Python float vs numpy.float32

Question

Using numpy.float32.

t = numpy.float32(.3)
x = numpy.float32(1)
r = numpy.float32(-.3)
_t = t+x+r
_t == 1 # -> False

Using regular Python float.

t = .3
x = 1
r = -.3
_t = t+x+r
_t == 1 # -> True

Why?

Answer 1

Floating point values are inherently non-exact on computers. The python default float is a what's called a double precision floating point number on most machines according to https://docs.python.org/2/tutorial/floatingpoint.html. numpy.float32 is a single precision float. It's double precision counterpart is numpy.float64. This could explain the difference in this case.

In general floating point numbers shouldn't be compared directly using ==. You can use numpy.isclose to deal with the small errors caused by non-exact floating point representations.

Answer 2

Python float is a C double type: documentation:

Floating point numbers are usually implemented using double in C; information about the precision and internal representation of floating point numbers for the machine on which your program is running is available in sys.float_info.

Therefore, you are comparing 32 and 64 precision floating point numbers. The following will work:

t = numpy.float64(.3)
x = numpy.float64(1)
r = numpy.float64(-.3)
_t = t+x+r
_t == 1