Let's say we have a matrix (or an arbitrary 2+D list) x and an integer vector l of the same length. What is the most elegant way to get a list {x[[1,l[[1]]]], x[[2, l[[2]]]],...}? The best solution I can think about is
x[[#, l[[#]]]]&/@Range@Length@x , but I'm sure there should be a much easier/more elegant way.
This might not be the most elegant way to do that, but it is descently efficient to use Extract:
m = 1000; n = 2000; x = RandomReal[{-1, 1}, {m, n}]; l = RandomInteger[{1, n}, m]; a = x[[#, l[[#]]]] & /@ Range@Length@x; // AbsoluteTiming // First b = Flatten@MapIndexed[x[[#2, #1]] &, l]; // AbsoluteTiming // First c = Extract[x, Transpose[{Range[Length[l]], l}]]; // AbsoluteTiming // First a == b == c 0.140761
0.002416
0.000067
True
MapIndexed[x[[#2, #1]] &, l] An example:
SeedRandom[90]; x = RandomReal[{0., 1.}, {3, 3}] l = {1, 2, 2}; MapIndexed[x[[#2, #1]] &, l] {{0.647385, 0.307327, 0.702473}, {0.656437, 0.675096, 0.0759657}, {0.733796, 0.139355, 0.0910026}}
{{0.647385}, {0.675096}, {0.139355}}
How about
MapThread[#1[[#2]]&, {x,l}] or equivalently
MapThread[ With[{f = #1, x = #2}, Construct[f[[#]] &, x]] &, {x, l}] With ReplacePart
list = {{9, 7, 8}, {3, 9, 4}, {8, 2, 5}}; p = {1, 2, 2}; ReplacePart[x, i_ :> list[[i, p[[i]]]]] {9, 9, 2}
xis an arbitrary matrix, e.g.x = RandomReal[{0., 1.}, {3, 3}].lis a list of integers with the same length asx, e.g.l={1,2,2}. Yes, this additional condition is required if we want the code to be evaluated without errors. $\endgroup$