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 2 hours ago
corey979
20.9k64382
asked 2 hours ago
SmerdjakovSmerdjakov
1305
$endgroup$
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 2 hours ago
corey979
20.9k64382
asked 2 hours ago
SmerdjakovSmerdjakov
1305
$endgroup$
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 2 hours ago
corey979
20.9k64382
asked 2 hours ago
SmerdjakovSmerdjakov
1305
$endgroup$
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 2 hours ago
corey979
20.9k64382
asked 2 hours ago
SmerdjakovSmerdjakov
1305
edited 2 hours ago
corey979
20.9k64382
asked 2 hours ago
SmerdjakovSmerdjakov
1305
edited 2 hours ago
corey979
20.9k64382
edited 2 hours ago
corey979
20.9k64382
edited 2 hours ago
corey979
20.9k64382
20.9k64382
asked 2 hours ago
SmerdjakovSmerdjakov
1305
asked 2 hours ago
SmerdjakovSmerdjakov
1305
asked 2 hours ago
SmerdjakovSmerdjakov
1305
1305
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 2 hours ago
Sjoerd SmitSjoerd Smit
4,610817
$endgroup$
1
$begingroup$
OrOuter[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1]so you don't need to unravel#1manually.
$endgroup$
– Roman
2 hours 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 2 hours 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 1 hour 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 2 hours ago
Sjoerd SmitSjoerd Smit
4,610817
$endgroup$
1
$begingroup$
OrOuter[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1]so you don't need to unravel#1manually.
$endgroup$
– Roman
2 hours 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 2 hours ago
Sjoerd SmitSjoerd Smit
4,610817
$endgroup$
1
$begingroup$
OrOuter[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1]so you don't need to unravel#1manually.
$endgroup$
– Roman
2 hours 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 2 hours ago
Sjoerd SmitSjoerd Smit
4,610817
$endgroup$
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 2 hours ago
Sjoerd SmitSjoerd Smit
4,610817
answered 2 hours ago
Sjoerd SmitSjoerd Smit
4,610817
answered 2 hours ago
Sjoerd SmitSjoerd Smit
4,610817
answered 2 hours ago
Sjoerd SmitSjoerd Smit
4,610817
4,610817
1
$begingroup$
OrOuter[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1]so you don't need to unravel#1manually.
$endgroup$
– Roman
2 hours 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
2 hours 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
2 hours ago
$begingroup$
Or
Outer[f[Sequence @@ #1, #2] &, Transpose@l1, l2, l3, 1] so you don't need to unravel #1 manually.$endgroup$
– Roman
2 hours 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 2 hours 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 2 hours 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 2 hours 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 2 hours ago
corey979corey979
20.9k64382
answered 2 hours ago
corey979corey979
20.9k64382
answered 2 hours ago
corey979corey979
20.9k64382
answered 2 hours 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
Carl WollCarl Woll
75.9k3100198
answered 1 hour ago
Carl WollCarl Woll
75.9k3100198
answered 1 hour ago
Carl WollCarl Woll
75.9k3100198
answered 1 hour ago
Carl WollCarl Woll
75.9k3100198
75.9k3100198
add a comment |
add a comment |
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