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Multiplying elements of a list

Replace elements in a listList of lengths of runs of elementsMaking a list of rules from the list of elementsMultiplying columns of a list of matricesHow to list ArrayFilter?Selecting elements from lists of a listAdding $y$ values based on the value of $x$ in a listExtracting the elements from a loop of sets/ listsAdding each element in a list to the previous onecontrariwise multiplying a list into each raw of a matrix

4

$begingroup$

Given I have a following list of numbers:

l={2,3,4,5,6}

I wonder how to multiply each number by the one before of it,
such that I get

{2*3,2*3*4,2*3*4*5,...}

Also how can I chose a level for this operation?
by level I mean,

level 1 : 2*3

level 2 : 2*3*4

and so on.

list-manipulation multiplcation

share|improve this question

asked 3 hours ago

WilliamWilliam

79248

$endgroup$

add a comment | 

4

$begingroup$

Given I have a following list of numbers:

l={2,3,4,5,6}

I wonder how to multiply each number by the one before of it,
such that I get

{2*3,2*3*4,2*3*4*5,...}

Also how can I chose a level for this operation?
by level I mean,

level 1 : 2*3

level 2 : 2*3*4

and so on.

list-manipulation multiplcation

share|improve this question

asked 3 hours ago

WilliamWilliam

79248

$endgroup$

add a comment | 

$begingroup$

Given I have a following list of numbers:

l={2,3,4,5,6}

I wonder how to multiply each number by the one before of it,
such that I get

{2*3,2*3*4,2*3*4*5,...}

Also how can I chose a level for this operation?
by level I mean,

level 1 : 2*3

level 2 : 2*3*4

and so on.

list-manipulation multiplcation

share|improve this question

asked 3 hours ago

WilliamWilliam

79248

$endgroup$

l={2,3,4,5,6}

{2*3,2*3*4,2*3*4*5,...}

share|improve this question

asked 3 hours ago

WilliamWilliam

79248

share|improve this question

asked 3 hours ago

WilliamWilliam

79248

asked 3 hours ago

WilliamWilliam

79248

asked 3 hours ago

WilliamWilliam

79248

4 Answers
4

active

oldest

votes

5

$begingroup$

FoldList[Times, l] (* or *)

FoldList[Times]@l

{2, 6, 24, 120, 720}

Also:

Exp @ Accumulate @ Log @ l

{2, 6, 24, 120, 720}

share|improve this answer

edited 2 hours ago

answered 3 hours ago

kglrkglr

185k10202420

$endgroup$

add a comment | 

3

$begingroup$

list = Range[2, 6]



(* {2, 3, 4, 5, 6} *)

To keep the factors separate, use Inactive with kglr's solution

list2 = Rest@FoldList[Inactive@Times, list[[1]], Rest@list]

Activate produces the result

list3 = list2 // Activate



(* {6, 24, 120, 720} *)

Or use NonCommutativeMultiply to hold the factors

list4 = Rest@FoldList[NonCommutativeMultiply, list[[1]], Rest@list]



(* {2 ** 3, 2 ** 3 ** 4, 2 ** 3 ** 4 ** 5, 2 ** 3 ** 4 ** 5 ** 6} *)

Then Apply Times to get the final result

list5 = Times @@@ list4



(* {6, 24, 120, 720} *)

For this specific case (sequential numbers), the result is just Factorial

list3 == list5 == Factorial /@ Rest@list



(* True *)

Use Part to access any element of the result, e.g.,

list3[[1]]



(* 6 *)

share|improve this answer

answered 1 hour ago

Bob HanlonBob Hanlon

60.5k33597

$endgroup$

add a comment | 

1

$begingroup$

This can also be solved using recursion. The function cumProd (cumulative product) can be defined as:

list = Range[10];

cumProd[n_] := cumProd[n - 1]*list[[n]];

cumProd[1] = list[[1]];

To use:

cumProd[6]

720

gives the 6th "level" of the product. Of course, list can be any set of numbers. Applying this to the whole list:

cumProd/@Range[Length[list]]

{1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800}

share|improve this answer

answered 1 hour ago

bill sbill s

53.5k376152

$endgroup$

add a comment | 

0

$begingroup$

here is your level function

level[x_] := Times @@ l[[;; x + 1]];



level[1]

level[2]   

