复制嵌套列表并用连续的数字填充它

你可以使用 itertools.count 使用嵌套列表理解:

from itertools import count

foo = [["apple", "cherry"], ["banana"], ["pear", "raspberry", "pineapple"]]

c = count()  # 0 start is default, e.g. count(1) will start from 1
res = [[next(c) for _ in lst] for lst in foo]

print(res)
# [[0, 1], [2], [3, 4, 5]]

你可以使用 list comprehension ,而不是内部循环:

foo = [["apple", "cherry"], ["banana"], ["pear", "raspberry", "pineapple"]]

foo_numbers = []
count = 0

for i, sublist in enumerate(foo):
    foo_numbers.append([j for j in range(count, len(sublist) + count)])
    count += len(sublist)

print(foo_numbers)

产量

[[0, 1], [2], [3, 4, 5]]

一般来说,列表理解比列表创建的循环要快。另一种方法是将range对象转换为列表,如下所示:

foo_numbers.append(list(range(count, len(sublist) + count)))

比较这些解决方案:

import itertools

def f1(foo):
    """original version in question"""
    foo_numbers = []
    count = 0
    for i, sublist in enumerate(foo):
        foo_numbers.append([])
        for item in sublist:
            foo_numbers[i].append(count)
            count += 1
    return foo_numbers

def f2(foo):
    """@jpp answer"""
    c = itertools.count()
    return [[next(c) for _ in range(len(lst))] for lst in foo]

def f3(foo):
    """@DanielMesejo answer"""
    foo_numbers = []
    count = 0
    for i, sublist in enumerate(foo):
        foo_numbers.append([j for j in range(count, len(sublist) + count)])
        count += len(sublist)
    return foo_numbers

给我们这个:

>>> import timeit
>>>
>>> timeit.timeit('f(foo)', 'from __main__ import f1 as f, foo')
1.2990377170499414
>>> timeit.timeit('f(foo)', 'from __main__ import f2 as f, foo')
2.260929798008874
>>> timeit.timeit('f(foo)', 'from __main__ import f3 as f, foo')
2.1552230638917536

原来的版本似乎更快(2倍)。

如果你真的 do not 要导入任何内容:

foo = [["apple", "cherry"], ["banana"], ["pear", "raspberry", "pineapple"]]

count = -1

def indexer(fruits):
    global count
    count += 1
    return count


foo_new = list(map(lambda fruits: list(map(indexer, fruits)), foo))

print(foo_new)              

输出:

[[0, 1], [2], [3, 4, 5]]