Reputation: 500437
Is there an idiomatic way to compare two NumPy arrays that would treat NaNs as being equal to each other (but not equal to anything other than a NaN).
For example, I want the following two arrays to compare equal:
np.array([1.0, np.NAN, 2.0])
np.array([1.0, np.NAN, 2.0])
and the following two arrays to compare unequal:
np.array([1.0, np.NAN, 2.0])
np.array([1.0, 0.0, 2.0])
I am looking for a method that would produce a scalar Boolean outcome.
The following would do it:
np.all((a == b) | (np.isnan(a) & np.isnan(b)))
but it's clunky and creates all those intermediate arrays.
Is there a way that's easier on the eye and makes better use of memory?
P.S. If it helps, the arrays are known to have the same shape and dtype.
Upvotes: 24
Views: 5982
Reputation: 139172
Numpy 1.10 added the equal_nan
keyword to np.allclose
(https://docs.scipy.org/doc/numpy/reference/generated/numpy.allclose.html).
So you can do now:
In [24]: np.allclose(np.array([1.0, np.NAN, 2.0]),
np.array([1.0, np.NAN, 2.0]), equal_nan=True)
Out[24]: True
Upvotes: 10
Reputation: 8538
If you really care about memory use (e.g. have very large arrays), then you should use numexpr and the following expression will work for you:
np.all(numexpr.evaluate('(a==b)|((a!=a)&(b!=b))'))
I've tested it on very big arrays with length of 3e8, and the code has the same performance on my machine as
np.all(a==b)
and uses the same amount of memory
Upvotes: 18
Reputation: 353179
Disclaimer: I don't recommend this for regular use, and I wouldn't use it myself, but I could imagine rare circumstances under which it might be useful.
If the arrays have the same shape and dtype, you could consider using the low-level memoryview
:
>>> import numpy as np
>>>
>>> a0 = np.array([1.0, np.NAN, 2.0])
>>> ac = a0 * (1+0j)
>>> b0 = np.array([1.0, np.NAN, 2.0])
>>> b1 = np.array([1.0, np.NAN, 2.0, np.NAN])
>>> c0 = np.array([1.0, 0.0, 2.0])
>>>
>>> memoryview(a0)
<memory at 0x85ba1bc>
>>> memoryview(a0) == memoryview(a0)
True
>>> memoryview(a0) == memoryview(ac) # equal but different dtype
False
>>> memoryview(a0) == memoryview(b0) # hooray!
True
>>> memoryview(a0) == memoryview(b1)
False
>>> memoryview(a0) == memoryview(c0)
False
But beware of subtle problems like this:
>>> zp = np.array([0.0])
>>> zm = -1*zp
>>> zp
array([ 0.])
>>> zm
array([-0.])
>>> zp == zm
array([ True], dtype=bool)
>>> memoryview(zp) == memoryview(zm)
False
which happens because the binary representations differ even though they compare equal (they have to, of course: that's how it knows to print the negative sign)
>>> memoryview(zp)[0]
'\x00\x00\x00\x00\x00\x00\x00\x00'
>>> memoryview(zm)[0]
'\x00\x00\x00\x00\x00\x00\x00\x80'
On the bright side, it short-circuits the way you might hope it would:
In [47]: a0 = np.arange(10**7)*1.0
In [48]: a0[-1] = np.NAN
In [49]: b0 = np.arange(10**7)*1.0
In [50]: b0[-1] = np.NAN
In [51]: timeit memoryview(a0) == memoryview(b0)
10 loops, best of 3: 31.7 ms per loop
In [52]: c0 = np.arange(10**7)*1.0
In [53]: c0[0] = np.NAN
In [54]: d0 = np.arange(10**7)*1.0
In [55]: d0[0] = 0.0
In [56]: timeit memoryview(c0) == memoryview(d0)
100000 loops, best of 3: 2.51 us per loop
and for comparison:
In [57]: timeit np.all((a0 == b0) | (np.isnan(a0) & np.isnan(b0)))
1 loops, best of 3: 296 ms per loop
In [58]: timeit np.all((c0 == d0) | (np.isnan(c0) & np.isnan(d0)))
1 loops, best of 3: 284 ms per loop
Upvotes: 8
Reputation: 771
Not sure this is any better, but a thought...
import numpy
class FloatOrNaN(numpy.float_):
def __eq__(self, other):
return (numpy.isnan(self) and numpy.isnan(other)) or super(FloatOrNaN,self).__eq__(other)
a = [1., np.nan, 2.]
one = numpy.array([FloatOrNaN(val) for val in a], dtype=object)
two = numpy.array([FloatOrNaN(val) for val in a], dtype=object)
print one == two # yields array([ True, True, True], dtype=bool)
This pushes the ugliness into the dtype, at the expense of making the inner workings python instead of c (Cython/etc would fix this). It does, however, greatly reduce memory costs.
Still kinda ugly though :(
Upvotes: 0