How to apply differences on part of a list and keep the rest?Is there a good way to map a function over a list to lists exclusively of a certain depth?Monitor the list for changesHow to apply a function of several arguments to a list?Split according to List and apply ruleSetting the value of a list itemCombining Map with DropHow to combine Nest and list PartDifferences applied to a list of matricesApplying Rest and Most to sublists of listMultiplying a list a vectors by the same matrix
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How to apply differences on part of a list and keep the rest?
Is there a good way to map a function over a list to lists exclusively of a certain depth?Monitor the list for changesHow to apply a function of several arguments to a list?Split according to List and apply ruleSetting the value of a list itemCombining Map with DropHow to combine Nest and list PartDifferences applied to a list of matricesApplying Rest and Most to sublists of listMultiplying a list a vectors by the same matrix
$begingroup$
I have a list,
l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y
and want to apply differences on the third parts and keep the parts right of the numerals collected.
My result should be
l2 = 2, c, k, 7, k, m, -11, m, y
I tried Map and MapAt, but I could not get anywhere. I could work around split things up and connect again. But is there a better way to do it?
list-manipulation
edited May 2 at 3:53
user64494
3,65111222
asked May 1 at 14:21
user57467user57467
563
$endgroup$
add a comment |
$begingroup$
I have a list,
l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y
and want to apply differences on the third parts and keep the parts right of the numerals collected.
My result should be
l2 = 2, c, k, 7, k, m, -11, m, y
I tried Map and MapAt, but I could not get anywhere. I could work around split things up and connect again. But is there a better way to do it?
list-manipulation
edited May 2 at 3:53
user64494
3,65111222
asked May 1 at 14:21
user57467user57467
563
$endgroup$
add a comment |
$begingroup$
I have a list,
l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y
and want to apply differences on the third parts and keep the parts right of the numerals collected.
My result should be
l2 = 2, c, k, 7, k, m, -11, m, y
I tried Map and MapAt, but I could not get anywhere. I could work around split things up and connect again. But is there a better way to do it?
list-manipulation
edited May 2 at 3:53
user64494
3,65111222
asked May 1 at 14:21
user57467user57467
563
$endgroup$
I have a list,
l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y
and want to apply differences on the third parts and keep the parts right of the numerals collected.
My result should be
l2 = 2, c, k, 7, k, m, -11, m, y
I tried Map and MapAt, but I could not get anywhere. I could work around split things up and connect again. But is there a better way to do it?
list-manipulation
list-manipulation
edited May 2 at 3:53
user64494
3,65111222
asked May 1 at 14:21
user57467user57467
563
edited May 2 at 3:53
user64494
3,65111222
asked May 1 at 14:21
user57467user57467
563
edited May 2 at 3:53
user64494
3,65111222
edited May 2 at 3:53
user64494
3,65111222
edited May 2 at 3:53
user64494
3,65111222
3,65111222
asked May 1 at 14:21
user57467user57467
563
asked May 1 at 14:21
user57467user57467
563
asked May 1 at 14:21
user57467user57467
563
563
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Perhaps this?:
l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y;
l2 = Differences[l1[[All, 3 ;;]]] /. b_ - a_ :> Sequence[a, b]
(* 2, c, k, 7, k, m, -11, m, y *)
It assumes the letter symbols are simple and not complicated expressions.
This is more complicated, but more robust:
Flatten /@
Transpose@
MapAt[Differences,
Partition[Transpose@l1[[All, 3 ;;]], 1, 2, 1, 1], 1, All, 1]
answered May 1 at 14:32
Michael E2Michael E2
151k12204485
$endgroup$
$begingroup$
Thank you. I have to digest your answer.
$endgroup$
– user57467
May 1 at 15:03
$begingroup$
How would you do it if l1= z,a, b, 3, c, w,e, f, 5, k, q,n, k, 12, m, p,s, t, 1, y;
$endgroup$
– user57467
May 1 at 15:36
$begingroup$
@user57467 I would change the third column to the fourth:l1[[All, 4 ;;]]. You can programmatically get the column withcol = Position[First@l1, _?NumericQ][[1, 1]]orcol = Position[First@l1, _Integer][[1, 1]]; then usel1[[All, col ;;]].
