Compare array values with a value in Julia

Question

I have an array a = [1, 2, 3, 4];. I want to compare each element of array a with a number and return a new array contains True/False elements in Julia as few steps as possible. I try result = a < 2 and expected array is result = [True, False, False, False] but it's not working. Hope your help

Probably a duplicate of stackoverflow.com/questions/54204198/… — High Performance Mark
– High Performance Mark, Commented Jan 6, 2022 at 10:21

niczky12 · Accepted Answer · 2022-01-06 14:13:02Z

5

You need to vectorize (broadcast) the comparison operator so it operates on Vectors. You can do this by adding a dot . to your code.

julia> a = [1, 2, 3, 4]
4-element Vector{Int64}:
 1
 2
 3
 4

julia> a .<= 2
4-element BitVector:
 1
 1
 0
 0

Read more about broadcasting here.

Note that Python's numpy will do this for you automatically, but there are cases where an operation might be ambiguous - do you want it to be element wise or a matrix multiplication? So Julia solves this by explicitly broadcasting any operation with the . command.

edited Jan 6, 2022 at 14:13

niczky12

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answered Jan 6, 2022 at 10:46

Przemyslaw Szufel

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