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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 1 hour ago
nanjunnanjun
43229
$endgroup$
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 1 hour ago
nanjunnanjun
43229
$endgroup$
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 1 hour ago
nanjunnanjun
43229
$endgroup$
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 1 hour ago
nanjunnanjun
43229
asked 1 hour ago
nanjunnanjun
43229
asked 1 hour ago
nanjunnanjun
43229
asked 1 hour ago
nanjunnanjun
43229
asked 1 hour ago
nanjunnanjun
43229
43229
add a comment |
add a comment |
3 Answers
3
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 1 hour ago
marchmarch
17.5k22769
$endgroup$
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 32 mins ago
Jason B.Jason B.
48.9k389196
$endgroup$
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 22 mins ago
Henrik SchumacherHenrik Schumacher
59.5k582166
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
marchmarch
17.5k22769
answered 1 hour ago
marchmarch
17.5k22769
answered 1 hour ago
marchmarch
17.5k22769
answered 1 hour ago
marchmarch
17.5k22769
17.5k22769
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 32 mins 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 32 mins 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 32 mins 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 32 mins ago
Jason B.Jason B.
48.9k389196
answered 32 mins ago
Jason B.Jason B.
48.9k389196
answered 32 mins ago
Jason B.Jason B.
48.9k389196
answered 32 mins ago
Jason B.Jason B.
48.9k389196
48.9k389196
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 22 mins ago
Henrik SchumacherHenrik Schumacher
59.5k582166
$endgroup$
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 22 mins ago
Henrik SchumacherHenrik Schumacher
59.5k582166
$endgroup$
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 22 mins ago
Henrik SchumacherHenrik Schumacher
59.5k582166
$endgroup$
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 22 mins ago
Henrik SchumacherHenrik Schumacher
59.5k582166
answered 22 mins ago
Henrik SchumacherHenrik Schumacher
59.5k582166
answered 22 mins ago
Henrik SchumacherHenrik Schumacher
59.5k582166
answered 22 mins ago
Henrik SchumacherHenrik Schumacher
59.5k582166
59.5k582166
add a comment |
add a comment |
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