How to make FoldListIndexed ie FoldList with an index or iterator?
4
$begingroup$
Is there a functional or inbuilt way to add an index to FoldList?
i.e.:
FoldListIndex[f,x,{a,b,...}]
gives
{x,f[x,a,1],f[f[x,a,1],b,2],...}
My current method with a loop is quite unsatisfactory
list-manipulation function-construction functional-style
asked Jan 30 at 7:55
user62657user62657
232
$endgroup$
add a comment |
4
$begingroup$
Is there a functional or inbuilt way to add an index to FoldList?
i.e.:
FoldListIndex[f,x,{a,b,...}]
gives
{x,f[x,a,1],f[f[x,a,1],b,2],...}
My current method with a loop is quite unsatisfactory
list-manipulation function-construction functional-style
asked Jan 30 at 7:55
user62657user62657
232
$endgroup$
add a comment |
4
4
4
2
$begingroup$
Is there a functional or inbuilt way to add an index to FoldList?
i.e.:
FoldListIndex[f,x,{a,b,...}]
gives
{x,f[x,a,1],f[f[x,a,1],b,2],...}
My current method with a loop is quite unsatisfactory
list-manipulation function-construction functional-style
asked Jan 30 at 7:55
user62657user62657
232
$endgroup$
Is there a functional or inbuilt way to add an index to FoldList?
i.e.:
FoldListIndex[f,x,{a,b,...}]
gives
{x,f[x,a,1],f[f[x,a,1],b,2],...}
My current method with a loop is quite unsatisfactory
list-manipulation function-construction functional-style
list-manipulation function-construction functional-style
asked Jan 30 at 7:55
user62657user62657
232
asked Jan 30 at 7:55
user62657user62657
232
asked Jan 30 at 7:55
user62657user62657
232
asked Jan 30 at 7:55
user62657user62657
232
asked Jan 30 at 7:55
user62657user62657
232
232
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
3
$begingroup$
foldIndexedList = Module[{i = 1, f = #}, FoldList[f[##, i++] &, ##2]] &;
foldIndexedList[f, x, {a, b, c, d}]
{x, f[x, a, 1], f[f[x, a, 1], b, 2], f[f[f[x, a, 1], b, 2], c, 3],
f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
foldIndexedList2 = Module[{f = #},
FoldList[f[#, ## & @@ #2] &, #2, MapIndexed[{#, #2[[1]]} &]@#3]] &;
foldIndexedList2[f, x, {a, b, c, d}]
{x, f[x, a, 1], f[f[x, a, 1], b, 2], f[f[f[x, a, 1], b, 2], c, 3],
f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
answered Jan 30 at 9:00
kglrkglr
188k10204422
$endgroup$
$begingroup$
Thanks rafalc and kglr for the answers. I selected kglr as it resulted in a faster function for my specific implementation and it worked with foldIndexedList[f,{a,b,c,d}] as well
$endgroup$
– user62657
Jan 30 at 9:37
$begingroup$
@user62657, thank you for the accept; and welcome to mma.se.
$endgroup$
– kglr
Jan 30 at 9:50
add a comment |
3
$begingroup$
You can try
FoldListIndexed[f_, x_, lst_] :=
FoldList[
Function[{a, b}, f[a, Sequence @@ b]],
x,
Transpose[{lst, Range @ Length @ lst}]
]
and then
In[4]:= FoldListIndexed[f, x, {a, b, c, d}]
Out[4]= {x, f[x, a, 1], f[f[x, a, 1], b, 2],
f[f[f[x, a, 1], b, 2], c, 3], f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
answered Jan 30 at 8:22
rafalcrafalc
698212
$endgroup$
$begingroup$
Nice solution, but I would rather avoid usingBlock, especially for functions living inGlobal`. Imagine thatghas been defined globally, and is called byf- then this code will break it in a very non-obvious way. I would rather use a pure function forg, andWithinstead ofBlock.
