Apply MapThread to all but one variableHow do you efficiently return all of a List but one element?All values for a function with two arguments without OuterEfficiently finding the maximum value of a column in a matrixnested use of Apply/Map/MapThread in pure functionsMapThread AlternativesFinding neighbors from listMapThread problemapply binary operation to all adjacent pairsFlip sign of one variable in listFind numbers from Mean, Variance and Correlation coefficient
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Apply MapThread to all but one variable
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Apply MapThread to all but one variable
How do you efficiently return all of a List but one element?All values for a function with two arguments without OuterEfficiently finding the maximum value of a column in a matrixnested use of Apply/Map/MapThread in pure functionsMapThread AlternativesFinding neighbors from listMapThread problemapply binary operation to all adjacent pairsFlip sign of one variable in listFind numbers from Mean, Variance and Correlation coefficient
$begingroup$
I would like to know what is the most efficient to implement the following computation. Given three lists
a = a_1,a_2, a_3, …, a_n
b = b_1,b_2, b_3, …, b_n
c = c_1,c_2, c_3, …, c_n
and a function $f(x_1,x_2,x_3)$, obtain
f(a_1,b_1,c_1) f(a_1,b_1,c_2) ..... f(a_1,b_1,c_n)
f(a_2,b_2,c_1) f(a_2,b_2,c_2) ..... f(a_2,b_2,c_n)
..... ..... ..... .....
f(a_n,b_n,c_1) f(a_n,b_n,c_2) ..... f(a_n,b_n,c_n)
I cannot find a solution not using For.
list-manipulation
edited 1 hour ago
corey979
20.9k64382
asked 2 hours ago
SmerdjakovSmerdjakov
1255
$endgroup$
add a comment |
$begingroup$
I would like to know what is the most efficient to implement the following computation. Given three lists
a = a_1,a_2, a_3, …, a_n
b = b_1,b_2, b_3, …, b_n
c = c_1,c_2, c_3, …, c_n
and a function $f(x_1,x_2,x_3)$, obtain
f(a_1,b_1,c_1) f(a_1,b_1,c_2) ..... f(a_1,b_1,c_n)
f(a_2,b_2,c_1) f(a_2,b_2,c_2) ..... f(a_2,b_2,c_n)
..... ..... ..... .....
f(a_n,b_n,c_1) f(a_n,b_n,c_2) ..... f(a_n,b_n,c_n)
I cannot find a solution not using For.
list-manipulation
edited 1 hour ago
corey979
20.9k64382
asked 2 hours ago
SmerdjakovSmerdjakov
1255
$endgroup$
add a comment |
$begingroup$
I would like to know what is the most efficient to implement the following computation. Given three lists
a = a_1,a_2, a_3, …, a_n
b = b_1,b_2, b_3, …, b_n
c = c_1,c_2, c_3, …, c_n
and a function $f(x_1,x_2,x_3)$, obtain
f(a_1,b_1,c_1) f(a_1,b_1,c_2) ..... f(a_1,b_1,c_n)
f(a_2,b_2,c_1) f(a_2,b_2,c_2) ..... f(a_2,b_2,c_n)
..... ..... ..... .....
f(a_n,b_n,c_1) f(a_n,b_n,c_2) ..... f(a_n,b_n,c_n)
I cannot find a solution not using For.
list-manipulation
edited 1 hour ago
corey979
20.9k64382
asked 2 hours ago
SmerdjakovSmerdjakov
1255
$endgroup$
I would like to know what is the most efficient to implement the following computation. Given three lists
a = a_1,a_2, a_3, …, a_n
b = b_1,b_2, b_3, …, b_n
c = c_1,c_2, c_3, …, c_n
and a function $f(x_1,x_2,x_3)$, obtain
f(a_1,b_1,c_1) f(a_1,b_1,c_2) ..... f(a_1,b_1,c_n)
f(a_2,b_2,c_1) f(a_2,b_2,c_2) ..... f(a_2,b_2,c_n)
..... ..... ..... .....
f(a_n,b_n,c_1) f(a_n,b_n,c_2) ..... f(a_n,b_n,c_n)
I cannot find a solution not using For.
