Detect if a NumPy array contains at least one non-numeric value?

Solution 1:

This should be faster than iterating and will work regardless of shape.

numpy.isnan(myarray).any()

Edit: 30x faster:

import timeit
s = 'import numpy;a = numpy.arange(10000.).reshape((100,100));a[10,10]=numpy.nan'
ms = [
    'numpy.isnan(a).any()',
    'any(numpy.isnan(x) for x in a.flatten())']
for m in ms:
    print "  %.2f s" % timeit.Timer(m, s).timeit(1000), m

Results:

  0.11 s numpy.isnan(a).any()
  3.75 s any(numpy.isnan(x) for x in a.flatten())

Bonus: it works fine for non-array NumPy types:

>>> a = numpy.float64(42.)
>>> numpy.isnan(a).any()
False
>>> a = numpy.float64(numpy.nan)
>>> numpy.isnan(a).any()
True

Solution 2:

If infinity is a possible value, I would use numpy.isfinite

numpy.isfinite(myarray).all()

If the above evaluates to True, then myarray contains none of numpy.nan, numpy.inf or -numpy.inf.

numpy.isnan will be OK with numpy.inf values, for example:

In [11]: import numpy as np

In [12]: b = np.array([[4, np.inf],[np.nan, -np.inf]])

In [13]: np.isnan(b)
Out[13]: 
array([[False, False],
       [ True, False]], dtype=bool)

In [14]: np.isfinite(b)
Out[14]: 
array([[ True, False],
       [False, False]], dtype=bool)

Solution 3:

Pfft! Microseconds! Never solve a problem in microseconds that can be solved in nanoseconds.

Note that the accepted answer:

iterates over the whole data, regardless of whether a nan is found
creates a temporary array of size N, which is redundant.

A better solution is to return True immediately when NAN is found:

import numba
import numpy as np

NAN = float("nan")

@numba.njit(nogil=True)
def _any_nans(a):
    for x in a:
        if np.isnan(x): return True
    return False

@numba.jit
def any_nans(a):
    if not a.dtype.kind=='f': return False
    return _any_nans(a.flat)

array1M = np.random.rand(1000000)
assert any_nans(array1M)==False
%timeit any_nans(array1M)  # 573us

array1M[0] = NAN
assert any_nans(array1M)==True
%timeit any_nans(array1M)  # 774ns  (!nanoseconds)

and works for n-dimensions:

array1M_nd = array1M.reshape((len(array1M)/2, 2))
assert any_nans(array1M_nd)==True
%timeit any_nans(array1M_nd)  # 774ns

Compare this to the numpy native solution:

def any_nans(a):
    if not a.dtype.kind=='f': return False
    return np.isnan(a).any()

array1M = np.random.rand(1000000)
assert any_nans(array1M)==False
%timeit any_nans(array1M)  # 456us

array1M[0] = NAN
assert any_nans(array1M)==True
%timeit any_nans(array1M)  # 470us

%timeit np.isnan(array1M).any()  # 532us

The early-exit method is 3 orders or magnitude speedup (in some cases). Not too shabby for a simple annotation.

Solution 4:

With numpy 1.3 or svn you can do this

In [1]: a = arange(10000.).reshape(100,100)

In [3]: isnan(a.max())
Out[3]: False

In [4]: a[50,50] = nan

In [5]: isnan(a.max())
Out[5]: True

In [6]: timeit isnan(a.max())
10000 loops, best of 3: 66.3 µs per loop

The treatment of nans in comparisons was not consistent in earlier versions.

Solution 5:

(np.where(np.isnan(A)))[0].shape[0] will be greater than 0 if A contains at least one element of nan, A could be an n x m matrix.

Example:

import numpy as np

A = np.array([1,2,4,np.nan])

if (np.where(np.isnan(A)))[0].shape[0]: 
    print "A contains nan"
else:
    print "A does not contain nan"