我有一个列表X
,其中有来自1-10
的数字,还有一个函数,它使用这个列表创建一个随机值为1-10的新列表。我想调用这个函数100次,并计算同一个值在嵌套列表的同一位置上出现的次数。你知道吗
我创建了函数lists
来实现这一点。我的代码返回正确的答案,但我认为必须有一个更简单的方法来编写它:
从集合导入计数器 随机导入
X = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
def randomX(l):
randomX = [random.choice(X) for i in range(len(data))]
return randomX
def lists(times):
lists = [randomX(X) for i in range(times)]
x1 = Counter(sublist[0] for sublist in lists)
x2 = Counter(sublist[1] for sublist in lists)
x3 = Counter(sublist[2] for sublist in lists)
x4 = Counter(sublist[3] for sublist in lists)
x5 = Counter(sublist[4] for sublist in lists)
x6 = Counter(sublist[5] for sublist in lists)
x7 = Counter(sublist[6] for sublist in lists)
x8 = Counter(sublist[7] for sublist in lists)
x9 = Counter(sublist[8] for sublist in lists)
x10 = Counter(sublist[9] for sublist in lists)
return x1, x2, x3, x4, x5, x6, x7, x8, x9, x10
print(lists(100))
这个打印的结果看起来像这样,这是我想要的,只是我要对它进行排序。然而,似乎没有必要连续写10次几乎相同的东西,但是列表切片不能与counter一起工作。你知道吗
(Counter({7: 14, 8: 12, 6: 12, 3: 11, 9: 11, 1: 10, 10: 9, 4: 9, 5: 7, 2: 5}),
Counter({5: 16, 3: 14, 7: 11, 4: 11, 10: 9, 1: 9, 6: 8, 8: 8, 2: 7, 9: 7}),
Counter({3: 14, 2: 14, 7: 13, 8: 13, 4: 10, 6: 10, 5: 10, 1: 8, 9: 5, 10: 3}),
Counter({3: 15, 6: 15, 8: 12, 7: 11, 1: 11, 4: 11, 2: 10, 9: 7, 5: 5, 10: 3}),
Counter({8: 20, 3: 15, 6: 13, 4: 11, 7: 10, 10: 10, 2: 7, 1: 7, 5: 4, 9: 3}),
Counter({9: 15, 6: 13, 10: 12, 4: 11, 2: 10, 8: 9, 3: 8, 5: 8, 1: 8, 7: 6}),
Counter({6: 17, 7: 13, 9: 11, 2: 11, 8: 11, 5: 10, 3: 8, 10: 8, 4: 6, 1: 5}),
Counter({6: 20, 5: 11, 10: 11, 1: 11, 9: 10, 2: 9, 4: 8, 3: 8, 7: 7, 8: 5}),
Counter({8: 13, 10: 13, 4: 13, 5: 11, 9: 11, 1: 10, 3: 9, 2: 8, 7: 6, 6: 6}),
Counter({10: 14, 8: 13, 1: 10, 2: 10, 3: 10, 5: 10, 7: 9, 4: 9, 9: 9, 6: 6}))
有人对我如何简化这段代码有什么建议吗?你知道吗
您可以使用zip()将创建的列表转换为列,并对这些列进行计数。你知道吗
关于zip()的作用的一个小例子:
输出:
您的代码使用zip()和生成器:
输出:
备注:
即使使用相同的
random.seed(42)
,我们的随机数也不会匹配。使用n次random.choice()
更改内部random Mersenne_Twister状态,与使用random.choices(.., k=n)
不同-如果切换到random.choices
,您将获得相同的输出:立即输出:
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