Python 随机选择两个列表的某些索引以创建两个新列表

假设我有两个列表：

A = ['cat', 'dog', 'cow', 'pig', 'monkey']
B = ['Felix', 'Fido', 'Moo', 'Trump', 'King Kong']

现在我想通过选择3个随机索引（不替换）创建

列表A

和

列表B

，同时保持

A

和

B

的值之间的关系

比如说,

随机选择的索引：

4,0,3

所以,

A = ['monkey', 'cat', 'pig']
B = ['King Kong', 'Felix', 'Trump']

---

有没有一种方法可以在不编写for循环的情况下执行此操作，for循环会迭代3次以选择随机索引？

Python有一个示例函数，该函数不需要重新执行选择操作。您可以从索引中取样，并将取样应用于您的输入

from random import sample

A = ['cat', 'dog', 'cow', 'pig', 'monkey']
B = ['Felix', 'Fido', 'Moo', 'Trump', 'King Kong']
k = 3

samp = sample(range(len(A)), k)
A_p = [A[i] for i in samp]
B_p = [B[i] for i in samp]

您可以选择列表，然后使用选择3个随机对，最后再次将这些对拆分为单独的列表：

import random

pairs = list(zip(A, B))  # make pairs out of the two lists
pairs = random.sample(pairs, 3)  # pick 3 random pairs
A1, B1 = zip(*pairs)  # separate the pairs

这是一步一步发生的：

>>> list(zip(A, B))
[('cat', 'Felix'), ('dog', 'Fido'), ('cow', 'Moo'), ('pig', 'Trump'), ('monkey', 'King Kong')]
>>> random.sample(_, 3)
[('monkey', 'King Kong'), ('pig', 'Trump'), ('dog', 'Fido')]
>>> list(zip(*_))
[('monkey', 'pig', 'dog'), ('King Kong', 'Trump', 'Fido')]

您可以使用：

import numpy as np

indx = np.random.choice(len(A),3,False)

np.array(A)[indx]
Out[593]: array(['cow', 'pig', 'monkey'], dtype='<U6')

np.array(B)[indx]
Out[594]: array(['Moo', 'Trump', 'King Kong'], dtype='<U9')

将numpy导入为np
indx=np.随机选择（len（A），3，False）
np.数组（A）[indx]
Out[593]：数组（['cow'，'pig'，'monkey'，]，dtype='我想我最喜欢zip答案，但我会使用列表理解的方法

尽管可以将循环委托给函数或将其装扮成列表，但无法避免循环。
不过，您可以在每个列表上循环一次

#lets assume your randomly selected numbers are 4, 0 and 3
indices = (4,0,3)
# lets assign an ordinal to each index
indices_ordinals = [(ind, ord) for ind, ord in enumerate(indices)]
# now we can re sort the indices, since we have recorded their original sequence
indices_ordinals.sort(key = lambda x: x[0])
# now go through the lists (for simplicity we assume they have equal number of elemwnts) and fill the output dicts
csi = 0 # csi is the index that we are looking for currently
A1 = {}
B1 = {}
for i, elem in enumerate(A):
    if i == indices_ordinals[csi][0]:
        A1[indices_ordinals[1]] : elem
        csi += 1
csi =0
for i, elem in enumerate(B):
    if i == indices_ordinals[csi][0]:
        B1[indices_ordinals[1]] : elem
        csi += 1
#now we only need to sort A1 and B1 to get desired output
A2=[A1[i] for i in sorted(A1)]
B2=[B1[i] for i in sorted(B1)]

是的，有一种方法可以做到这一点，但是作为一个新来的人，你应该检查提问策略，因为它将帮助你获得答案。如果你用一些你已经尝试过的代码提问会容易得多，因为这将允许mw给你一个具体的答案。你能包括一些代码吗？你看过python的随机库吗？谢谢！这正是我要找的
~/python/stackoverflow/9.23$ python3.7 loop.py 
['monkey', 'cow', 'cat']
['King Kong', 'Moo', 'Felix']

#lets assume your randomly selected numbers are 4, 0 and 3
indices = (4,0,3)
# lets assign an ordinal to each index
indices_ordinals = [(ind, ord) for ind, ord in enumerate(indices)]
# now we can re sort the indices, since we have recorded their original sequence
indices_ordinals.sort(key = lambda x: x[0])
# now go through the lists (for simplicity we assume they have equal number of elemwnts) and fill the output dicts
csi = 0 # csi is the index that we are looking for currently
A1 = {}
B1 = {}
for i, elem in enumerate(A):
    if i == indices_ordinals[csi][0]:
        A1[indices_ordinals[1]] : elem
        csi += 1
csi =0
for i, elem in enumerate(B):
    if i == indices_ordinals[csi][0]:
        B1[indices_ordinals[1]] : elem
        csi += 1
#now we only need to sort A1 and B1 to get desired output
A2=[A1[i] for i in sorted(A1)]
B2=[B1[i] for i in sorted(B1)]