重复的阵列的不同时间的每个值

问题描述

假设 A = [0.1，0.2，0.3，0.4，0.5，0.6] 和 S = [3,3，9,3， 6,3] 。我正在寻找最好的方式来重复 A [I] 究竟 S [I] 次，然后有一个在的形式扁平化数组b = [0.1，0.1，0.1，0.2，0.2，0.2，0.3，0.3，0.3，0.3，...] 。

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, ... ].

我想，因为我必须做了很多次这样做尽可能快。我使用Python和numpy的和数组定义为numpy.ndarray。我四处搜寻，并了解重复，瓷砖和 column_stack 它可以很好地用于 N 次重复每一个元素，但我想他们每个人重复不同的时间。

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.

做到这一点的方法之一是：

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()

我想知道是否有更好的方法来做到这一点。

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

推荐答案

这正是的做：

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])

在纯Python，你可以这样做：

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]

但是，当然，这将是慢的方式比numpy的等效的：

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

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