
asked Apr 9 at 20:32
nanjunnanjun
44629
asked Apr 9 at 20:32
nanjunnanjun
44629
asked Apr 9 at 20:32
nanjunnanjun
44629
44629
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
13
$begingroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, {2 ;; ;; 2}]
(* {{1, 2}, {4, 3}, {5, 6}, {8, 7}} *)
answered Apr 9 at 20:33
marchmarch
17.6k22870
$endgroup$
add a comment |
9
$begingroup$
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 Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k583169
$endgroup$
2
$begingroup$
A bit cleaner, s/b comparable in speed:Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
$endgroup$
– ciao
Apr 10 at 21:00
add a comment |
4
$begingroup$
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 Apr 9 at 21:14
Jason B.Jason B.
49k390197
$endgroup$
add a comment |
3
$begingroup$
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 Apr 9 at 23:16
cphyscphys
815
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.
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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
13
$begingroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, {2 ;; ;; 2}]
(* {{1, 2}, {4, 3}, {5, 6}, {8, 7}} *)
answered Apr 9 at 20:33
marchmarch
17.6k22870
$endgroup$
add a comment |
13
$begingroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, {2 ;; ;; 2}]
(* {{1, 2}, {4, 3}, {5, 6}, {8, 7}} *)
answered Apr 9 at 20:33
marchmarch
17.6k22870
$endgroup$
add a comment |
13
13
13
$begingroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, {2 ;; ;; 2}]
(* {{1, 2}, {4, 3}, {5, 6}, {8, 7}} *)
answered Apr 9 at 20:33
marchmarch
17.6k22870
$endgroup$
Use MapAt, which accepts the same syntax as Part:
MapAt[Reverse, list, {2 ;; ;; 2}]
(* {{1, 2}, {4, 3}, {5, 6}, {8, 7}} *)
answered Apr 9 at 20:33
marchmarch
17.6k22870
answered Apr 9 at 20:33
marchmarch
17.6k22870
answered Apr 9 at 20:33
marchmarch
17.6k22870
answered Apr 9 at 20:33
marchmarch
17.6k22870
17.6k22870
add a comment |
add a comment |
9
$begingroup$
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 Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k583169
$endgroup$
2
$begingroup$
A bit cleaner, s/b comparable in speed:Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
$endgroup$
– ciao
Apr 10 at 21:00
add a comment |
9
$begingroup$
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 Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k583169
$endgroup$
2
$begingroup$
A bit cleaner, s/b comparable in speed:Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
$endgroup$
– ciao
Apr 10 at 21:00
add a comment |
9
9
9
$begingroup$
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 Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k583169
$endgroup$
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 Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k583169
edited Apr 10 at 6:12
answered Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k583169
answered Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k583169
answered Apr 9 at 21:24
Henrik SchumacherHenrik Schumacher
60.2k583169
60.2k583169
2
$begingroup$
A bit cleaner, s/b comparable in speed:Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
$endgroup$
– ciao
Apr 10 at 21:00
add a comment |
2
$begingroup$
A bit cleaner, s/b comparable in speed:Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]
$endgroup$
– ciao
Apr 10 at 21:00
2
2
$begingroup$
A bit cleaner, s/b comparable in speed:
Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]$endgroup$
– ciao
Apr 10 at 21:00
$begingroup$
A bit cleaner, s/b comparable in speed:
Riffle[list[[;; ;; 2]], list[[2 ;; ;; 2, -1 ;; 1 ;; -1]]]$endgroup$
– ciao
Apr 10 at 21:00
add a comment |
4
$begingroup$
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 Apr 9 at 21:14
Jason B.Jason B.
49k390197
$endgroup$
add a comment |
4
$begingroup$
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 Apr 9 at 21:14
Jason B.Jason B.
49k390197
$endgroup$
add a comment |
4
4
4
$begingroup$
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 Apr 9 at 21:14
Jason B.Jason B.
49k390197
$endgroup$
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 Apr 9 at 21:14
Jason B.Jason B.
49k390197
answered Apr 9 at 21:14
Jason B.Jason B.
49k390197
answered Apr 9 at 21:14
Jason B.Jason B.
49k390197
answered Apr 9 at 21:14
Jason B.Jason B.
49k390197
49k390197
add a comment |
add a comment |
3
$begingroup$
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 Apr 9 at 23:16
cphyscphys
815
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 |
3
$begingroup$
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 Apr 9 at 23:16
cphyscphys
815
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 |
3
3
3
$begingroup$
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 Apr 9 at 23:16
cphyscphys
815
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 Apr 9 at 23:16
cphyscphys
815
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 Apr 9 at 23:16
cphyscphys
815
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 Apr 9 at 23:16
cphyscphys
815
answered Apr 9 at 23:16
cphyscphys
815
815
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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