Question

让我们假设在python中我们有数字列表，如下所示：

[1, 3, 4, 5, 6, 7, 8, 9]

将这个数字列表转换为列表中的随机系列列表的最简单方法是什么？

像这样：

[[1, 2, 3], [4], 5, 6, [7, 8], 9]

Answer 1

使用random.randint：

import random

def random_series(lst, size=3):
    start = end = 0
    n = len(lst)
    while end < n:
        end += random.randint(1, size)
        if end - start == 1:
            yield lst[start]
        else:
            yield lst[start:end]
        start = end

使用示例：

>>> lst = [1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> list(random_series(lst))
[[1, 2, 3], 4, 5, [6, 7], [8, 9]]
>>> list(random_series(lst))
[[1, 2], [3, 4], [5, 6], 7, 8, 9]
>>> list(random_series(lst))
[[1, 2], [3, 4, 5], [6, 7], [8, 9]]
>>> list(random_series(lst))
[[1, 2, 3], [4, 5, 6], 7, [8, 9]]

Answer 2

我的代码（我假设2在输入列表中）：

from random import randint

L = [1, 2, 3, 4, 5, 6, 7, 8, 9]
print L
l, length = 0, len(L)
L2 = []
while l < length:
    r = randint(0, length - l)  # rnum < length_remaining  
    x = L[l: l + r] # taking next `r`  numbers from original L 
    if len(x) > 1:   
        L2.append(x)  # append as list        
    if len(x) == 1:
        L2.append(x[0]) # append as single element 
    l += r    
print L2

一些执行：

$ python  x.py
[1, 2, 3, 4, 5, 6, 7, 8, 9]
[[1, 2, 3, 4], 5, [6, 7, 8], 9]
$ python  x.py
[1, 2, 3, 4, 5, 6, 7, 8, 9]
[[1, 2, 3, 4], [5, 6, 7, 8, 9]]
$ python  x.py
[1, 2, 3, 4, 5, 6, 7, 8, 9]
[[1, 2, 3, 4, 5, 6], [7, 8], 9]
$ python  x.py
[1, 2, 3, 4, 5, 6, 7, 8, 9]
[1, [2, 3], 4, [5, 6], [7, 8], 9]

不是很多，但有点改进：

L = [1, 2, 3, 4, 5, 6, 7, 8, 9]
print L
l, length = 0, len(L)
L2 = []
while l < length:
    r = randint(1, length - l)  # rnum < length_remaining  
    x = L[l: l + r] # taking next `r`  numbers from original L 
    if len(x) > 1:   
        L2.append(x)  # append as list        
    else:
        L2.extend(x) # append as single element 
    l += r    
print L2

将列表转换为列表python中的随机列表

2 个答案: