Question

代码：

import random

x = ['A','B','C','D','E','F',
     'G','H','I','J','K','L',
     'M','N','O','P','Q','R',
     'S','T','U','V','W','X',
     'Y','Z']

y1 = random.sample(x, 2)
y2 = random.sample(x, 2)
y3 = random.sample(x, 2)
y4 = random.sample(x, 2)
y5 = random.sample(x, 2)

查询

如上所示，我选择了5个随机样本组合，并在变量y'x'下声明它们。

为了改进我的代码，我想这样做，但要确保列表中的项目在所有变量输出中不会出现多次，其中所有组合都是不同且不重复的。我最好希望实现这一点，而不必从列表中删除项目，因为它将在代码中稍后重用。

预期输出（示例）：

>>> y1
['A', 'Q']
>>> y2
['E', 'K']
>>> y3
['C', 'O']
>>> y4
['Z', 'X']
>>> y5
['P', 'L']

Answer 1

你可以随机播放列表的副本（你说你想重复使用它，因此需要复制，因为shuffle就地工作）然后只为每个样本取两个元素：

import random

x_copy = x[:]  # copy
random.shuffle(x_copy)
y1 = x[:2]
y2 = x[2:4]
y3 = x[4:6]
y4 = x[6:8]
y5 = x[8:10]

或者如果您不想对yi s进行硬编码：

x_copy = x[:]  # copy
random.shuffle(x_copy)
y = [x_copy[i*2: (i+1)*2] for i in range(5)]
print(y)
# [['W', 'Z'], ['A', 'Q'], ['B', 'J'], ['O', 'D'], ['X', 'E']]

Answer 2

您可以使用numpy.random.choice。它的目的是从类似数组的对象（也适用于您的列表）中选择（replace=True）或不选择（replace=False）：

import numpy as np
x = ['A','B','C','D','E','F',
     'G','H','I','J','K','L',
     'M','N','O','P','Q','R',
     'S','T','U','V','W','X',
     'Y','Z']
np.random.choice(x, size=(5, 2), replace=False)

结果：

array([['Y', 'Q'],
       ['W', 'R'],
       ['O', 'H'],
       ['Z', 'G'],
       ['L', 'M']], 
      dtype='<U1')

这将返回一个包含5行的数组，每行包含一个大小为2的样本。

Answer 3

您可以简单地构建生成值的“缓存” - 因此不会删除x的元素：

import random

class SampleCache():
    x = ['A','B','C','D','E','F',
         'G','H','I','J','K','L',
         'M','N','O','P','Q','R',
         'S','T','U','V','W','X',
         'Y','Z']

    def __init__(self):
        self.cache = []

    def get(self):
        _iterations = 0
        while 1:
            sample = random.sample(self.x, 2)
            if not sample in self.cache:
                self.cache.append(sample)
                return sample

            if _iterations > 1000: # just to prevent NOT to run into an infinite loop
                break


s = SampleCache()
for x in range(25):
    print(s.get())

Answer 4

random.sample是正确的方法，你需要用10个字母调用一次而不是用2个字母调用5次：

import random
import string


def random_letters(m=5, n=2):
    letters = random.sample(string.ascii_uppercase, m * n)
    return [letters[n * i:n * (i + 1)] for i in range(m)]

print(random_letters())
# [['I', 'X'], ['J', 'U'], ['O', 'W'], ['G', 'C'], ['D', 'F']]
print(random_letters())
# [['J', 'X'], ['N', 'P'], ['A', 'C'], ['O', 'Z'], ['B', 'H']]
print(random_letters())
# [['U', 'T'], ['J', 'N'], ['C', 'H'], ['D', 'I'], ['K', 'P']]
print(random_letters())
# [['U', 'G'], ['L', 'V'], ['A', 'R'], ['J', 'F'], ['S', 'C']]
print(random_letters())
# [['Y', 'C'], ['R', 'B'], ['E', 'I'], ['S', 'T'], ['H', 'X']]

Answer 5

使用<div class="background" style="background-image: url('link-1-bkg.jpg');"> <a href="#" data-bkgimg="https://rlv.zcache.com/crash_test_dummy_marker_classic_round_sticker-r8b36e2d5c0f74a4c843565523094b867_v9wth_8byvr_324.jpg" class="is-active">Link 1</a> <a href="#" data-bkgimg="https://dummyimage.com/300.png/09f/fff">Link 2</a> </div>生成初始列表的混洗副本，并使用生成器根据需要生成混洗值。

random.sample

输出;

def random_sample(x, n):
    shuffled = random.sample(x, k=len(x))
    for val in range(0, len(x), n):
        yield shuffled[val: val+n]

print([sample for sample in random_sample(x, 2)])

如果你想要五个随机值，那么使用它;

[['I', 'O'], ['V', 'T'], ['U', 'J'], ['L', 'A'], 
 ['E', 'G'], ['Q', 'F'], ['M', 'H'], ['B', 'K'], 
 ['R', 'P'], ['W', 'N'], ['D', 'S'], ['Z', 'Y'], 
 ['X', 'C']]

如果一次想要一个，那么使用，

samples = random_sample(x,  2)
five_samples = [next(samples) for _ in range(5)]
print(five_samples)

Answer 6

您可以循环生成的示例并从x中删除元素：

x = ['A','B','C','D','E','F',
 'G','H','I','J','K','L',
 'M','N','O','P','Q','R',
 'S','T','U','V','W','X',
 'Y','Z']

new_x = x[:]

import random
final_list = []
for i in range(5):
   the_sample = random.sample(new_x, 2)
   final_list.append(the_sample)
   for b in the_sample:
       new_x.remove(b)

输出：

[['C', 'R'], ['L', 'V'], ['W', 'Y'], ['D', 'O'], ['J', 'Q']]

从列表中随机选择一对不同的项目

6 个答案: