How to make FoldListIndexed ie FoldList with an index or iterator?

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

asked 4 hours ago

user62657

182

New contributor

add a comment |

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

asked 4 hours ago

user62657

182

New contributor

add a comment |

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

asked 4 hours ago

user62657

182

New contributor

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 4 hours ago

user62657

182

New contributor

asked 4 hours ago

user62657

182

New contributor

asked 4 hours ago

user62657

182

New contributor

asked 4 hours ago

user62657

182

asked 4 hours ago

user62657

182

New contributor

user62657 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
2

active

oldest

votes

foldIndexedList = Module[{i = 1, f = #}, FoldList[f[##, i++] &, ##2]] &;

foldIndexedList[f, x, {a, b, c, d}]

foldIndexedList2 =  Module[{f = #}, 

    FoldList[f[#, ## & @@ #2] &, #2, MapIndexed[{#, #2[[1]]} &]@#3]] &;

foldIndexedList2[f, x, {a, b, c, d}]

edited 3 hours ago

answered 3 hours ago

kglr

181k10200414

$begingroup$
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
2 hours ago

$begingroup$
@user62657, thank you for the accept; and welcome to mma.se.
$endgroup$
– kglr
2 hours ago

add a comment |

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]}

edited 3 hours ago

answered 3 hours ago

rafalc

563211

$begingroup$
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
3 hours ago

$begingroup$
@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
3 hours ago

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

foldIndexedList = Module[{i = 1, f = #}, FoldList[f[##, i++] &, ##2]] &;

foldIndexedList[f, x, {a, b, c, d}]

foldIndexedList2 =  Module[{f = #}, 

    FoldList[f[#, ## & @@ #2] &, #2, MapIndexed[{#, #2[[1]]} &]@#3]] &;

foldIndexedList2[f, x, {a, b, c, d}]

edited 3 hours ago

answered 3 hours ago

kglr

181k10200414

$begingroup$
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
2 hours ago

$begingroup$
@user62657, thank you for the accept; and welcome to mma.se.
$endgroup$
– kglr
2 hours ago

add a comment |

foldIndexedList = Module[{i = 1, f = #}, FoldList[f[##, i++] &, ##2]] &;

foldIndexedList[f, x, {a, b, c, d}]

foldIndexedList2 =  Module[{f = #}, 

    FoldList[f[#, ## & @@ #2] &, #2, MapIndexed[{#, #2[[1]]} &]@#3]] &;

foldIndexedList2[f, x, {a, b, c, d}]

edited 3 hours ago

answered 3 hours ago

kglr

181k10200414

$begingroup$
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
2 hours ago

$begingroup$
@user62657, thank you for the accept; and welcome to mma.se.
$endgroup$
– kglr
2 hours ago

add a comment |

foldIndexedList = Module[{i = 1, f = #}, FoldList[f[##, i++] &, ##2]] &;

foldIndexedList[f, x, {a, b, c, d}]

foldIndexedList2 =  Module[{f = #}, 

    FoldList[f[#, ## & @@ #2] &, #2, MapIndexed[{#, #2[[1]]} &]@#3]] &;

foldIndexedList2[f, x, {a, b, c, d}]

edited 3 hours ago

answered 3 hours ago

kglr

181k10200414

foldIndexedList = Module[{i = 1, f = #}, FoldList[f[##, i++] &, ##2]] &;

foldIndexedList[f, x, {a, b, c, d}]

foldIndexedList2 =  Module[{f = #}, 

    FoldList[f[#, ## & @@ #2] &, #2, MapIndexed[{#, #2[[1]]} &]@#3]] &;

foldIndexedList2[f, x, {a, b, c, d}]

edited 3 hours ago

answered 3 hours ago

kglr

181k10200414

edited 3 hours ago

answered 3 hours ago

kglr

181k10200414

answered 3 hours ago

kglr

181k10200414

answered 3 hours ago

kglr

181k10200414

$begingroup$
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
2 hours ago

$begingroup$
@user62657, thank you for the accept; and welcome to mma.se.
$endgroup$
– kglr
2 hours ago

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
2 hours ago

$begingroup$
@user62657, thank you for the accept; and welcome to mma.se.
$endgroup$
– kglr
2 hours ago

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

– user62657
2 hours ago

@user62657, thank you for the accept; and welcome to mma.se.

– kglr
2 hours ago

add a comment |

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]}

edited 3 hours ago

answered 3 hours ago

rafalc

563211

$begingroup$
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
3 hours ago

$begingroup$
@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
3 hours ago

add a comment |

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]}

edited 3 hours ago

answered 3 hours ago

rafalc

563211

$begingroup$
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
3 hours ago

$begingroup$
@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
3 hours ago

add a comment |

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]}

edited 3 hours ago

answered 3 hours ago

rafalc

563211

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]}

edited 3 hours ago

answered 3 hours ago

rafalc

563211

edited 3 hours ago

answered 3 hours ago

rafalc

563211

answered 3 hours ago

rafalc

563211

answered 3 hours ago

rafalc

563211

$begingroup$
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
3 hours ago

$begingroup$
@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
3 hours ago

add a comment |

$begingroup$
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
3 hours ago

$begingroup$
@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
3 hours ago

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.

– Leonid Shifrin
3 hours ago

@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)

– rafalc
3 hours ago

add a comment |

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