numpy.ufunc.reduceat

原文:https://docs.scipy.org/doc/numpy/reference/generated/numpy.ufunc.reduceat.html

译者:飞龙 UsyiyiCN

校对:(虚位以待)

ufunc.reduceat(a, indices, axis=0, dtype=None, out=None)

使用单个轴上的指定切片执行(局部)缩减。

For i in range(len(indices)), reduceat computes ufunc.reduce(a[indices[i]:indices[i+1]]), which becomes the i-th generalized “row” parallel to axis in the final result (i.e., in a 2-D array, for example, if axis = 0, it becomes the i-th row, but if axis = 1, it becomes the i-th column). 这里有三个例外:

输出的形状取决于indices的大小,并且可能大于a(如果len(indices) > a.shape [axis])。

参数:

a:array_like

要执行的数组。

indices:array_like

成对索引,逗号分隔(不是冒号),指定要减少的切片。

axis:int,可选

沿着哪条轴应用缩小。

dtype:数据类型代码,可选

用于表示中间结果的类型。默认为输出数组(如果提供)的数据类型,或输入数组(如果未提供输出数组)的数据类型。

out:ndarray,可选

存储结果的位置。如果未提供,则返回新分配的数组。

返回:

r:ndarray

减小的值。如果提供out,则r是对out的引用。

笔记

一个描述性示例:

If a is 1-D, the function ufunc.accumulate(a) is the same as ufunc.reduceat(a, indices)[::2] where indices is range(len(array) - 1) with a zero placed in every other element: indices = zeros(2 * len(a) - 1), indices[1::2] = range(1, len(a)).

不要被此属性的名称欺骗:reduceat(a)不一定小于a

例子

要获取四个连续值的运行总和:

>>> np.add.reduceat(np.arange(8),[0,4, 1,5, 2,6, 3,7])[::2]
array([ 6, 10, 14, 18])

2-D示例:

>>> x = np.linspace(0, 15, 16).reshape(4,4)
>>> x
array([[  0.,   1.,   2.,   3.],
       [  4.,   5.,   6.,   7.],
       [  8.,   9.,  10.,  11.],
       [ 12.,  13.,  14.,  15.]])
# reduce such that the result has the following five rows:
# [row1 + row2 + row3]
# [row4]
# [row2]
# [row3]
# [row1 + row2 + row3 + row4]
>>> np.add.reduceat(x, [0, 3, 1, 2, 0])
array([[ 12.,  15.,  18.,  21.],
       [ 12.,  13.,  14.,  15.],
       [  4.,   5.,   6.,   7.],
       [  8.,   9.,  10.,  11.],
       [ 24.,  28.,  32.,  36.]])
# reduce such that result has the following two columns:
# [col1 * col2 * col3, col4]
>>> np.multiply.reduceat(x, [0, 3], 1)
array([[    0.,     3.],
       [  120.,     7.],
       [  720.,    11.],
       [ 2184.,    15.]])