python - 如何计算Python中numpy数组中n个元素的总和？

Question

假设我有一个 n x d python 数组，例如a=np.array([[1,2,3],[4,5,6], [7,8,9], [10,11,12], [13,14,15]])

所以在这种情况下 n=5，d=3 并想象我有一些小于或等于 n 的数字 c，我想要计算如下:

独立考虑每一列并计算每个c值的总和；例如如果c=2，则解为

solution=np.array([[1+4, 2+5, 3+6], [7+10,8+11,9+12]])

最后一行被跳过，因为 5 mod 2 = 1，所以我们需要在最后省略一行；

如果 c=1，解将是原始数组，如果例如c=3 解决方案是

solution=np.array([[1+4+7, 2+5+8, 3+6+9]]) ，而最后两行被省略；

现在最优雅、最有效的解决方案是什么？网上查了很多，没有找到类似的问题

Answer 1

这是一种方法 -

def sum_in_blocks(a, c):
    # Get extent of each col for summing
    l = c*(len(a)//c)

    # Reshape to 3D considering first l rows, and "cutting" after each c rows
    # Then sum along second axis
    return a[:l].reshape(-1,c,a.shape[1]).sum(1)

有关第二步的更多信息 - General idea for nd to nd transformation .

示例运行 -

In [79]: a
Out[79]: 
array([[ 1,  2,  3],
       [ 4,  5,  6],
       [ 7,  8,  9],
       [10, 11, 12],
       [13, 14, 15]])

In [80]: sum_in_blocks(a, c=1)
Out[80]: 
array([[ 1,  2,  3],
       [ 4,  5,  6],
       [ 7,  8,  9],
       [10, 11, 12],
       [13, 14, 15]])

In [81]: sum_in_blocks(a, c=2)
Out[81]: 
array([[ 5,  7,  9],
       [17, 19, 21]])

In [82]: sum_in_blocks(a, c=3)
Out[82]: array([[12, 15, 18]])

<小时/>

给定示例的说明

In [84]: a
Out[84]: 
array([[ 1,  2,  3],
       [ 4,  5,  6],
       [ 7,  8,  9],
       [10, 11, 12],
       [13, 14, 15]])

In [85]: c = 2

In [87]: l = c*(len(a)//c) # = 4; Get extent of each col for summing

In [89]: a[:l] # hence not relevant rows are skipped
Out[89]: 
array([[ 1,  2,  3],
       [ 4,  5,  6],
       [ 7,  8,  9],
       [10, 11, 12]])

# Reshape to 3D "cutting" after every c=2 rows
In [90]: a[:l].reshape(-1,c,a.shape[1])
Out[90]: 
array([[[ 1,  2,  3],
        [ 4,  5,  6]],

       [[ 7,  8,  9],
        [10, 11, 12]]])

# Sum along axis=1 for final o/p
In [91]: a[:l].reshape(-1,c,a.shape[1]).sum(axis=1)
Out[91]: 
array([[ 5,  7,  9],
       [17, 19, 21]])