Question

我想将列表转换为列表，并且在这些列表中，列表仅转换为列表。例如：

我的代码：

a = [[], [], [['around_the_world']], [['around_the_globe']], [], [], [], []]
aaa = len(a)
aa = [[] for i in range(aaa)]
for i, x in enumerate(a):
    if len(x) != 0:
        for xx in x:
            for xxx in xx:
                aa[i].append(xxx)
print(aa)

当前：

a = [[], [], [['around_the_world']], [['around_the_globe']], [], [], [], []]

符合预期：

[[], [], ['around_the_world'], ['around_the_globe'], [], [], [], []]

我当前的代码可用于查找预期的输出。但是，我必须使用太多的for循环及其太深的内容。有没有像一两行这样的更短方法呢？

Answer 1

您可以通过列表理解来做到这一点，只需检查嵌套列表是否为空，如果不是，则将内部列表替换为内部列表（按索引）。

data = [[], [], [['around_the_world']], [['around_the_globe']], [], [], [], []]

result = [d[0] if d else d for d in data]
print(result)

# OUTPUT
# [[], [], ['around_the_world'], ['around_the_globe'], [], [], [], []]

Answer 2

我为此使用了itertools。了解更多信息flatten list of list in python

import itertools
a=[[], [], [['around_the_world']], [['around_the_globe']], [], [], [], []]
a = [list(itertools.chain(*li)) for li in a]
print(a)

输出

[[], [], ['around_the_world'], ['around_the_globe'], [], [], [], []]

Answer 3

尝试next和iter：

a = [[], [], [['around_the_world']], [['around_the_globe']], [], [], [], []]
print([next(iter(i),[]) for i in a])

输出：

[[], [], ['around_the_world'], ['around_the_globe'], [], [], [], []]

Answer 4

这里似乎解决了所有短方法，是对大多数这些过程中正在发生的事情的扩展解释。您几乎是unpacking的嵌套列表，并且没有碰到空列表。

a = [[], [], [['around_the_world']], [['around_the_globe']], [], [], [], []]
result = []

for i in a:
    if i == []:
        result.append(i)
    else:
        result.append(*i)

print(result)
# [[], [], ['around_the_world'], ['around_the_globe'], [], [], [], []]

Answer 5

由于您要展平内部子列表，因此可以在列表推导中使用列表推导：

[[i for s in l for i in s] for l in a]

这将返回：

[[], [], ['around_the_world'], ['around_the_globe'], [], [], [], []]

将列表列表/嵌套列表转换为列表列表而不进行嵌套

5 个答案: