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I have a list of points $$L = \{\{x_1,y_1\}, \{x_2,y_2\}, \{x_3,y_3\},\cdots\{x_n,y_n\}\}$$ and I want to create a list $$L1 = \{f(x_1,y_1), f(x_2,y_2), \cdots, f(x_n,y_n)\},$$ where $f$ is some function. Let us take the example $f(x,y) = x^i y^j.$

Is there a simple direct method? In my example, would it be possible to get the list $\{x_1^iy_1^j, x_2^iy_2^j,\cdots x_n^iy_n^j\}$ directly without defining $f(x,y) = x^i y^j$?

This might be a trivial question, but not for people who used to use Python.

A. PI
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  • If alist = {{x1, y1}, {x2, y2}, {x3, y3}};, then f @@@ alist will do this. Or the more verbose way is to do: Map[Apply@f, alist, {1}]. – Syed Feb 01 '23 at 15:03
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    f /@ points if f takes a list e.g f[{x,y}], otherwise just f@@@points. This is all covered in the documentation so I'm voting to close. If you want to also pass an index, then use MapIndexed. – flinty Feb 01 '23 at 15:11
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    #1^i #2^j & @@@ {{x1, y1}, {x2, y2}, {x3, y3}} – Syed Feb 01 '23 at 15:21
  • I’m voting to close this question because the OP needs to spend time learning the tool and show minimal knowledge of Mathematica before posting questions. I also wish the OP good luck. – Syed Feb 01 '23 at 15:23

2 Answers2

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Clear["Global`*"]

Format indexed variables as subscripts

(Format[#[n_]] := Subscript[#, n]) & /@ {x, y};

pts = Array[{x[#], y[#]} &, 5]

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Map

f /@ pts

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ex1 = #[[1]]^i*#[[2]]^j & /@ pts

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MapApply

f @@@ pts

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ex2 = #1^i*#2^j & @@@ pts

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ex1 === ex2

(* True *)
Bob Hanlon
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4

Another approach (which I learned from this post by bmf) is to use ArrayReduce

m = Array[{Subscript[x, #],Subscript[y,#]} &, 5]

{{x1, y1}, {x2, y2}, {x3, y3}, {x4, y4 }, {x5, y5}}
ArrayReduce[f, m, 2]

{f[{x1, y1}], f[{x2, y2}], f[{x3, y3}], f[{x4, y4}], f[{x5, y6}]} 
ArrayReduce[Times@@{#[[1]]^i,#[[2]]^j}&, m,2]

enter image description here

user1066
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