Cleanest way to take a[b[c]] to a[b][c]
2
$begingroup$
As indicated in the title I'm looking for the fastest way to transform a[b[c]] into a[b][c], and the natural generalization to an arbitrary chaining of arguments. I'm sure there's got to be a convenient way that I've overlooked.
In my cases a, b, and c can be any expression with any complicated internal structure they like.
function-construction
edited 22 mins ago
David G. Stork
24.1k22153
asked 1 hour ago
b3m2a1b3m2a1
27.2k257156
$endgroup$
add a comment |
2
$begingroup$
As indicated in the title I'm looking for the fastest way to transform a[b[c]] into a[b][c], and the natural generalization to an arbitrary chaining of arguments. I'm sure there's got to be a convenient way that I've overlooked.
In my cases a, b, and c can be any expression with any complicated internal structure they like.
function-construction
edited 22 mins ago
David G. Stork
24.1k22153
asked 1 hour ago
b3m2a1b3m2a1
27.2k257156
$endgroup$
1
$begingroup$
The solution likely would useOperate.
$endgroup$
– QuantumDot
56 mins ago
$begingroup$
f = Curry[Replace][a_[b_[c_]] :> a[b][c]]also works, so thata[b[c]] // fgives the desired result.
$endgroup$
– Shredderroy
47 mins ago
$begingroup$
How about generalizing the question to takinga[b[c[d[...]]]]toa[b][c][d]...?
$endgroup$
– David G. Stork
35 mins ago
$begingroup$
@DavidG.Stork sorry I thought that was implicit
$endgroup$
– b3m2a1
30 mins ago
$begingroup$
@b3m2a1: Oh.... well I recommend you alter the question... and seem my new solution.
$endgroup$
– David G. Stork
27 mins ago
add a comment |
2
2
2
$begingroup$
As indicated in the title I'm looking for the fastest way to transform a[b[c]] into a[b][c], and the natural generalization to an arbitrary chaining of arguments. I'm sure there's got to be a convenient way that I've overlooked.
In my cases a, b, and c can be any expression with any complicated internal structure they like.
function-construction
edited 22 mins ago
David G. Stork
24.1k22153
asked 1 hour ago
b3m2a1b3m2a1
27.2k257156
$endgroup$
As indicated in the title I'm looking for the fastest way to transform a[b[c]] into a[b][c], and the natural generalization to an arbitrary chaining of arguments. I'm sure there's got to be a convenient way that I've overlooked.
In my cases a, b, and c can be any expression with any complicated internal structure they like.
function-construction
function-construction
edited 22 mins ago
David G. Stork
24.1k22153
asked 1 hour ago
b3m2a1b3m2a1
27.2k257156
edited 22 mins ago
David G. Stork
24.1k22153
asked 1 hour ago
b3m2a1b3m2a1
27.2k257156
edited 22 mins ago
David G. Stork
24.1k22153
edited 22 mins ago
David G. Stork
24.1k22153
edited 22 mins ago
David G. Stork
24.1k22153
24.1k22153
asked 1 hour ago
b3m2a1b3m2a1
27.2k257156
asked 1 hour ago
b3m2a1b3m2a1
27.2k257156
asked 1 hour ago
b3m2a1b3m2a1
27.2k257156
27.2k257156
1
$begingroup$
The solution likely would useOperate.
$endgroup$
– QuantumDot
56 mins ago
$begingroup$
f = Curry[Replace][a_[b_[c_]] :> a[b][c]]also works, so thata[b[c]] // fgives the desired result.
$endgroup$
– Shredderroy
47 mins ago
$begingroup$
How about generalizing the question to takinga[b[c[d[...]]]]toa[b][c][d]...?
$endgroup$
– David G. Stork
35 mins ago
$begingroup$
@DavidG.Stork sorry I thought that was implicit
$endgroup$
– b3m2a1
30 mins ago
$begingroup$
@b3m2a1: Oh.... well I recommend you alter the question... and seem my new solution.
$endgroup$
– David G. Stork
27 mins ago
add a comment |
1
$begingroup$
The solution likely would useOperate.
