Question

我有

 list = [a, b, c, d]

和

numbers = [2, 4, 3, 1]

我想获得一个类型列表：

new_list = [a, a, b, b, b, b, c, c, c, d]

这是我到目前为止所做的：

new_list=[] 
for i in numbers: 
    for x in list: 
        for i in range(1,i+1): 
            new_list.append(x)

Answer 1

以下是使用zip，字符串乘法和列表理解进行此操作的一种方法：

lst = ['a', 'b', 'c', 'd'] 
numbers = [2 , 4, 3, 1]

r = [x for i, j in zip(lst, numbers) for x in i*j]
print(r)
# ['a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

使用Python时要注意名称的选择。像list这样的名称会使内置列表函数无法使用。

如果lst中的项目不是字符串，您只需在range上使用嵌套理解就可以复制列表中的项目。

Answer 2

嵌套列表理解有效：

L = ['a','b','c','d']
numbers = [2, 4, 3, 1]

>>> [x for x, number in zip(L, numbers) for _ in range(number)]
['a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

＆＃34;子循环＆＃34; for _ in range(number)重复number次值。这里L可以保存任何对象，而不仅仅是字符串。

示例：

L = [[1, 2, 3],'b','c', 'd']
numbers = [2, 4, 3, 1]
[x for x, number in zip(L, numbers) for _ in range(number)]
[[1, 2, 3], [1, 2, 3], 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

但这会使子列表变平：

[x for i, j in zip(L, numbers) for x in i*j]
[1, 2, 3, 1, 2, 3, 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

不完全是理想的结果。

Answer 3

作为任何对象（不仅仅是字符串）的一般方法，您可以在生成器表达式中使用itertools.repeat()：

def repeat_it(lst, numbers):
    return chain.from_iterable(repeat(i, j) for i, j in zip(lst, numbers))

演示：

In [13]: from itertools import repeat, chain

In [21]: lst=[5,4,6,0]

In [22]: list(repeat_it(lst, numbers))
Out[22]: [5, 5, 4, 4, 4, 4, 6, 6, 6, 0]

In [23]: lst=['a','b','c','d']

In [24]: list(repeat_it(lst, numbers))
Out[24]: ['a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

以下是3种主要方法的基准。请注意，最后一个onley适用于字符串：

In [49]: lst = lst * 1000

In [50]: numbers = numbers * 1000

In [51]: %timeit list(chain.from_iterable(repeat(i, j) for i, j in zip(lst, numbers)))
1 loops, best of 3: 8.8 s per loop

In [52]: %timeit [x for x, number in zip(lst, numbers) for _ in range(number)]
1 loops, best of 3: 12.4 s per loop

In [53]: %timeit [x for i, j in zip(lst, numbers) for x in i*j]
1 loops, best of 3: 7.2 s per loop

Answer 4

您可以使用numpy.repeat()作为另一种选择：

import numpy as np
np.repeat(lst, numbers).tolist()

# ['a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

Answer 5

使用循环执行此操作的另一种方法是：

new_list = []
for number, item in zip(numbers, l):
    for i in range(number):
        new_list.append(item)

现在我们有：

new_list = ['a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

Answer 6

如果您不确定列表推导是如何工作的，

myList=['a','b','c','d'] # not a good idea to use list as a name for your variable
numbers=[2,4,3,1]
new_list=[]
for i in range(len(myList)):     
    for j in range(numbers[i]):          
        new_list.append(myList[i])

print(new_list)

Answer 7

无论a，b，c和d是变量还是字符串，这都会起作用：

a = 1
b = 2.0
c = "cheese"
d = ["c", "a", "k", "e"]

lst = [a, b, c, d]
numbers = [2, 4, 3, 1]

# if len(lst) == len(numbers):
new_lst = [i for i, j in zip(lst, numbers) for k in range(j)]

您可能想要取消注释if语句（并缩进下面的行）以检查列表是否具有相同的长度，否则new_lst将只包含与较短列表一样多的项目。

This，this和the documentation section on nested list comprehensions值得一读。

Answer 8

这是我的解决方案，只是为了添加另一个。

l = ['a', 'b', 'c', 'd']
n = [2, 4, 3, 1]
r = []
for i,v in enumerate(l):
    r += list(v*n[i])

>>> r
    ['a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd']

Answer 9

假设两个列表长度相同而第二个列表始终是数字列表，这里是一个不使用zip或任何导入的解决方案：

lst = ['a', 'b', 'c', 'd']
numbers = [2,4,3,1]

result = sum([[lst[i]]*numbers[i] for i in range(len(lst))],[])

在列表中创建重复项

9 个答案: