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使用Python中的频率列表创建列表

[db:作者]

3月 7, 2023
age = [19, 20, 21, 22, 23, 24, 25]
frequency = [2, 1, 1, 3, 2, 1, 1]
output_age = [19, 19, 20, 21, 22, 22, 22, 23, 23, 24, 25]

我们如何创建一个新列表,根据另一个列表多次添加一个列表中的项目?在

谢谢

Tags:

项目列表outputagefrequency3条回答网友

1楼 ·

编辑于 2023-03-07 00:10:43

使用列表理解:

output_age = [i for l in ([a]*f for a, f in zip(age, frequency)) for i in l]
#[19, 19, 20, 21, 22, 22, 22, 23, 23, 24, 25]

为什么?

我们首先zipage和{}列表放在一起,这样我们就可以一致地迭代它们。因此:

^{pr2}$

给出:

19 2
20 1
21 1
22 3
23 2
24 1
25 1

然后我们要重复每个元素,a,重复次数由f决定。这可以通过创建一个列表并将其相乘来完成。就像:

[4] * 3
#[4, 4, 4]

然后我们需要解压这些值,这样我们就把这个表达式包装在一个生成器中(用括号表示)并在上面迭代。这会使列表变平。注意,有alternative ways来实现这一点(例如使用itertools.chain.from_iterable)。在


另一种方法是通过迭代一个range对象来重复这个数字a,而不是通过乘以一个列表来获得重复次数。在

此方法类似于:

output_age = [a for a, f in zip(age, frequency) for _ in range(f)]
#[19, 19, 20, 21, 22, 22, 22, 23, 23, 24, 25]

网友

2楼 ·

编辑于 2023-03-07 00:10:43

下面是一个使用ziprange的解决方案

>>> age = [19, 20, 21, 22, 23, 24, 25]
>>> frequency = [2, 1, 1, 3, 2, 1, 1]
>>> [a for a,f in zip(age, frequency) for _ in range(f)]
[19, 19, 20, 21, 22, 22, 22, 23, 23, 24, 25]

网友

3楼 ·

编辑于 2023-03-07 00:10:43

使用itertoolszip

例如:

from itertools import chain
age = [19, 20, 21, 22, 23, 24, 25]
frequency = [2, 1, 1, 3, 2, 1, 1]

print( list(chain.from_iterable([[i] * v for i,v in zip(age, frequency)])) )

输出:

^{pr2}$

  • 注意:chain.from_iterable将列表展平。在

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