$endgroup$
– QuantumDot
56 mins ago
$begingroup$
f = Curry[Replace][a_[b_[c_]] :> a[b][c]]also works, so thata[b[c]] // fgives the desired result.
$endgroup$
– Shredderroy
47 mins ago
$begingroup$
How about generalizing the question to takinga[b[c[d[...]]]]toa[b][c][d]...?
$endgroup$
– David G. Stork
35 mins ago
$begingroup$
@DavidG.Stork sorry I thought that was implicit
$endgroup$
– b3m2a1
30 mins ago
$begingroup$
@b3m2a1: Oh.... well I recommend you alter the question... and seem my new solution.
$endgroup$
– David G. Stork
27 mins ago
1
1
$begingroup$
The solution likely would use
Operate.$endgroup$
– QuantumDot
56 mins ago
$begingroup$
The solution likely would use
Operate.$endgroup$
– QuantumDot
56 mins ago
$begingroup$
f = Curry[Replace][a_[b_[c_]] :> a[b][c]] also works, so that a[b[c]] // f gives the desired result.$endgroup$
– Shredderroy
47 mins ago
$begingroup$
f = Curry[Replace][a_[b_[c_]] :> a[b][c]] also works, so that a[b[c]] // f gives the desired result.$endgroup$
– Shredderroy
47 mins ago
$begingroup$
How about generalizing the question to taking
a[b[c[d[...]]]] to a[b][c][d]...?$endgroup$
– David G. Stork
35 mins ago
$begingroup$
How about generalizing the question to taking
a[b[c[d[...]]]] to a[b][c][d]...?$endgroup$
– David G. Stork
35 mins ago
$begingroup$
@DavidG.Stork sorry I thought that was implicit
$endgroup$
– b3m2a1
30 mins ago
$begingroup$
@DavidG.Stork sorry I thought that was implicit
$endgroup$
– b3m2a1
30 mins ago
$begingroup$
@b3m2a1: Oh.... well I recommend you alter the question... and seem my new solution.
$endgroup$
– David G. Stork
27 mins ago
$begingroup$
@b3m2a1: Oh.... well I recommend you alter the question... and seem my new solution.
$endgroup$
– David G. Stork
27 mins ago
add a comment |
2 Answers
2
active
oldest
votes
2
$begingroup$
Operate[#[[0]], First@#] &[a[b[c]]]
a[b][c]
ClearAll[deCompose]
deCompose = Nest[Operate[#[[0]], First@#] &, #, Depth[#] - 2] &;
deCompose@a[b[c]]
a[b][c]
exp = Compose[a, b, c, d, e, f, g]
a[b[c[d[e[f[g]]]]]]
deCompose @ exp
a[b][c][d][e][f][g]
answered 31 mins ago
kglrkglr
180k9200413
$endgroup$
add a comment |
2
$begingroup$
Not particularly "clean," but it works:
Operate[Head[#], Level[#, 2][[2]]] & @ a[b[c]]
For the full generalization:
Nest[Operate[Head[#], Level[#, 2][[2]]] & , #, Depth[#] -2] & @
a[b[c[d[e]]]]
answered 48 mins ago
David G. StorkDavid G. Stork
24.1k22153
$endgroup$
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
2
$begingroup$
Operate[#[[0]], First@#] &[a[b[c]]]
a[b][c]
ClearAll[deCompose]
deCompose = Nest[Operate[#[[0]], First@#] &, #, Depth[#] - 2] &;
deCompose@a[b[c]]
a[b][c]
exp = Compose[a, b, c, d, e, f, g]
a[b[c[d[e[f[g]]]]]]
deCompose @ exp
a[b][c][d][e][f][g]
answered 31 mins ago
kglrkglr
180k9200413
$endgroup$
add a comment |
2
$begingroup$
Operate[#[[0]], First@#] &[a[b[c]]]
a[b][c]
ClearAll[deCompose]
deCompose = Nest[Operate[#[[0]], First@#] &, #, Depth[#] - 2] &;
deCompose@a[b[c]]
a[b][c]