$endgroup$
– Leonid Shifrin
Jan 30 at 8:35
$begingroup$
@LeonidShifrin thank you, I have edited my answer following your suggestions (got rid of Block entirely - I think the code is still readable even without the helper variable)
$endgroup$
– rafalc
Jan 30 at 8:47
add a comment |
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2 Answers
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2 Answers
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3
$begingroup$
foldIndexedList = Module[{i = 1, f = #}, FoldList[f[##, i++] &, ##2]] &;
foldIndexedList[f, x, {a, b, c, d}]
{x, f[x, a, 1], f[f[x, a, 1], b, 2], f[f[f[x, a, 1], b, 2], c, 3],
f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
foldIndexedList2 = Module[{f = #},
FoldList[f[#, ## & @@ #2] &, #2, MapIndexed[{#, #2[[1]]} &]@#3]] &;
foldIndexedList2[f, x, {a, b, c, d}]
{x, f[x, a, 1], f[f[x, a, 1], b, 2], f[f[f[x, a, 1], b, 2], c, 3],
f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
answered Jan 30 at 9:00
kglrkglr
188k10204422
$endgroup$
$begingroup$
Thanks rafalc and kglr for the answers. I selected kglr as it resulted in a faster function for my specific implementation and it worked with foldIndexedList[f,{a,b,c,d}] as well
$endgroup$
– user62657
Jan 30 at 9:37
$begingroup$
@user62657, thank you for the accept; and welcome to mma.se.
$endgroup$
– kglr
Jan 30 at 9:50
add a comment |
3
$begingroup$
foldIndexedList = Module[{i = 1, f = #}, FoldList[f[##, i++] &, ##2]] &;
foldIndexedList[f, x, {a, b, c, d}]
{x, f[x, a, 1], f[f[x, a, 1], b, 2], f[f[f[x, a, 1], b, 2], c, 3],
f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
foldIndexedList2 = Module[{f = #},
FoldList[f[#, ## & @@ #2] &, #2, MapIndexed[{#, #2[[1]]} &]@#3]] &;
foldIndexedList2[f, x, {a, b, c, d}]
{x, f[x, a, 1], f[f[x, a, 1], b, 2], f[f[f[x, a, 1], b, 2], c, 3],
f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
answered Jan 30 at 9:00
kglrkglr
188k10204422
$endgroup$
$begingroup$
Thanks rafalc and kglr for the answers. I selected kglr as it resulted in a faster function for my specific implementation and it worked with foldIndexedList[f,{a,b,c,d}] as well
$endgroup$
– user62657
Jan 30 at 9:37
$begingroup$
@user62657, thank you for the accept; and welcome to mma.se.
$endgroup$
– kglr
Jan 30 at 9:50
add a comment |
3
3
3
$begingroup$
foldIndexedList = Module[{i = 1, f = #}, FoldList[f[##, i++] &, ##2]] &;
foldIndexedList[f, x, {a, b, c, d}]
{x, f[x, a, 1], f[f[x, a, 1], b, 2], f[f[f[x, a, 1], b, 2], c, 3],
f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
foldIndexedList2 = Module[{f = #},
FoldList[f[#, ## & @@ #2] &, #2, MapIndexed[{#, #2[[1]]} &]@#3]] &;
foldIndexedList2[f, x, {a, b, c, d}]
{x, f[x, a, 1], f[f[x, a, 1], b, 2], f[f[f[x, a, 1], b, 2], c, 3],
f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
answered Jan 30 at 9:00
kglrkglr
188k10204422
$endgroup$
foldIndexedList = Module[{i = 1, f = #}, FoldList[f[##, i++] &, ##2]] &;
foldIndexedList[f, x, {a, b, c, d}]
{x, f[x, a, 1], f[f[x, a, 1], b, 2], f[f[f[x, a, 1], b, 2], c, 3],
f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
foldIndexedList2 = Module[{f = #},
FoldList[f[#, ## & @@ #2] &, #2, MapIndexed[{#, #2[[1]]} &]@#3]] &;
foldIndexedList2[f, x, {a, b, c, d}]
{x, f[x, a, 1], f[f[x, a, 1], b, 2], f[f[f[x, a, 1], b, 2], c, 3],
f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
answered Jan 30 at 9:00
kglrkglr
188k10204422
edited Jan 30 at 9:08
answered Jan 30 at 9:00
kglrkglr
188k10204422
answered Jan 30 at 9:00
kglrkglr
188k10204422
answered Jan 30 at 9:00
kglrkglr
188k10204422
188k10204422
$begingroup$
Thanks rafalc and kglr for the answers. I selected kglr as it resulted in a faster function for my specific implementation and it worked with foldIndexedList[f,{a,b,c,d}] as well
$endgroup$
– user62657
Jan 30 at 9:37
$begingroup$
@user62657, thank you for the accept; and welcome to mma.se.
$endgroup$
– kglr
Jan 30 at 9:50
add a comment |
$begingroup$
Thanks rafalc and kglr for the answers. I selected kglr as it resulted in a faster function for my specific implementation and it worked with foldIndexedList[f,{a,b,c,d}] as well
$endgroup$
– user62657
Jan 30 at 9:37
$begingroup$
@user62657, thank you for the accept; and welcome to mma.se.
$endgroup$
– kglr
Jan 30 at 9:50
$begingroup$
Thanks rafalc and kglr for the answers. I selected kglr as it resulted in a faster function for my specific implementation and it worked with foldIndexedList[f,{a,b,c,d}] as well
$endgroup$
– user62657
Jan 30 at 9:37
$begingroup$
Thanks rafalc and kglr for the answers. I selected kglr as it resulted in a faster function for my specific implementation and it worked with foldIndexedList[f,{a,b,c,d}] as well
$endgroup$
– user62657
Jan 30 at 9:37
$begingroup$
@user62657, thank you for the accept; and welcome to mma.se.
$endgroup$
– kglr
Jan 30 at 9:50
$begingroup$
@user62657, thank you for the accept; and welcome to mma.se.
$endgroup$
– kglr
Jan 30 at 9:50
add a comment |
3
$begingroup$
You can try
FoldListIndexed[f_, x_, lst_] :=
FoldList[
Function[{a, b}, f[a, Sequence @@ b]],
x,
Transpose[{lst, Range @ Length @ lst}]
]
and then
In[4]:= FoldListIndexed[f, x, {a, b, c, d}]
Out[4]= {x, f[x, a, 1], f[f[x, a, 1], b, 2],
f[f[f[x, a, 1], b, 2], c, 3], f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
answered Jan 30 at 8:22
rafalcrafalc
698212
$endgroup$
$begingroup$
Nice solution, but I would rather avoid usingBlock, especially for functions living inGlobal`. Imagine thatghas been defined globally, and is called byf- then this code will break it in a very non-obvious way. I would rather use a pure function forg, andWithinstead ofBlock.
$endgroup$
– Leonid Shifrin
Jan 30 at 8:35
$begingroup$
@LeonidShifrin thank you, I have edited my answer following your suggestions (got rid of Block entirely - I think the code is still readable even without the helper variable)
$endgroup$
– rafalc
Jan 30 at 8:47
add a comment |
3
$begingroup$
You can try
FoldListIndexed[f_, x_, lst_] :=
FoldList[
Function[{a, b}, f[a, Sequence @@ b]],
x,
Transpose[{lst, Range @ Length @ lst}]
]
and then
In[4]:= FoldListIndexed[f, x, {a, b, c, d}]
Out[4]= {x, f[x, a, 1], f[f[x, a, 1], b, 2],
f[f[f[x, a, 1], b, 2], c, 3], f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
answered Jan 30 at 8:22
rafalcrafalc
698212
$endgroup$
$begingroup$
Nice solution, but I would rather avoid usingBlock, especially for functions living inGlobal`. Imagine thatghas been defined globally, and is called byf- then this code will break it in a very non-obvious way. I would rather use a pure function forg, andWithinstead ofBlock.
