在列表中创建重复项

我有

 list = [a, b, c, d]

和

numbers = [2, 4, 3, 1]

我想得到一个类型列表:

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

这是我到目前为止:

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

Moses Koledo.. 10

这是使用zip字符串乘法和列表理解的一种方法:

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.一个名称list使内置列表功能无法使用.

如果项目lst不是字符串,您只需使用嵌套理解range,即可复制列表中的项目.

1> Moses Koledo..：

---

这是使用zip字符串乘法和列表理解的一种方法:

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.一个名称list使内置列表功能无法使用.

如果项目lst不是字符串,您只需使用嵌套理解range,即可复制列表中的项目.

---

OP没有指定`a`,`b`等是字符串.您的解决方案仅基于"`list`"包含一系列序列的假设.

2> Mike Müller..：

---

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']

"子循环" for _ in range(number)重复值number时间.这里L可以容纳任何对象,而不仅仅是字符串.

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.

3> Kasramvd..：

---

作为任何对象（不仅是字符串）的通用方法，可以 itertools.repeat()在生成器表达式中使用：

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']

这是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