exp = Compose[a, b, c, d, e, f, g]
a[b[c[d[e[f[g]]]]]]
deCompose @ exp
a[b][c][d][e][f][g]
answered 31 mins ago
kglrkglr
180k9200413
$endgroup$
add a comment |
2
2
2
$begingroup$
Operate[#[[0]], First@#] &[a[b[c]]]
a[b][c]
ClearAll[deCompose]
deCompose = Nest[Operate[#[[0]], First@#] &, #, Depth[#] - 2] &;
deCompose@a[b[c]]
a[b][c]
exp = Compose[a, b, c, d, e, f, g]
a[b[c[d[e[f[g]]]]]]
deCompose @ exp
a[b][c][d][e][f][g]
answered 31 mins ago
kglrkglr
180k9200413
$endgroup$
Operate[#[[0]], First@#] &[a[b[c]]]
a[b][c]
ClearAll[deCompose]
deCompose = Nest[Operate[#[[0]], First@#] &, #, Depth[#] - 2] &;
deCompose@a[b[c]]
a[b][c]
exp = Compose[a, b, c, d, e, f, g]
a[b[c[d[e[f[g]]]]]]
deCompose @ exp
a[b][c][d][e][f][g]
answered 31 mins ago
kglrkglr
180k9200413
edited 23 mins ago
answered 31 mins ago
kglrkglr
180k9200413
answered 31 mins ago
kglrkglr
180k9200413
answered 31 mins ago
kglrkglr
180k9200413
180k9200413
add a comment |
add a comment |
2
$begingroup$
Not particularly "clean," but it works:
Operate[Head[#], Level[#, 2][[2]]] & @ a[b[c]]
For the full generalization:
Nest[Operate[Head[#], Level[#, 2][[2]]] & , #, Depth[#] -2] & @
a[b[c[d[e]]]]
answered 48 mins ago
David G. StorkDavid G. Stork
24.1k22153
$endgroup$
add a comment |
2
$begingroup$
Not particularly "clean," but it works:
Operate[Head[#], Level[#, 2][[2]]] & @ a[b[c]]
For the full generalization:
Nest[Operate[Head[#], Level[#, 2][[2]]] & , #, Depth[#] -2] & @
a[b[c[d[e]]]]
answered 48 mins ago
David G. StorkDavid G. Stork
24.1k22153
$endgroup$
add a comment |
2
2
2
$begingroup$
Not particularly "clean," but it works:
Operate[Head[#], Level[#, 2][[2]]] & @ a[b[c]]
For the full generalization:
Nest[Operate[Head[#], Level[#, 2][[2]]] & , #, Depth[#] -2] & @
a[b[c[d[e]]]]
answered 48 mins ago
David G. StorkDavid G. Stork
24.1k22153
$endgroup$
Not particularly "clean," but it works:
Operate[Head[#], Level[#, 2][[2]]] & @ a[b[c]]
For the full generalization:
Nest[Operate[Head[#], Level[#, 2][[2]]] & , #, Depth[#] -2] & @
a[b[c[d[e]]]]
answered 48 mins ago
David G. StorkDavid G. Stork
24.1k22153
edited 23 mins ago
answered 48 mins ago
David G. StorkDavid G. Stork
24.1k22153
answered 48 mins ago
David G. StorkDavid G. Stork
24.1k22153
answered 48 mins ago
David G. StorkDavid G. Stork
24.1k22153
24.1k22153
add a comment |
add a comment |
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1
$begingroup$
The solution likely would use
Operate.$endgroup$
– QuantumDot
56 mins ago
$begingroup$
f = Curry[Replace][a_[b_[c_]] :> a[b][c]]also works, so thata[b[c]] // fgives the desired result.$endgroup$
– Shredderroy
47 mins ago
$begingroup$
How about generalizing the question to taking
a[b[c[d[...]]]]toa[b][c][d]...?$endgroup$
– David G. Stork
35 mins ago
$begingroup$
@DavidG.Stork sorry I thought that was implicit
$endgroup$
– b3m2a1
30 mins ago
$begingroup$
@b3m2a1: Oh.... well I recommend you alter the question... and seem my new solution.
$endgroup$
– David G. Stork
27 mins ago