这正是 numpy.repeat 确实:

>>>a = np.array([0.1, 0.2, 0.3, 0.4, 0.5, 0.6])>>>s = np.array([3, 3, 9, 3, 6, 3])>>>np.repeat(a, s)数组([ 0.1, 0.1, 0.1, 0.2, 0.2, 0.2, 0.3, 0.3, 0.3, 0.3, 0.3,0.3, 0.3, 0.3, 0.3, 0.4, 0.4, 0.4, 0.5, 0.5, 0.5, 0.5,0.5, 0.5, 0.6, 0.6, 0.6])

在纯 Python 中，您可以执行以下操作:

>>>从 itertools 导入重复、链、imap>>>列表(chain.from_iterable(imap(repeat, a, s)))[0.1, 0.1, 0.1, 0.2, 0.2, 0.2, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3, 0.4, 0.4, 0.4, 0.5, 0.5, 0.5, .5, 0.5, 0.5, 0.5, 0.5, 0.6, 0.6]

但当然它会比它的 NumPy 等价物慢得多:

>>>s = [3, 3, 9, 3, 6, 3]*1000>>>a = [0.1, 0.2, 0.3, 0.4, 0.5, 0.6]*1000>>>%timeit 列表(chain.from_iterable(imap(repeat, a, s)))1000 个循环，***的 3 个:每个循环 1.21 毫秒>>>%timeit np.repeat(a_a, s_a) #a_a 和 s_a 是与 a 和 b 大小相同的 NumPy 数组10000 个循环，***的 3 个:每个循环 202 µs

Suppose a = [0.1, 0.2, 0.3, 0.4, 0.5, 0.6] and s = [3, 3, 9, 3, 6, 3]. I'm looking for the best way to repeat a[i] exactly s[i] times and then have a flatten array in the form of b = [0.1, 0.1, 0.1, 0.2, 0.2, 0.2, 0.3, 0.3, 0.3, 0.3, ... ].

I want to do this as fast as possible since I have to do it many times. I'm using Python and numpy and the arrays are defined as numpy.ndarray. I searched around and find out about repeat, tile and column_stack which can be used nicely to repeat each element n times but I wanted to repeat each of them different times.

One way to do this is:

a = hsplit(a, 6)
for i in range(len(a)):
    a[i] = repeat(a[i], s[i])
a = a.flatten()

I am wondering if there is a better way to do it.

That's exactly what numpy.repeat does:

>>> a = np.array([0.1, 0.2, 0.3, 0.4, 0.5, 0.6])
>>> s = np.array([3, 3, 9, 3, 6, 3])
>>> np.repeat(a, s)
array([ 0.1,  0.1,  0.1,  0.2,  0.2,  0.2,  0.3,  0.3,  0.3,  0.3,  0.3,
        0.3,  0.3,  0.3,  0.3,  0.4,  0.4,  0.4,  0.5,  0.5,  0.5,  0.5,
        0.5,  0.5,  0.6,  0.6,  0.6])

In pure Python you can do something like:

>>> from itertools import repeat, chain, imap
>>> list(chain.from_iterable(imap(repeat, a, s)))
[0.1, 0.1, 0.1, 0.2, 0.2, 0.2, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3, 0.4, 0.4, 0.4, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.6, 0.6, 0.6]

But of course it is going to be way slower than its NumPy equivalent:

>>> s = [3, 3, 9, 3, 6, 3]*1000
>>> a = [0.1, 0.2, 0.3, 0.4, 0.5, 0.6]*1000
>>> %timeit list(chain.from_iterable(imap(repeat, a, s)))
1000 loops, best of 3: 1.21 ms per loop
>>> %timeit np.repeat(a_a, s_a) #a_a and s_a are NumPy arrays of same size as a and b
10000 loops, best of 3: 202 µs per loop