python - 将列表列表/嵌套列表转换为没有嵌套的列表列表

Question

我想转换我的列表列表，并且在这些列表中有列表变成列表列表。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)

目前:

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

预期:

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

我当前的代码可以找到预期的输出。但是，我不得不使用太多的 for 循环而且它太深了。有没有更短的方法，比如一行或两行？

Answer 1

您可以使用列表推导来做到这一点，只需检查嵌套列表是否为空，如果不是，则用内部列表(按索引)替换外部列表。

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'], [], [], [], []]