6

24

share|improve this answer

answered 3 hours ago

J42161217J42161217

3,792321

$endgroup$

add a comment | 

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5

$begingroup$

FoldList[Times, l] (* or *)

FoldList[Times]@l

{2, 6, 24, 120, 720}

Also:

Exp @ Accumulate @ Log @ l

{2, 6, 24, 120, 720}

share|improve this answer

edited 2 hours ago

answered 3 hours ago

kglrkglr

185k10202420

$endgroup$

add a comment | 

5

$begingroup$

FoldList[Times, l] (* or *)

FoldList[Times]@l

{2, 6, 24, 120, 720}

Also:

Exp @ Accumulate @ Log @ l

{2, 6, 24, 120, 720}

share|improve this answer

edited 2 hours ago

answered 3 hours ago

kglrkglr

185k10202420

$endgroup$

add a comment | 

$begingroup$

FoldList[Times, l] (* or *)

FoldList[Times]@l

{2, 6, 24, 120, 720}

Also:

Exp @ Accumulate @ Log @ l

{2, 6, 24, 120, 720}

share|improve this answer

edited 2 hours ago

answered 3 hours ago

kglrkglr

185k10202420

$endgroup$

FoldList[Times, l] (* or *)

FoldList[Times]@l

Exp @ Accumulate @ Log @ l

share|improve this answer

edited 2 hours ago

answered 3 hours ago

kglrkglr

185k10202420

answered 3 hours ago

kglrkglr

185k10202420

answered 3 hours ago

kglrkglr

185k10202420

3

$begingroup$

list = Range[2, 6]



(* {2, 3, 4, 5, 6} *)

To keep the factors separate, use Inactive with kglr's solution

list2 = Rest@FoldList[Inactive@Times, list[[1]], Rest@list]

Activate produces the result

list3 = list2 // Activate



(* {6, 24, 120, 720} *)

Or use NonCommutativeMultiply to hold the factors

list4 = Rest@FoldList[NonCommutativeMultiply, list[[1]], Rest@list]



(* {2 ** 3, 2 ** 3 ** 4, 2 ** 3 ** 4 ** 5, 2 ** 3 ** 4 ** 5 ** 6} *)

Then Apply Times to get the final result

list5 = Times @@@ list4



(* {6, 24, 120, 720} *)

For this specific case (sequential numbers), the result is just Factorial

list3 == list5 == Factorial /@ Rest@list



(* True *)

Use Part to access any element of the result, e.g.,

list3[[1]]



(* 6 *)

share|improve this answer

answered 1 hour ago

Bob HanlonBob Hanlon

60.5k33597

$endgroup$

add a comment | 

3

$begingroup$

list = Range[2, 6]



(* {2, 3, 4, 5, 6} *)

To keep the factors separate, use Inactive with kglr's solution

list2 = Rest@FoldList[Inactive@Times, list[[1]], Rest@list]

Activate produces the result

list3 = list2 // Activate



(* {6, 24, 120, 720} *)

Or use NonCommutativeMultiply to hold the factors

list4 = Rest@FoldList[NonCommutativeMultiply, list[[1]], Rest@list]



(* {2 ** 3, 2 ** 3 ** 4, 2 ** 3 ** 4 ** 5, 2 ** 3 ** 4 ** 5 ** 6} *)

Then Apply Times to get the final result

list5 = Times @@@ list4



(* {6, 24, 120, 720} *)

For this specific case (sequential numbers), the result is just Factorial

list3 == list5 == Factorial /@ Rest@list



(* True *)

Use Part to access any element of the result, e.g.,

list3[[1]]



(* 6 *)

share|improve this answer

answered 1 hour ago

Bob HanlonBob Hanlon

60.5k33597

$endgroup$

add a comment | 

$begingroup$

list = Range[2, 6]



(* {2, 3, 4, 5, 6} *)

To keep the factors separate, use Inactive with kglr's solution

list2 = Rest@FoldList[Inactive@Times, list[[1]], Rest@list]

Activate produces the result

list3 = list2 // Activate



(* {6, 24, 120, 720} *)

Or use NonCommutativeMultiply to hold the factors

list4 = Rest@FoldList[NonCommutativeMultiply, list[[1]], Rest@list]



(* {2 ** 3, 2 ** 3 ** 4, 2 ** 3 ** 4 ** 5, 2 ** 3 ** 4 ** 5 ** 6} *)