$endgroup$
– Leonid Shifrin
Jan 30 at 8:35
$begingroup$
@LeonidShifrin thank you, I have edited my answer following your suggestions (got rid of Block entirely - I think the code is still readable even without the helper variable)
$endgroup$
– rafalc
Jan 30 at 8:47
add a comment |
3
3
3
$begingroup$
You can try
FoldListIndexed[f_, x_, lst_] :=
FoldList[
Function[{a, b}, f[a, Sequence @@ b]],
x,
Transpose[{lst, Range @ Length @ lst}]
]
and then
In[4]:= FoldListIndexed[f, x, {a, b, c, d}]
Out[4]= {x, f[x, a, 1], f[f[x, a, 1], b, 2],
f[f[f[x, a, 1], b, 2], c, 3], f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
answered Jan 30 at 8:22
rafalcrafalc
698212
$endgroup$
You can try
FoldListIndexed[f_, x_, lst_] :=
FoldList[
Function[{a, b}, f[a, Sequence @@ b]],
x,
Transpose[{lst, Range @ Length @ lst}]
]
and then
In[4]:= FoldListIndexed[f, x, {a, b, c, d}]
Out[4]= {x, f[x, a, 1], f[f[x, a, 1], b, 2],
f[f[f[x, a, 1], b, 2], c, 3], f[f[f[f[x, a, 1], b, 2], c, 3], d, 4]}
answered Jan 30 at 8:22
rafalcrafalc
698212
edited Jan 30 at 8:43
answered Jan 30 at 8:22
rafalcrafalc
698212
answered Jan 30 at 8:22
rafalcrafalc
698212
answered Jan 30 at 8:22
rafalcrafalc
698212
698212
$begingroup$
Nice solution, but I would rather avoid usingBlock, especially for functions living inGlobal`. Imagine thatghas been defined globally, and is called byf- then this code will break it in a very non-obvious way. I would rather use a pure function forg, andWithinstead ofBlock.
$endgroup$
– Leonid Shifrin
Jan 30 at 8:35
$begingroup$
@LeonidShifrin thank you, I have edited my answer following your suggestions (got rid of Block entirely - I think the code is still readable even without the helper variable)
$endgroup$
– rafalc
Jan 30 at 8:47
add a comment |
$begingroup$
Nice solution, but I would rather avoid usingBlock, especially for functions living inGlobal`. Imagine thatghas been defined globally, and is called byf- then this code will break it in a very non-obvious way. I would rather use a pure function forg, andWithinstead ofBlock.
$endgroup$
– Leonid Shifrin
Jan 30 at 8:35
$begingroup$
@LeonidShifrin thank you, I have edited my answer following your suggestions (got rid of Block entirely - I think the code is still readable even without the helper variable)
$endgroup$
– rafalc
Jan 30 at 8:47
$begingroup$
Nice solution, but I would rather avoid using
Block, especially for functions living in Global` . Imagine that g has been defined globally, and is called by f - then this code will break it in a very non-obvious way. I would rather use a pure function for g, and With instead of Block.$endgroup$
– Leonid Shifrin
Jan 30 at 8:35
$begingroup$
Nice solution, but I would rather avoid using
Block, especially for functions living in Global` . Imagine that g has been defined globally, and is called by f - then this code will break it in a very non-obvious way. I would rather use a pure function for g, and With instead of Block.$endgroup$
– Leonid Shifrin
Jan 30 at 8:35
$begingroup$
@LeonidShifrin thank you, I have edited my answer following your suggestions (got rid of Block entirely - I think the code is still readable even without the helper variable)
$endgroup$
– rafalc
Jan 30 at 8:47
$begingroup$
@LeonidShifrin thank you, I have edited my answer following your suggestions (got rid of Block entirely - I think the code is still readable even without the helper variable)
$endgroup$
– rafalc
Jan 30 at 8:47
add a comment |
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