$endgroup$
– Michael E2
May 1 at 15:44
add a comment |
$begingroup$
You can also use BlockMap as follows:
BlockMap[#[[3]].-1, 1, ## & @@ Flatten@#[[4 ;;]] &@* Transpose, l1, 2, 1]
2, c, k, 7, k, m, -11, m, y
or
BlockMap[#[[1]].-1, 1, ## & @@ Flatten@ #[[2 ;;]] &@*Transpose, l1[[All, 3 ;;]], 2, 1]
2, c, k, 7, k, m, -11, m, y
answered May 1 at 15:49
kglrkglr
191k10211430
$endgroup$
$begingroup$
I am impressed!
$endgroup$
– user57467
May 1 at 16:08
add a comment |
$begingroup$
This is very similar to kglr's first solution but picks the relevant quantities a bit more explicitly:
l2 = BlockMap[#[[2, 3]] - #[[1, 3]], #[[1, 4]], #[[2, 4]] &, l1, 2, 1]
2, c, k, 7, k, m, -11, m, y
With a parameter to change the symbolic column quickly:
l2 = With[col = 3,
BlockMap[#[[2,col]] - #[[1,col]], #[[1,col+1]], #[[2,col+1]] &, l1, 2, 1]]
2, c, k, 7, k, m, -11, m, y
answered May 1 at 16:31
RomanRoman
7,15511134
$endgroup$
$begingroup$
Wow! Good! I enjoy the clarity!
$endgroup$
– user57467
2 days ago
add a comment |
$begingroup$
A solution with MapThread on an offset Partition.
MapThread[Sequence @@ #@#2 &, Differences, Identity, Transpose@#] & /@
Partition[l1[[All, 3 ;;]], 2, 1]
2, c, k, 7, k, m, -11, m, y
Differences is applied to the integers while Identity preserves the form of the symbols.
Hope this helps.
answered May 1 at 21:11
EdmundEdmund
26.9k330103
$endgroup$
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
$begingroup$
Perhaps this?:
l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y;
l2 = Differences[l1[[All, 3 ;;]]] /. b_ - a_ :> Sequence[a, b]
(* 2, c, k, 7, k, m, -11, m, y *)
It assumes the letter symbols are simple and not complicated expressions.
This is more complicated, but more robust:
Flatten /@
Transpose@
MapAt[Differences,
Partition[Transpose@l1[[All, 3 ;;]], 1, 2, 1, 1], 1, All, 1]
answered May 1 at 14:32
Michael E2Michael E2
151k12204485
$endgroup$
$begingroup$
Thank you. I have to digest your answer.
$endgroup$
– user57467
May 1 at 15:03
$begingroup$
How would you do it if l1= z,a, b, 3, c, w,e, f, 5, k, q,n, k, 12, m, p,s, t, 1, y;
$endgroup$
– user57467
May 1 at 15:36
$begingroup$
@user57467 I would change the third column to the fourth:l1[[All, 4 ;;]]. You can programmatically get the column withcol = Position[First@l1, _?NumericQ][[1, 1]]orcol = Position[First@l1, _Integer][[1, 1]]; then usel1[[All, col ;;]].
$endgroup$
– Michael E2
May 1 at 15:44
add a comment |
$begingroup$
Perhaps this?:
l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y;
l2 = Differences[l1[[All, 3 ;;]]] /. b_ - a_ :> Sequence[a, b]
(* 2, c, k, 7, k, m, -11, m, y *)
It assumes the letter symbols are simple and not complicated expressions.
This is more complicated, but more robust:
Flatten /@
Transpose@
MapAt[Differences,
Partition[Transpose@l1[[All, 3 ;;]], 1, 2, 1, 1], 1, All, 1]
answered May 1 at 14:32
Michael E2Michael E2
151k12204485
$endgroup$
$begingroup$
Thank you. I have to digest your answer.
$endgroup$
– user57467
May 1 at 15:03
$begingroup$
How would you do it if l1= z,a, b, 3, c, w,e, f, 5, k, q,n, k, 12, m, p,s, t, 1, y;
$endgroup$
– user57467
May 1 at 15:36
$begingroup$
@user57467 I would change the third column to the fourth:l1[[All, 4 ;;]]. You can programmatically get the column withcol = Position[First@l1, _?NumericQ][[1, 1]]orcol = Position[First@l1, _Integer][[1, 1]]; then usel1[[All, col ;;]].
