Python:如何创建由随机数组成的，长度不断增加的列表列表? [英] Python: how to create a list of lists with increasing length, made of random numbers?

问题描述

说我有200个称为masterlist的正，唯一，随机整数的列表.

我想生成一个名为container的10个列表的列表，这样:l1有两个来自masterlist的随机数，不包括重复； l2有4个元素，l3有6个元素，依此类推.

我知道我可以这样创建我的container列表:

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

，然后我可以使用random.choice()从列表中选择一个随机值.

嵌套这10个列表的填充过程的最好的Python方法是什么，以便我创建一个列表，附加正确数量的值以检查没有重复，然后继续下一个?

编辑

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))

这有什么问题?

解决方案

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)
]

容器由列表理解组成，使用range()调用将变量l设置为2、4、6 ... 18、20.在理解的每个循环"中，内置的random.sample()调用会执行您要进行的无替换采样.

Say I have a list of 200 positive, unique, random integers called masterlist.

I want to generate a list of 10 lists called container so that: l1 has 2 random numbers coming from masterlist, repetitions excluded; l2 has 4 elements, l3 has 6 elements, and so forth.

I know I can create my container list like this:

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

and that I can select a random value from a list using random.choice().

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?

EDIT

This is my attempt:

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?

解决方案

This should do the trick:

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. Within each 'loop' of the comprehension, the built in random.sample() call does the sampling-without-replacement that you're after.

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