Question

我正在寻找一种方法来实现与此问题R: repeat elements of a list based on another list中的问题类似的方法，但是在python中。

基本上我有三个相同长度的列表：

a = [0.9935,0.1955,1.3165,0.0975,0.2995,-0.0075,1.5015,1.0055,0.4525,0.2235,1.3815,0.4195,1.3685,0.0325,0.0055,0.6175,0.0615,1.8115]
b = [4.01,5.207,-0.245,5.312,0.841,1.204,-0.413,4.398,5.309,1.149, -0.295,1.903,-0.851,1.236,1.39,3.48,-0.034,4.286]
c = [0.221, 0.423, 0.125, 0.228, 0.233, 0.235, 0.244, 0.249, 0.265, 0.265, 0.268, 0.268, 0.275, 0.299, 0.301, 0.316, 0.318, 0.329]

我希望根据a次10中存储的浮点数（舍入到最接近的整数），在b和c中生成新的重复项。例如，a和b中的第一个元素必须重复2次，因为c中的第一个元素是0.221和

int(round(c[0]*10.),0) = 2

因此，新a_2和b_2列表中的前两个元素如下所示：

a_2 = [0.9935, 0.9935, ...]
b_2 = [4.01, 4.01, ...]

将此应用于a和b中的前三项会导致：

a_2 = [0.9935, 0.9935, 0.1955, 0.1955, 0.1955, 0.1955, 1.3165, ...]
b_2 = [4.01, 4.01, 5.207, 5.207, 5.207, 5.207, -0.245, ...]

自c[1]=0.423和c[2]=0.125，这意味着a和b中的第二项和第三项需要重复4和1次分别

添加

为了完整起见，我会提到这个问题与另一个问题How to obtain a weighted gaussian filter中给出的答案有关。因此，这是使其他问题的答案有效的一个组成部分。

Answer 1

您可以使用zip()将a和c以及b和c与itertools.chain.from_iterable()结合使用，以生成新序列：

from itertools import chain

a_2 = list(chain.from_iterable([i] * int(round(j * 10)) for i, j in zip(a, c)))
b_2 = list(chain.from_iterable([i] * int(round(j * 10)) for i, j in zip(b, c)))

演示：

>>> from itertools import chain
>>> a = [0.9935,0.1955,1.3165,0.0975,0.2995,-0.0075,1.5015,1.0055,0.4525,0.2235,1.3815,0.4195,1.3685,0.0325,0.0055,0.6175,0.0615,1.8115]
>>> b = [4.01,5.207,-0.245,5.312,0.841,1.204,-0.413,4.398,5.309,1.149, -0.295,1.903,-0.851,1.236,1.39,3.48,-0.034,4.286]
>>> c = [0.221, 0.423, 0.125, 0.228, 0.233, 0.235, 0.244, 0.249, 0.265, 0.265, 0.268, 0.268, 0.275, 0.299, 0.301, 0.316, 0.318, 0.329]
>>> list(chain.from_iterable([i] * int(round(j * 10)) for i, j in zip(a, c)))
[0.9935, 0.9935, 0.1955, 0.1955, 0.1955, 0.1955, 1.3165, 0.0975, 0.0975, 0.2995, 0.2995, -0.0075, -0.0075, 1.5015, 1.5015, 1.0055, 1.0055, 0.4525, 0.4525, 0.4525, 0.2235, 0.2235, 0.2235, 1.3815, 1.3815, 1.3815, 0.4195, 0.4195, 0.4195, 1.3685, 1.3685, 1.3685, 0.0325, 0.0325, 0.0325, 0.0055, 0.0055, 0.0055, 0.6175, 0.6175, 0.6175, 0.0615, 0.0615, 0.0615, 1.8115, 1.8115, 1.8115]
>>> list(chain.from_iterable([i] * int(round(j * 10)) for i, j in zip(b, c)))
[4.01, 4.01, 5.207, 5.207, 5.207, 5.207, -0.245, 5.312, 5.312, 0.841, 0.841, 1.204, 1.204, -0.413, -0.413, 4.398, 4.398, 5.309, 5.309, 5.309, 1.149, 1.149, 1.149, -0.295, -0.295, -0.295, 1.903, 1.903, 1.903, -0.851, -0.851, -0.851, 1.236, 1.236, 1.236, 1.39, 1.39, 1.39, 3.48, 3.48, 3.48, -0.034, -0.034, -0.034, 4.286, 4.286, 4.286]

如果a，b和c特别大，或者对于某些c次10的值会导致大量重复并且您只需要处理这些值一个接一个（例如，您不需要在内存中实现整个a_2和b_2列表），您可以在此处使用更多itertools个函数：

from itertools import chain, repeat, izip

a_2_generator = chain.from_iterable(repeat(i, int(round(j * 10))) for i, j in izip(a, c))
b_2_generator = chain.from_iterable(repeat(i, int(round(j * 10))) for i, j in izip(b, c))

请注意此处缺少list();现在每个阶段都由迭代器处理，只产生足以一次满足一个步骤，节省内存。

如果您需要，只需要迭代输出一次就可以使用它;如果您的绘图库采用一般迭代来绘制这些应该可以正常工作。

Answer 2

您可以使用chain，izip和repeat：

from itertools import chain, repeat, izip
new_a = list(chain.from_iterable(repeat(el, int(round(n * 10))) for el, n in izip(a, c)))

基于另一个列表的元素在列表中生成重复的元素

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2 个答案: