Test element-wise for Not a Number (NaN), return result as a bool array. Input array. For scalar input, the result is a new boolean with value True if the input is NaN; otherwise the value is False.
Using Numpy array, we can easily find whether specific values are present or not. For this purpose, we use the “in” operator. “in” operator is used to check whether certain element and values are present in a given sequence and hence return Boolean values 'True” and “False“.
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
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)
Pfft! Microseconds! Never solve a problem in microseconds that can be solved in nanoseconds.
Note that the accepted answer:
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.
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.
(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"
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