numpy.sum(a, axis=None, dtype=None, out=None, keepdims=<no value>, initial=<no value>, where=<no value>)
[source]
Sum of array elements over a given axis.
Parameters: |
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Returns: |
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See also
ndarray.sum
add.reduce
add
.cumsum
trapz
Arithmetic is modular when using integer types, and no error is raised on overflow.
The sum of an empty array is the neutral element 0:
>>> np.sum([]) 0.0
For floating point numbers the numerical precision of sum (and np.add.reduce
) is in general limited by directly adding each number individually to the result causing rounding errors in every step. However, often numpy will use a numerically better approach (partial pairwise summation) leading to improved precision in many use-cases. This improved precision is always provided when no axis
is given. When axis
is given, it will depend on which axis is summed. Technically, to provide the best speed possible, the improved precision is only used when the summation is along the fast axis in memory. Note that the exact precision may vary depending on other parameters. In contrast to NumPy, Python’s math.fsum
function uses a slower but more precise approach to summation. Especially when summing a large number of lower precision floating point numbers, such as float32
, numerical errors can become significant. In such cases it can be advisable to use dtype=”float64”
to use a higher precision for the output.
>>> np.sum([0.5, 1.5]) 2.0 >>> np.sum([0.5, 0.7, 0.2, 1.5], dtype=np.int32) 1 >>> np.sum([[0, 1], [0, 5]]) 6 >>> np.sum([[0, 1], [0, 5]], axis=0) array([0, 6]) >>> np.sum([[0, 1], [0, 5]], axis=1) array([1, 5]) >>> np.sum([[0, 1], [np.nan, 5]], where=[False, True], axis=1) array([1., 5.])
If the accumulator is too small, overflow occurs:
>>> np.ones(128, dtype=np.int8).sum(dtype=np.int8) -128
You can also start the sum with a value other than zero:
>>> np.sum([10], initial=5) 15
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https://docs.scipy.org/doc/numpy-1.17.0/reference/generated/numpy.sum.html