CODEWITHSUNDEEP
booleanjuliabitarray
00koeffers
00koeffers

Reputation: 327

Julia prints different results when calculating with variables

The problem is difficult to describe (hence the bad titel), it is easier to show it:

matr =
[
1 1 3
2 1 10
3 2 3
13 8 10]

i=1

print(matr[:,2].==i .* matr[:,3].!=i)
x = matr[:,2].==i
y = matr[:,3].!=i
print(x .* y)

The first and the second print() should be equivalent. Regardless, the first one prints Bool[false, false, false, false] the second one prints Bool[true, true, false, false].

Since the first result is wrong, I am wondering how this problem arises and how it can be prevented.

Upvotes: 0

Views: 71

Answers (1)

hckr
hckr

Reputation: 5583

This is an operator precedence issue. What you write in the first print is not what you intended.

print(matr[:,2].==i .* matr[:,3].!=i)

This is equivalent to (see the parentheses)

print(matr[:,2] .== (i .* matr[:,3]) .!=i)

which is a chaining comparison, so that both matr[k,2] == (i .* matr[k,3])) and matr[k,2] == (i .* matr[k,3])) must be true for the kth index in order for the chaining comparison to return true for the kth index.

Instead you can write the same term with explicit parentheses to take care of the precedence issue.

print((matr[:,2].==i) .* (matr[:,3].!=i))

Instead of multiplication, you can broadcast &.

print((matr[:,2].==i) .& (matr[:,3].!=i))

Both of these should give you the right answer.

Upvotes: 2

Related Questions