$endgroup$
– Michael E2
May 1 at 15:44
add a comment |
$begingroup$
Perhaps this?:
l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y;
l2 = Differences[l1[[All, 3 ;;]]] /. b_ - a_ :> Sequence[a, b]
(* 2, c, k, 7, k, m, -11, m, y *)
It assumes the letter symbols are simple and not complicated expressions.
This is more complicated, but more robust:
Flatten /@
Transpose@
MapAt[Differences,
Partition[Transpose@l1[[All, 3 ;;]], 1, 2, 1, 1], 1, All, 1]
answered May 1 at 14:32
Michael E2Michael E2
151k12204485
$endgroup$
Perhaps this?:
l1 = a, b, 3, c, e, f, 5, k, n, k, 12, m, s, t, 1, y;
l2 = Differences[l1[[All, 3 ;;]]] /. b_ - a_ :> Sequence[a, b]
(* 2, c, k, 7, k, m, -11, m, y *)
It assumes the letter symbols are simple and not complicated expressions.
This is more complicated, but more robust:
Flatten /@
Transpose@
MapAt[Differences,
Partition[Transpose@l1[[All, 3 ;;]], 1, 2, 1, 1], 1, All, 1]
answered May 1 at 14:32
Michael E2Michael E2
151k12204485
edited May 1 at 14:39
answered May 1 at 14:32
Michael E2Michael E2
151k12204485
answered May 1 at 14:32
Michael E2Michael E2
151k12204485
answered May 1 at 14:32
Michael E2Michael E2
151k12204485
151k12204485
$begingroup$
Thank you. I have to digest your answer.
$endgroup$
– user57467
May 1 at 15:03
$begingroup$
How would you do it if l1= z,a, b, 3, c, w,e, f, 5, k, q,n, k, 12, m, p,s, t, 1, y;
$endgroup$
– user57467
May 1 at 15:36
$begingroup$
@user57467 I would change the third column to the fourth:l1[[All, 4 ;;]]. You can programmatically get the column withcol = Position[First@l1, _?NumericQ][[1, 1]]orcol = Position[First@l1, _Integer][[1, 1]]; then usel1[[All, col ;;]].
$endgroup$
– Michael E2
May 1 at 15:44
add a comment |
$begingroup$
Thank you. I have to digest your answer.
$endgroup$
– user57467
May 1 at 15:03
$begingroup$
How would you do it if l1= z,a, b, 3, c, w,e, f, 5, k, q,n, k, 12, m, p,s, t, 1, y;
$endgroup$
– user57467
May 1 at 15:36
$begingroup$
@user57467 I would change the third column to the fourth:l1[[All, 4 ;;]]. You can programmatically get the column withcol = Position[First@l1, _?NumericQ][[1, 1]]orcol = Position[First@l1, _Integer][[1, 1]]; then usel1[[All, col ;;]].
$endgroup$
– Michael E2
May 1 at 15:44
$begingroup$
Thank you. I have to digest your answer.
$endgroup$
– user57467
May 1 at 15:03
$begingroup$
Thank you. I have to digest your answer.
$endgroup$
– user57467
May 1 at 15:03
$begingroup$
How would you do it if l1= z,a, b, 3, c, w,e, f, 5, k, q,n, k, 12, m, p,s, t, 1, y;
$endgroup$
– user57467
May 1 at 15:36
$begingroup$
How would you do it if l1= z,a, b, 3, c, w,e, f, 5, k, q,n, k, 12, m, p,s, t, 1, y;
$endgroup$
– user57467
May 1 at 15:36
$begingroup$
@user57467 I would change the third column to the fourth:
l1[[All, 4 ;;]]. You can programmatically get the column with col = Position[First@l1, _?NumericQ][[1, 1]] or col = Position[First@l1, _Integer][[1, 1]]; then use l1[[All, col ;;]].$endgroup$
– Michael E2
May 1 at 15:44
$begingroup$
@user57467 I would change the third column to the fourth:
l1[[All, 4 ;;]]. You can programmatically get the column with col = Position[First@l1, _?NumericQ][[1, 1]] or col = Position[First@l1, _Integer][[1, 1]]; then use l1[[All, col ;;]].$endgroup$
– Michael E2
May 1 at 15:44
add a comment |
$begingroup$
You can also use BlockMap as follows:
BlockMap[#[[3]].-1, 1, ## & @@ Flatten@#[[4 ;;]] &@* Transpose, l1, 2, 1]
2, c, k, 7, k, m, -11, m, y
or
BlockMap[#[[1]].-1, 1, ## & @@ Flatten@ #[[2 ;;]] &@*Transpose, l1[[All, 3 ;;]], 2, 1]
2, c, k, 7, k, m, -11, m, y
answered May 1 at 15:49
kglrkglr
191k10211430
$endgroup$
$begingroup$
I am impressed!
