NumPy：创建类似于“重复”的布尔数组，但具有多个维度

Question

I'm looking for sort of the opposite of sum() I guess. Here we go:

x = array([
    [False, False, False, False, False],
    [ True, False, False, False, False],
    [ True,  True, False, False, False],
    [ True,  True,  True, False, False]])

x.sum(axis=1)
Out: array([0, 1, 2, 3])

So I want to go the opposite direction: from [0,1,2,3] to an array like x (I can specify the number of columns I want in x, of course, above it's 5).

The solution should ideally work for higher dimensions too, and I don't want to loop in Python of course, because the input could be longer than this example. That said, here's a solution using a loop:

s = np.array([0, 1, 2, 3])
y = np.zeros((len(s), 5), np.bool)
for row,col in enumerate(s):
    y[row,0:col] = True

Answer 1

IIUC -- and I'm not sure that I do -- you could use arange and a broadcasting comparison:

>>> v = np.array([0,1,3,2])
>>> np.arange(5) < v[...,None]
array([[False, False, False, False, False],
       [ True, False, False, False, False],
       [ True,  True,  True, False, False],
       [ True,  True, False, False, False]], dtype=bool)

or in 2D:

>>> v = np.array([[1,2],[0,2]])
>>> np.arange(5) < v[...,None]
array([[[ True, False, False, False, False],
        [ True,  True, False, False, False]],

       [[False, False, False, False, False],
        [ True,  True, False, False, False]]], dtype=bool)
>>> ((np.arange(5) < v[...,None]).sum(2) == v).all()
True