Then Apply Times to get the final result

list5 = Times @@@ list4



(* {6, 24, 120, 720} *)

For this specific case (sequential numbers), the result is just Factorial

list3 == list5 == Factorial /@ Rest@list



(* True *)

Use Part to access any element of the result, e.g.,

list3[[1]]



(* 6 *)

share|improve this answer

answered 1 hour ago

Bob HanlonBob Hanlon

60.5k33597

$endgroup$

list = Range[2, 6]



(* {2, 3, 4, 5, 6} *)

list2 = Rest@FoldList[Inactive@Times, list[[1]], Rest@list]

list3 = list2 // Activate



(* {6, 24, 120, 720} *)

list4 = Rest@FoldList[NonCommutativeMultiply, list[[1]], Rest@list]



(* {2 ** 3, 2 ** 3 ** 4, 2 ** 3 ** 4 ** 5, 2 ** 3 ** 4 ** 5 ** 6} *)

list5 = Times @@@ list4



(* {6, 24, 120, 720} *)

list3 == list5 == Factorial /@ Rest@list



(* True *)

list3[[1]]



(* 6 *)

share|improve this answer

answered 1 hour ago

Bob HanlonBob Hanlon

60.5k33597

answered 1 hour ago

Bob HanlonBob Hanlon

60.5k33597

answered 1 hour ago

Bob HanlonBob Hanlon

60.5k33597

1

$begingroup$

This can also be solved using recursion. The function cumProd (cumulative product) can be defined as:

list = Range[10];

cumProd[n_] := cumProd[n - 1]*list[[n]];

cumProd[1] = list[[1]];

To use:

cumProd[6]

720

gives the 6th "level" of the product. Of course, list can be any set of numbers. Applying this to the whole list:

cumProd/@Range[Length[list]]

{1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800}

share|improve this answer

answered 1 hour ago

bill sbill s

53.5k376152

$endgroup$

add a comment | 

1

$begingroup$

This can also be solved using recursion. The function cumProd (cumulative product) can be defined as:

list = Range[10];

cumProd[n_] := cumProd[n - 1]*list[[n]];

cumProd[1] = list[[1]];

To use:

cumProd[6]

720

gives the 6th "level" of the product. Of course, list can be any set of numbers. Applying this to the whole list:

cumProd/@Range[Length[list]]

{1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800}

share|improve this answer

answered 1 hour ago

bill sbill s

53.5k376152

$endgroup$

add a comment | 

$begingroup$

This can also be solved using recursion. The function cumProd (cumulative product) can be defined as:

list = Range[10];

cumProd[n_] := cumProd[n - 1]*list[[n]];

cumProd[1] = list[[1]];

To use:

cumProd[6]

720

gives the 6th "level" of the product. Of course, list can be any set of numbers. Applying this to the whole list:

cumProd/@Range[Length[list]]

{1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800}

share|improve this answer

answered 1 hour ago

bill sbill s

53.5k376152

$endgroup$

list = Range[10];

cumProd[n_] := cumProd[n - 1]*list[[n]];

cumProd[1] = list[[1]];

cumProd[6]

720

cumProd/@Range[Length[list]]

{1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800}

share|improve this answer

answered 1 hour ago

bill sbill s

53.5k376152

answered 1 hour ago

bill sbill s

53.5k376152

answered 1 hour ago

bill sbill s

53.5k376152

0

$begingroup$

here is your level function

level[x_] := Times @@ l[[;; x + 1]];



level[1]

level[2]   

6

24

share|improve this answer

answered 3 hours ago

J42161217J42161217

3,792321

$endgroup$

add a comment | 

0

$begingroup$

here is your level function

level[x_] := Times @@ l[[;; x + 1]];



level[1]

level[2]   

6

24

share|improve this answer

answered 3 hours ago

J42161217J42161217

3,792321

$endgroup$

add a comment | 

$begingroup$

here is your level function

level[x_] := Times @@ l[[;; x + 1]];



level[1]

level[2]   

6

24

share|improve this answer

answered 3 hours ago

J42161217J42161217

3,792321

$endgroup$

level[x_] := Times @@ l[[;; x + 1]];



level[1]

level[2]   

share|improve this answer

answered 3 hours ago

J42161217J42161217

3,792321

answered 3 hours ago

J42161217J42161217

3,792321

answered 3 hours ago

J42161217J42161217

3,792321