Multiplying elements of a list
$begingroup$
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
asked Feb 21 at 13:11
WilliamWilliam
84658
$endgroup$
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
asked Feb 21 at 13:11
WilliamWilliam
84658
$endgroup$
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
asked Feb 21 at 13:11
WilliamWilliam
84658
$endgroup$
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
list-manipulation multiplcation
asked Feb 21 at 13:11
WilliamWilliam
84658
asked Feb 21 at 13:11
WilliamWilliam
84658
asked Feb 21 at 13:11
WilliamWilliam
84658
asked Feb 21 at 13:11
WilliamWilliam
84658
asked Feb 21 at 13:11
WilliamWilliam
84658
84658
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
FoldList[Times, l] (* or *)
FoldList[Times]@l
{2, 6, 24, 120, 720}
Also:
Exp @ Accumulate @ Log @ l
{2, 6, 24, 120, 720}
answered Feb 21 at 13:18
kglrkglr
189k10206424
$endgroup$
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 *)
answered Feb 21 at 14:58
Bob HanlonBob Hanlon
61k33597
$endgroup$
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}
answered Feb 21 at 15:20
bill sbill s
54.7k377157
$endgroup$
add a comment |
$begingroup$
We can use the MapIndexed function
list = {2, 3, 4, 5, 6}
f[x_, {i_}] := Times @@ list[[1 ;; i]]
Rest[MapIndexed[f, list, {1}]]
(* 6, 24, 120, 720} *)
answered Feb 21 at 17:50
user63033user63033
211
$endgroup$
add a comment |
$begingroup$
here is your level function
level[x_] := Times @@ l[[;; x + 1]];
level[1]
level[2]
6
24
answered Feb 21 at 13:42
J42161217J42161217
3,968323
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
FoldList[Times, l] (* or *)
FoldList[Times]@l
{2, 6, 24, 120, 720}
Also:
Exp @ Accumulate @ Log @ l
{2, 6, 24, 120, 720}
answered Feb 21 at 13:18
kglrkglr
189k10206424
$endgroup$
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}
answered Feb 21 at 13:18
kglrkglr
189k10206424
$endgroup$
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}
answered Feb 21 at 13:18
kglrkglr
189k10206424
$endgroup$
FoldList[Times, l] (* or *)
FoldList[Times]@l
{2, 6, 24, 120, 720}
Also:
Exp @ Accumulate @ Log @ l
{2, 6, 24, 120, 720}
answered Feb 21 at 13:18
kglrkglr
189k10206424
edited Feb 21 at 14:02
answered Feb 21 at 13:18
kglrkglr
189k10206424
answered Feb 21 at 13:18
kglrkglr
189k10206424
answered Feb 21 at 13:18
kglrkglr
189k10206424
189k10206424
add a comment |
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 *)
answered Feb 21 at 14:58
Bob HanlonBob Hanlon
61k33597
$endgroup$
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 *)
answered Feb 21 at 14:58
Bob HanlonBob Hanlon
61k33597
$endgroup$
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 *)
answered Feb 21 at 14:58
Bob HanlonBob Hanlon
61k33597
$endgroup$
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 *)
answered Feb 21 at 14:58
Bob HanlonBob Hanlon
61k33597
answered Feb 21 at 14:58
Bob HanlonBob Hanlon
61k33597
answered Feb 21 at 14:58
Bob HanlonBob Hanlon
61k33597
answered Feb 21 at 14:58
Bob HanlonBob Hanlon
61k33597
61k33597
add a comment |
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}
answered Feb 21 at 15:20
bill sbill s
54.7k377157
$endgroup$
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}
answered Feb 21 at 15:20
bill sbill s
54.7k377157
$endgroup$
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}
answered Feb 21 at 15:20
bill sbill s
54.7k377157
$endgroup$
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}
answered Feb 21 at 15:20
bill sbill s
54.7k377157
answered Feb 21 at 15:20
bill sbill s
54.7k377157
answered Feb 21 at 15:20
bill sbill s
54.7k377157
answered Feb 21 at 15:20
bill sbill s
54.7k377157
54.7k377157
add a comment |
add a comment |
$begingroup$
We can use the MapIndexed function
list = {2, 3, 4, 5, 6}
f[x_, {i_}] := Times @@ list[[1 ;; i]]
Rest[MapIndexed[f, list, {1}]]
(* 6, 24, 120, 720} *)
answered Feb 21 at 17:50
user63033user63033
211
$endgroup$
add a comment |
$begingroup$
We can use the MapIndexed function
list = {2, 3, 4, 5, 6}
f[x_, {i_}] := Times @@ list[[1 ;; i]]
Rest[MapIndexed[f, list, {1}]]
(* 6, 24, 120, 720} *)
answered Feb 21 at 17:50
user63033user63033
211
$endgroup$
add a comment |
$begingroup$
We can use the MapIndexed function
list = {2, 3, 4, 5, 6}
f[x_, {i_}] := Times @@ list[[1 ;; i]]
Rest[MapIndexed[f, list, {1}]]
(* 6, 24, 120, 720} *)
answered Feb 21 at 17:50
user63033user63033
211
$endgroup$
We can use the MapIndexed function
list = {2, 3, 4, 5, 6}
f[x_, {i_}] := Times @@ list[[1 ;; i]]
Rest[MapIndexed[f, list, {1}]]
(* 6, 24, 120, 720} *)
answered Feb 21 at 17:50
user63033user63033
211
answered Feb 21 at 17:50
user63033user63033
211
answered Feb 21 at 17:50
user63033user63033
211
answered Feb 21 at 17:50
user63033user63033
211
211
add a comment |
add a comment |
$begingroup$
here is your level function
level[x_] := Times @@ l[[;; x + 1]];
level[1]
level[2]
6
24
answered Feb 21 at 13:42
J42161217J42161217
3,968323
$endgroup$
add a comment |
$begingroup$
here is your level function
level[x_] := Times @@ l[[;; x + 1]];
level[1]
level[2]
6
24
answered Feb 21 at 13:42
J42161217J42161217
3,968323
$endgroup$
add a comment |
$begingroup$
here is your level function
level[x_] := Times @@ l[[;; x + 1]];
level[1]
level[2]
6
24
answered Feb 21 at 13:42
J42161217J42161217
3,968323
$endgroup$
here is your level function
level[x_] := Times @@ l[[;; x + 1]];
level[1]
level[2]
6
24
answered Feb 21 at 13:42
J42161217J42161217
3,968323
answered Feb 21 at 13:42
J42161217J42161217
3,968323
answered Feb 21 at 13:42
J42161217J42161217
3,968323
answered Feb 21 at 13:42
J42161217J42161217
3,968323
3,968323
add a comment |
add a comment |
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