确定两个复数是否相等

Question

The following code causes the print statements to be executed:

import numpy as np
import math

foo = np.array([1/math.sqrt(2), 1/math.sqrt(2)], dtype=np.complex_)

total = complex(0, 0)
one = complex(1, 0)
for f in foo:
   total = total + pow(np.abs(f), 2)
   if(total != one):
      print str(total) + " vs " + str(one)
      print "NOT EQUAL"

However, my input of [1/math.sqrt(2), 1/math.sqrt(2)] results in the total being one :

(1+0j) vs (1+0j) NOT EQUAL

Is it something to do with mixing NumPy with Python's complex type?

Answer 1

When using floating point numbers it is important to keep in mind that working with these numbers is never accurate and thus computations are every time subject to rounding errors. This is caused by the design of floating point arithmetic and currently the most practicable way to do high arbitrary precision mathematics on computers with limited resources. You can't compute exactly using floats (means you have practically no alternative), as your numbers have to be cut off somewhere to fit in a reasonable amount of memory (in most cases at maximum 64 bits), this cut-off is done by rounding it (see below for an example).

To deal correctly with these shortcomings you should never compare to floats for equality, but for closeness. Numpy provides 2 functions for that: np.isclose for comparison of single values (or a item-wise comparison for arrays) and np.allclose for whole arrays. The latter is a np.all(np.isclose(a, b)) , so you get a single value for an array.

>>> np.isclose(np.float32('1.000001'), np.float32('0.999999'))
True

But sometimes the rounding is very practicable and matches with our analytical expectation, see for example:

>>> np.float(1) == np.square(np.sqrt(1))
True

After squaring the value will be reduced in size to fit in the given memory, so in this case it's rounded to what we would expect.

These two functions have built-in absolute and relative tolerances (you can also give then as parameter) that are use to compare two values. By default they are rtol=1e-05 and atol=1e-08 .

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Also, don't mix different packages with their types. If you use Numpy, use Numpy-Types and Numpy-Functions. This will also reduce your rounding errors.

Btw: Rounding errors have even more impact when working with numbers which differ in their exponent widely.

Answer 2

I guess, the same considerations as for real numbers are applicable: never assume they can be equal, but rather close enough:

eps = 0.000001
if abs(a - b) < eps:
    print "Equal"