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Python:如何创建由随机数组成的,长度增加的列表列表?

[英]Python: how to create a list of lists with increasing length, made of random numbers?

Say I have a list of 200 positive, unique, random integers called masterlist . 假设我有一个200个称为masterlist正,唯一,随机整数的masterlist

I want to generate a list of 10 lists called container so that: l1 has 2 random numbers coming from masterlist , repetitions excluded; 我想生成一个称为container的10个列表的列表,以便: l1有2个来自masterlist随机数,排除了重复项; l2 has 4 elements, l3 has 6 elements, and so forth. l2有4个元素, l3有6个元素,依此类推。

I know I can create my container list like this: 我知道我可以这样创建我的container列表:

comb=[[] for i in range(10)]

and that I can select a random value from a list using random.choice() . 而且我可以使用random.choice()从列表中选择一个随机值。

What is the best Pythonic way to nest the populating process of these 10 lists, so that I create one list, append the correct number of values checking that there are no repetitions, and proceed on to the next? 嵌套这10个列表的填充过程的最佳Pythonic方法是什么,以便我创建一个列表,追加正确数量的值以检查是否存在重复项,然后继续进行下一个?

EDIT 编辑

This is my attempt: 这是我的尝试:

comb=[[] for i in range(10)]
for j in range(1,11):
    for k in range(0,2*j):
        comb[j][k].append(random.choice(masterlist))

What is wrong with this? 这有什么问题?

This should do the trick: 这应该可以解决问题:

import random

masterlist = [i for i in range(200)]  # For example

container = [
    random.sample(masterlist, l)
    for l in range(2, 21, 2)
]

The container is made up of a list comprehension, setting the variable l to 2, 4, 6 ... 18, 20 using the range() call. 容器由列表推导组成,使用range()调用将变量l设置为2、4、6 ... 18、20。 Within each 'loop' of the comprehension, the built in random.sample() call does the sampling-without-replacement that you're after. 在理解的每个“循环”中,内置的random.sample()调用都会执行您要进行的无替换采样。

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