numpy.cumsum#

numpy.cumsum(a, axis=None, dtype=None, out=None)[source]#

Return the cumulative sum of the elements along a given axis.

Parameters:

aarray_like: Input array.
axisint, optional: Axis along which the cumulative sum is computed. The default (None) is to compute the cumsum over the flattened array.
dtypedtype, optional: Type of the returned array and of the accumulator in which the elements are summed. If dtype is not specified, it defaults to the dtype of a, unless a has an integer dtype with a precision less than that of the default platform integer. In that case, the default platform integer is used.
outndarray, optional: Alternative output array in which to place the result. It must have the same shape and buffer length as the expected output but the type will be cast if necessary. See Output type determination for more details.

Returns:

cumsum_along_axisndarray.: A new array holding the result is returned unless out is specified, in which case a reference to out is returned. The result has the same size as a, and the same shape as a if axis is not None or a is a 1-d array.

See also

sum: Sum array elements.
diff: Calculate the n-th discrete difference along given axis.

Notes

Arithmetic is modular when using integer types, and no error is raised on overflow.

cumsum(a)[-1] may not be equal to sum(a) for floating-point values since sum may use a pairwise summation routine, reducing the roundoff-error. See sum for more information.

Examples

>>> a = np.array([[1,2,3], [4,5,6]])
>>> a
array([[1, 2, 3],
       [4, 5, 6]])
>>> np.cumsum(a)
array([ 1,  3,  6, 10, 15, 21])
>>> np.cumsum(a, dtype=float)     # specifies type of output value(s)
array([  1.,   3.,   6.,  10.,  15.,  21.])

>>> np.cumsum(a,axis=0)      # sum over rows for each of the 3 columns
array([[1, 2, 3],
       [5, 7, 9]])
>>> np.cumsum(a,axis=1)      # sum over columns for each of the 2 rows
array([[ 1,  3,  6],
       [ 4,  9, 15]])

cumsum(b)[-1] may not be equal to sum(b)

>>> b = np.array([1, 2e-9, 3e-9] * 1000000)
>>> b.cumsum()[-1]
1000000.0050045159
>>> b.sum()
1000000.0050000029