If I have
list1 = {1, 2, 3, 4}
list2 = {5, 6, 7, 8}
How can I do so, that
function[#1, #2]
takes #1 from list1 and #2 from list2??
$\begingroup$
$\endgroup$
-
$\begingroup$ Duplicates: (10211), (15556) $\endgroup$ – Mr.Wizard♦ Jan 18 '15 at 15:26
$\begingroup$
$\endgroup$
MapThread[function[#1, #2] &, {list1, list2}]
answered Jan 18 '15 at 14:31
-
7
-
$\begingroup$ @David No, that's not the same, unless
functionhappens to be listable. $\endgroup$ – Mr.Wizard♦ Jan 18 '15 at 19:39
$\begingroup$
$\endgroup$
Since your question leaves a wide range of possible interpretations I suggest to examine these methods
Inner[ f, list1, list2, List]
{f[1, 5], f[2, 6], f[3, 7], f[4, 8]}
or
f @@@ Transpose[{list1, list2}]
or
Thread[ f[ Transpose[{list1, list2}]]]
Perhaps you could prefer Outer providing all possible pairs
Outer[ f, list1, list2]
{{f[1, 5], f[1, 6], f[1, 7], f[1, 8]}, {f[2, 5], f[2, 6], f[2, 7], 2, 8]}, {f[3, 5], f[3, 6], f[3, 7], f[3, 8]}, {f[4, 5], f[4, 6], f[4, 7], f[4, 8]}}
or
f @@@ Tuples[{list1, list2}]
{ f[1, 5], f[1, 6], f[1, 7], f[1, 8], f[2, 5], f[2, 6], f[2, 7], f[2, 8], 3, 5], f[3, 6], f[3, 7], f[3, 8], f[4, 5], f[4, 6], f[4, 7], f[4, 8]}
