How do I get the results from argmax of [0., 1e-8]?

Question

When I ran below code, I got x=1, y=0

a = np.array([0., 1e-8]).astype('float32') 
x = a.argmax()
y = (a+1).argmax()

I already know floating point expression. But, I don't know why I can get x=1, y=0.
I think that overflow or underflow can be related to the results. Please help to explain!

`1+1e-8 == 1` when calculated with `float32`, so the 0th element is also a maximum. — chtz, Mar 08 '22 at 13:53

score 2 · Accepted Answer · answered Mar 08 '22 at 14:08

When 1e-8 is converted to float32, the result is the nearest number representable in float32, which is 9.99999993922529029077850282192230224609375e-09, which is about 1.34•2⁻²⁷.

So, in real-number arithmetic, the sum of 1 and 9.99999993922529029077850282192230224609375e-09 is about 1 + 1.34•2⁻²⁷. The two numbers representable in float32 that are closest to that are 1 and 1+2⁻²³. Of those two, 1 is closer to 1 + 1.34•2⁻²⁷, so float32 addition produces 1 as a result. Thus, the elements of a+1 are 1 and 1. Both are tied candidates for the maximum, and argmax returns the index of the first one.

"Thus, the elements of a+1 are 1 and 1." needs re-wording to be clear. — chux - Reinstate Monica, Mar 08 '22 at 17:06

How do I get the results from argmax of [0., 1e-8]?

1 Answers1