python - 将 numpy 数组分解为二进制数组

Question

我有一个 numpy 数组

[0 0 0 0 0 0 0 1 1 2 2 2 2 2 1 1 0 0 0 0 0 0 0 0]

我想转换/溶解成

[[0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0]]

我目前的做法是首先使用 while 循环将数组拆分为 1，然后根据 np.where(x>0) 创建一个数组.然而，我认为这不是最有效和优雅的 numpy 解决方案。关于如何改进这一点的任何想法？

source = np.array([0., 0., 0., 0., 0., 0., 0., 1., 1., 2., 2., 2., 2.,
                   2., 1., 1., 0., 0., 0., 0., 0., 0., 0., 0.], dtype=np.int)

diss = None
while np.any(source):
    row = np.greater(source, 0).astype(np.int)
    if diss is None:
        diss = row
    else:
        diss = np.vstack([diss, row])
    source -= row

idx = np.where(diss > 0)
result = np.zeros((0,source.shape[0]), dtype=np.int)
for x, y in zip(*idx):
    row = np.zeros(source.shape, dtype=np.int)
    row[y] = 1
    result = np.vstack([result, row])

Answer 1

这是一种方法:

In [38]: x = np.array([0,0,0,0,0,1,1,2,2,2,1,1,0,0])

In [39]: n = x.sum()

In [40]: rows = np.arange(n)

In [41]: positions = np.nonzero(x)[0]

In [42]: cols = np.repeat(positions, x[positions])

In [43]: result = np.zeros((n, len(x)), dtype=int)

In [44]: result[rows, cols] = 1

In [45]: result
Out[45]: 
array([[0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0]])