list-manipulation
list-manipulation
edited 1 hour ago
corey979
20.9k64382
asked 2 hours ago
SmerdjakovSmerdjakov
1255
edited 1 hour ago
corey979
20.9k64382
asked 2 hours ago
SmerdjakovSmerdjakov
1255
edited 1 hour ago
corey979
20.9k64382
edited 1 hour ago
corey979
20.9k64382
edited 1 hour ago
corey979
20.9k64382
20.9k64382
asked 2 hours ago
SmerdjakovSmerdjakov
1255
asked 2 hours ago
SmerdjakovSmerdjakov
1255
asked 2 hours ago
SmerdjakovSmerdjakov
1255
1255
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here's one way to do it with Outer:
n = 3;
l1 = Array[a, n];
l2 = Array[b, n];
l3 = Array[c, n];
Outer[
f[#1[[1]], #1[[2]], #2] &,
Transpose @ l1, l2,
l3,
1
]
Out[25]= f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]
answered 1 hour ago
Sjoerd SmitSjoerd Smit
4,600817
$endgroup$
1
$begingroup$
OrOuter[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1]so you don't need to unravel#1manually.
$endgroup$
– Roman
1 hour ago
add a comment |
$begingroup$
a = a1, a2, a3, a4, a5;
b = b1, b2, b3, b4, b5;
c = c1, c2, c3, c4, c5;
Table[f[a[[j]], b[[j]], c[[k]]], j, 1, 5, k, 1, 5]
answered 1 hour ago
corey979corey979
20.9k64382
$endgroup$
add a comment |
$begingroup$
Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:
n = 3;
l1 = Array[a,n];
l2 = Array[b,n];
l3 = Array[c,n];
Using Thread:
Thread /@ Thread[f[l1, l2, l3], List, 2]
f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]
answered 46 mins ago
Carl WollCarl Woll
75.9k3100198
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's one way to do it with Outer:
n = 3;
l1 = Array[a, n];
l2 = Array[b, n];
l3 = Array[c, n];
Outer[
f[#1[[1]], #1[[2]], #2] &,
Transpose @ l1, l2,
l3,
1
]
Out[25]= f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]
answered 1 hour ago
Sjoerd SmitSjoerd Smit
4,600817
$endgroup$
1
$begingroup$
OrOuter[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1]so you don't need to unravel#1manually.
$endgroup$
– Roman
1 hour ago
add a comment |
$begingroup$
Here's one way to do it with Outer:
n = 3;
l1 = Array[a, n];
l2 = Array[b, n];
l3 = Array[c, n];
Outer[
f[#1[[1]], #1[[2]], #2] &,
Transpose @ l1, l2,
l3,
1
]
Out[25]= f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]
answered 1 hour ago
Sjoerd SmitSjoerd Smit
4,600817
$endgroup$
1
$begingroup$
OrOuter[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1]so you don't need to unravel#1manually.
$endgroup$
– Roman
1 hour ago
add a comment |
$begingroup$
Here's one way to do it with Outer:
n = 3;
l1 = Array[a, n];
l2 = Array[b, n];
l3 = Array[c, n];
Outer[
f[#1[[1]], #1[[2]], #2] &,
Transpose @ l1, l2,
l3,
1
]
Out[25]= f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]
answered 1 hour ago
Sjoerd SmitSjoerd Smit
4,600817
$endgroup$
Here's one way to do it with Outer:
n = 3;
l1 = Array[a, n];
l2 = Array[b, n];
l3 = Array[c, n];
Outer[
f[#1[[1]], #1[[2]], #2] &,
Transpose @ l1, l2,
l3,
1
]
Out[25]= f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]
answered 1 hour ago
Sjoerd SmitSjoerd Smit
4,600817
answered 1 hour ago
Sjoerd SmitSjoerd Smit
4,600817
answered 1 hour ago
Sjoerd SmitSjoerd Smit
4,600817
answered 1 hour ago
Sjoerd SmitSjoerd Smit
4,600817
4,600817
1
$begingroup$
OrOuter[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1]so you don't need to unravel#1manually.