$endgroup$
– user57467
May 1 at 16:08
add a comment |
$begingroup$
You can also use BlockMap as follows:
BlockMap[#[[3]].-1, 1, ## & @@ Flatten@#[[4 ;;]] &@* Transpose, l1, 2, 1]
2, c, k, 7, k, m, -11, m, y
or
BlockMap[#[[1]].-1, 1, ## & @@ Flatten@ #[[2 ;;]] &@*Transpose, l1[[All, 3 ;;]], 2, 1]
2, c, k, 7, k, m, -11, m, y
answered May 1 at 15:49
kglrkglr
191k10211430
$endgroup$
$begingroup$
I am impressed!
$endgroup$
– user57467
May 1 at 16:08
add a comment |
$begingroup$
You can also use BlockMap as follows:
BlockMap[#[[3]].-1, 1, ## & @@ Flatten@#[[4 ;;]] &@* Transpose, l1, 2, 1]
2, c, k, 7, k, m, -11, m, y
or
BlockMap[#[[1]].-1, 1, ## & @@ Flatten@ #[[2 ;;]] &@*Transpose, l1[[All, 3 ;;]], 2, 1]
2, c, k, 7, k, m, -11, m, y
answered May 1 at 15:49
kglrkglr
191k10211430
$endgroup$
You can also use BlockMap as follows:
BlockMap[#[[3]].-1, 1, ## & @@ Flatten@#[[4 ;;]] &@* Transpose, l1, 2, 1]
2, c, k, 7, k, m, -11, m, y
or
BlockMap[#[[1]].-1, 1, ## & @@ Flatten@ #[[2 ;;]] &@*Transpose, l1[[All, 3 ;;]], 2, 1]
2, c, k, 7, k, m, -11, m, y
answered May 1 at 15:49
kglrkglr
191k10211430
edited May 1 at 15:57
answered May 1 at 15:49
kglrkglr
191k10211430
answered May 1 at 15:49
kglrkglr
191k10211430
answered May 1 at 15:49
kglrkglr
191k10211430
191k10211430
$begingroup$
I am impressed!
$endgroup$
– user57467
May 1 at 16:08
add a comment |
$begingroup$
I am impressed!
$endgroup$
– user57467
May 1 at 16:08
$begingroup$
I am impressed!
$endgroup$
– user57467
May 1 at 16:08
$begingroup$
I am impressed!
$endgroup$
– user57467
May 1 at 16:08
add a comment |
$begingroup$
This is very similar to kglr's first solution but picks the relevant quantities a bit more explicitly:
l2 = BlockMap[#[[2, 3]] - #[[1, 3]], #[[1, 4]], #[[2, 4]] &, l1, 2, 1]
2, c, k, 7, k, m, -11, m, y
With a parameter to change the symbolic column quickly:
l2 = With[col = 3,
BlockMap[#[[2,col]] - #[[1,col]], #[[1,col+1]], #[[2,col+1]] &, l1, 2, 1]]
2, c, k, 7, k, m, -11, m, y
answered May 1 at 16:31
RomanRoman
7,15511134
$endgroup$
$begingroup$
Wow! Good! I enjoy the clarity!
$endgroup$
– user57467
2 days ago
add a comment |
$begingroup$
This is very similar to kglr's first solution but picks the relevant quantities a bit more explicitly:
l2 = BlockMap[#[[2, 3]] - #[[1, 3]], #[[1, 4]], #[[2, 4]] &, l1, 2, 1]
2, c, k, 7, k, m, -11, m, y
With a parameter to change the symbolic column quickly:
l2 = With[col = 3,
BlockMap[#[[2,col]] - #[[1,col]], #[[1,col+1]], #[[2,col+1]] &, l1, 2, 1]]
2, c, k, 7, k, m, -11, m, y
answered May 1 at 16:31
RomanRoman
7,15511134
$endgroup$
$begingroup$
Wow! Good! I enjoy the clarity!
