-
numpy.nansum(a, axis=None, dtype=None, out=None, keepdims=0)
[source] -
Return the sum of array elements over a given axis treating Not a Numbers (NaNs) as zero.
In Numpy versions <= 1.8 Nan is returned for slices that are all-NaN or empty. In later versions zero is returned.
Parameters: a : array_like
Array containing numbers whose sum is desired. If
a
is not an array, a conversion is attempted.axis : int, optional
Axis along which the sum is computed. The default is to compute the sum of the flattened array.
dtype : data-type, optional
The type of the returned array and of the accumulator in which the elements are summed. By default, the dtype of
a
is used. An exception is whena
has an integer type with less precision than the platform (u)intp. In that case, the default will be either (u)int32 or (u)int64 depending on whether the platform is 32 or 64 bits. For inexact inputs, dtype must be inexact.New in version 1.8.0.
out : ndarray, optional
Alternate output array in which to place the result. The default is
None
. If provided, it must have the same shape as the expected output, but the type will be cast if necessary. Seedoc.ufuncs
for details. The casting of NaN to integer can yield unexpected results.New in version 1.8.0.
keepdims : bool, optional
If True, the axes which are reduced are left in the result as dimensions with size one. With this option, the result will broadcast correctly against the original
arr
.New in version 1.8.0.
Returns: y : ndarray or numpy scalar
See also
Notes
If both positive and negative infinity are present, the sum will be Not A Number (NaN).
Numpy integer arithmetic is modular. If the size of a sum exceeds the size of an integer accumulator, its value will wrap around and the result will be incorrect. Specifying
dtype=double
can alleviate that problem.Examples
>>> np.nansum(1) 1 >>> np.nansum([1]) 1 >>> np.nansum([1, np.nan]) 1.0 >>> a = np.array([[1, 1], [1, np.nan]]) >>> np.nansum(a) 3.0 >>> np.nansum(a, axis=0) array([ 2., 1.]) >>> np.nansum([1, np.nan, np.inf]) inf >>> np.nansum([1, np.nan, np.NINF]) -inf >>> np.nansum([1, np.nan, np.inf, -np.inf]) # both +/- infinity present nan
numpy.nansum()
2017-01-10 18:16:14
Please login to continue.