$endgroup$
– Roman
1 hour ago
add a comment |
1
$begingroup$
OrOuter[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1]so you don't need to unravel#1manually.
$endgroup$
– Roman
1 hour ago
1
1
$begingroup$
Or
Outer[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1] so you don't need to unravel #1 manually.$endgroup$
– Roman
1 hour ago
$begingroup$
Or
Outer[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1] so you don't need to unravel #1 manually.$endgroup$
– Roman
1 hour ago
add a comment |
$begingroup$
a = a1, a2, a3, a4, a5;
b = b1, b2, b3, b4, b5;
c = c1, c2, c3, c4, c5;
Table[f[a[[j]], b[[j]], c[[k]]], j, 1, 5, k, 1, 5]
answered 1 hour ago
corey979corey979
20.9k64382
$endgroup$
add a comment |
$begingroup$
a = a1, a2, a3, a4, a5;
b = b1, b2, b3, b4, b5;
c = c1, c2, c3, c4, c5;
Table[f[a[[j]], b[[j]], c[[k]]], j, 1, 5, k, 1, 5]
answered 1 hour ago
corey979corey979
20.9k64382
$endgroup$
add a comment |
$begingroup$
a = a1, a2, a3, a4, a5;
b = b1, b2, b3, b4, b5;
c = c1, c2, c3, c4, c5;
Table[f[a[[j]], b[[j]], c[[k]]], j, 1, 5, k, 1, 5]
answered 1 hour ago
corey979corey979
20.9k64382
$endgroup$
a = a1, a2, a3, a4, a5;
b = b1, b2, b3, b4, b5;
c = c1, c2, c3, c4, c5;
Table[f[a[[j]], b[[j]], c[[k]]], j, 1, 5, k, 1, 5]
answered 1 hour ago
corey979corey979
20.9k64382
answered 1 hour ago
corey979corey979
20.9k64382
answered 1 hour ago
corey979corey979
20.9k64382
answered 1 hour ago
corey979corey979
20.9k64382
20.9k64382
add a comment |
add a comment |
$begingroup$
Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:
n = 3;
l1 = Array[a,n];
l2 = Array[b,n];
l3 = Array[c,n];
Using Thread:
Thread /@ Thread[f[l1, l2, l3], List, 2]
f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]
answered 46 mins ago
Carl WollCarl Woll
75.9k3100198
$endgroup$
add a comment |
$begingroup$
Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:
n = 3;
l1 = Array[a,n];
l2 = Array[b,n];
l3 = Array[c,n];
Using Thread:
Thread /@ Thread[f[l1, l2, l3], List, 2]
f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]
answered 46 mins ago
Carl WollCarl Woll
75.9k3100198
$endgroup$
add a comment |
$begingroup$
Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:
n = 3;
l1 = Array[a,n];
l2 = Array[b,n];
l3 = Array[c,n];
Using Thread:
Thread /@ Thread[f[l1, l2, l3], List, 2]
f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]
answered 46 mins ago
Carl WollCarl Woll
75.9k3100198
$endgroup$
Another possibility is to use the 3-arg version of Thread. With Sjoerd's example:
n = 3;
l1 = Array[a,n];
l2 = Array[b,n];
l3 = Array[c,n];
Using Thread:
Thread /@ Thread[f[l1, l2, l3], List, 2]
f[a[1], b[1], c[1]], f[a[1], b[1], c[2]],
f[a[1], b[1], c[3]], f[a[2], b[2], c[1]], f[a[2], b[2], c[2]],
f[a[2], b[2], c[3]], f[a[3], b[3], c[1]], f[a[3], b[3], c[2]],
f[a[3], b[3], c[3]]
answered 46 mins ago
Carl WollCarl Woll
75.9k3100198
answered 46 mins ago
Carl WollCarl Woll
75.9k3100198
answered 46 mins ago
Carl WollCarl Woll
75.9k3100198
answered 46 mins ago
Carl WollCarl Woll
75.9k3100198
75.9k3100198
add a comment |
add a comment |
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