$endgroup$
– user57467
2 days ago
add a comment |
$begingroup$
This is very similar to kglr's first solution but picks the relevant quantities a bit more explicitly:
l2 = BlockMap[#[[2, 3]] - #[[1, 3]], #[[1, 4]], #[[2, 4]] &, l1, 2, 1]
2, c, k, 7, k, m, -11, m, y
With a parameter to change the symbolic column quickly:
l2 = With[col = 3,
BlockMap[#[[2,col]] - #[[1,col]], #[[1,col+1]], #[[2,col+1]] &, l1, 2, 1]]
2, c, k, 7, k, m, -11, m, y
answered May 1 at 16:31
RomanRoman
7,15511134
$endgroup$
This is very similar to kglr's first solution but picks the relevant quantities a bit more explicitly:
l2 = BlockMap[#[[2, 3]] - #[[1, 3]], #[[1, 4]], #[[2, 4]] &, l1, 2, 1]
2, c, k, 7, k, m, -11, m, y
With a parameter to change the symbolic column quickly:
l2 = With[col = 3,
BlockMap[#[[2,col]] - #[[1,col]], #[[1,col+1]], #[[2,col+1]] &, l1, 2, 1]]
2, c, k, 7, k, m, -11, m, y
answered May 1 at 16:31
RomanRoman
7,15511134
edited May 1 at 17:09
answered May 1 at 16:31
RomanRoman
7,15511134
answered May 1 at 16:31
RomanRoman
7,15511134
answered May 1 at 16:31
RomanRoman
7,15511134
7,15511134
$begingroup$
Wow! Good! I enjoy the clarity!
$endgroup$
– user57467
2 days ago
add a comment |
$begingroup$
Wow! Good! I enjoy the clarity!
$endgroup$
– user57467
2 days ago
$begingroup$
Wow! Good! I enjoy the clarity!
$endgroup$
– user57467
2 days ago
$begingroup$
Wow! Good! I enjoy the clarity!
$endgroup$
– user57467
2 days ago
add a comment |
$begingroup$
A solution with MapThread on an offset Partition.
MapThread[Sequence @@ #@#2 &, Differences, Identity, Transpose@#] & /@
Partition[l1[[All, 3 ;;]], 2, 1]
2, c, k, 7, k, m, -11, m, y
Differences is applied to the integers while Identity preserves the form of the symbols.
Hope this helps.
answered May 1 at 21:11
EdmundEdmund
26.9k330103
$endgroup$
add a comment |
$begingroup$
A solution with MapThread on an offset Partition.
MapThread[Sequence @@ #@#2 &, Differences, Identity, Transpose@#] & /@
Partition[l1[[All, 3 ;;]], 2, 1]
2, c, k, 7, k, m, -11, m, y
Differences is applied to the integers while Identity preserves the form of the symbols.
Hope this helps.
answered May 1 at 21:11
EdmundEdmund
26.9k330103
$endgroup$
add a comment |
$begingroup$
A solution with MapThread on an offset Partition.
MapThread[Sequence @@ #@#2 &, Differences, Identity, Transpose@#] & /@
Partition[l1[[All, 3 ;;]], 2, 1]
2, c, k, 7, k, m, -11, m, y
Differences is applied to the integers while Identity preserves the form of the symbols.
Hope this helps.
answered May 1 at 21:11
EdmundEdmund
26.9k330103
$endgroup$
A solution with MapThread on an offset Partition.
MapThread[Sequence @@ #@#2 &, Differences, Identity, Transpose@#] & /@
Partition[l1[[All, 3 ;;]], 2, 1]
2, c, k, 7, k, m, -11, m, y
Differences is applied to the integers while Identity preserves the form of the symbols.
Hope this helps.
answered May 1 at 21:11
EdmundEdmund
26.9k330103
answered May 1 at 21:11
EdmundEdmund
26.9k330103
answered May 1 at 21:11
EdmundEdmund
26.9k330103
answered May 1 at 21:11
EdmundEdmund
26.9k330103
26.9k330103
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