Question

假设我有以下数组：

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

我想对第一行的前两个值求和：1+2 = 3，然后接下来的两个值：3+4 = 7，然后是5+6 = 11，依此类推每一行。我想要的输出是：

array([[ 3,  7, 11],
       [15, 19, 23],
       [ 8, 13, 17]])

我有以下解决方案：

def sum_chunks(x, chunk_size):
    rows, cols = x.shape
    x = x.reshape(x.size / chunk_size, chunk_size)
    return x.sum(axis=1).reshape(rows, cols/chunk_size)

但感觉不必要的复杂，有更好的方法吗？也许是内置的？

Answer 1

只需使用切片：

a[:,::2] + a[:,1::2]

这将获取每个偶数索引列（::2）形成的数组，并将其添加到由每个奇数索引列（1::2）形成的数组中。

Answer 2

当我必须做这种事情时，我更喜欢将2D阵列重塑为3D阵列，然后使用np.sum折叠额外的维度。将其推广到n维数组，你可以这样做：

def sum_chunk(x, chunk_size, axis=-1):
    shape = x.shape
    if axis < 0:
        axis += x.ndim
    shape = shape[:axis] + (-1, chunk_size) + shape[axis+1:]
    x = x.reshape(shape)
    return x.sum(axis=axis+1)

>>> a = np.arange(24).reshape(4, 6)
>>> a
array([[ 0,  1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10, 11],
       [12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23]])
>>> sum_chunk(a, 2)
array([[ 1,  5,  9],
       [13, 17, 21],
       [25, 29, 33],
       [37, 41, 45]])
>>> sum_chunk(a, 2, axis=0)
array([[ 6,  8, 10, 12, 14, 16],
       [30, 32, 34, 36, 38, 40]])

Answer 3

这是一种方式：

>>> a[:,::2] + a[:,1::2]
array([[ 3,  7, 11],
       [15, 19, 23],
       [ 8, 13, 17]])

Answer 4

>>> def sum_chunks(x, chunk_size):
        shape = list(x.shape)
        reshape = shape[:-1] + [-1, chunk_size]
        return x.reshape(*reshape).sum(axis=-1)
>>>
>>> # Example usage:
>>> y = np.arange(0, 18, 1).reshape(3, 6)
>>> y
array([[ 0,  1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10, 11],
       [12, 13, 14, 15, 16, 17]])
>>>
>>> sum_chunks(y, 3)
array([[ 3, 12],
       [21, 30],
       [39, 48]])
>>>
>>> sum_chunks(y, 2)
array([[ 1,  5,  9],
       [13, 17, 21],
       [25, 29, 33]])

numpy.array的部分内容

4 个答案: