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Say I have a list like this

a={-7.61575, 7.5, -6.84206, 6.64598, 5.7654, -5.64823, 4.69842, -4.5, \
-1.86729 + 1.90939 I, -1.86729 - 1.90939 I, 0.905621 + 1.98443 I, 
0.905621 - 1.98443 I, 0.631991 + 1.96227 I, 0.631991 - 1.96227 I, 
0.327801 + 1.9312 I, 0.327801 - 1.9312 I}

How do I define a function that counts the number of real and complex entries 0n this list?

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  • 2
    $\begingroup$ Not sure what format you want, but you might start with CountsBy[a,Head] $\endgroup$
    – lericr
    Sep 6, 2022 at 10:10
  • 1
    $\begingroup$ Try: Count[a, _Real] and Count[a, _Complex] $\endgroup$ Sep 6, 2022 at 10:39
  • 7
    $\begingroup$ Is 1. + 0. I a complex or real number for you? $\endgroup$ Sep 6, 2022 at 11:21
  • $\begingroup$ {Head@#[[1]], #[[2]]} & /@ Tally[a, Head[#1] === Head[#2] &] $\endgroup$
    – Bob Hanlon
    Sep 6, 2022 at 18:58

3 Answers 3

3
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Note that in Mathematica, 0 (the exact integer) and 0. (the floating-point zero) are equal: 0==0. returns True. We can use this trick to split the numbers into those with zero imaginary part ($\mathbb{R}$) and those with non-zero imaginary part ($\mathbb{C}\setminus\mathbb{R}$):

a = {1., 1. I, 1. + 0. I};
CountsBy[a, Im[#] == 0 &]
(*    <|True -> 2, False -> 1|>    *)

We thus have two real numbers (True) and one truly complex one (False). This result differs from that obtained by looking at the Head of the list entries: CountsBy[a, Head] returns <|Real -> 1, Complex -> 2|>, which seems wrong.

For the paranoid, a more general formulation uses PossibleZeroQ to check for zero-ness of the imaginary part:

CountsBy[a, PossibleZeroQ@*Im]
(*    <|True -> 2, False -> 1|>    *)
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Counts[Head /@ a]

<|Real -> 8, Complex -> 8|>

or

Length@Cases[a, #] & /@ {_Real, _Complex}

{8, 8}

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Using GroupBy:

MapAt[Length, GroupBy[a, Head], Outer[List, Range[2]]]
(*<|Real -> 8, Complex -> 8|>*)

Or, as pointed out, @Lukas Lang:

GroupBy[list, Head, Length]
(**<|Real -> 8, Complex -> 8|>**)
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  • 1
    $\begingroup$ You can also use the third argument of GroupBy to do the counting for you: GroupBy[a, Head, Length] $\endgroup$
    – Lukas Lang
    Sep 6, 2022 at 22:45
  • $\begingroup$ Thanks for pointing out that important detail, @Lukas Lang! :) $\endgroup$ Sep 6, 2022 at 22:47

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