忽略嵌套列表中的某些元素

Question

我有以下列表

[["abc","cdf","efgh","x","hijk","y","z"],["xyz","qwerty","uiop","x","asdf","y","z"]]

我想要以下输出

[["abc","cdf","efgh","hijk"],["xyz","qwerty","uiop","asdf"]]

这里如何进行拆分操作？ PS: 原始数据很大。原始数据：http://pasted.co/20f85ce5

Answer 1

我会为你的任务使用嵌套列表理解。

old = [["abc","cdf","efgh","x","hijk","y","z"],["xyz","qwerty","uiop","x","asdf","y","z"]]

droplist = ["x", "y", "z"]
new = [[item for item in sublist if item not in droplist] for sublist in old]
print(new)

注意 new 列表的创建。外部列表理解考虑了每个子列表。内部列表理解考虑了单个字符串。

过滤发生在if item not in droplist执行时。

if item not in droplist 可以替换为您可以编写代码的任何条件。例如：

new = [[item for item in sublist if len(item) >= 3] for sublist in old]

甚至：

def do_I_like_it(s):
    # Arbitrary code to decide if `s` is worth keeping
    return True
new = [[item for item in sublist if do_I_like_it(item)] for sublist in old]

如果您想按项目在子列表中的位置删除项目，请使用切片：

# Remove last 2 elements of each sublist
new = [sublist[:-2] for sublist in old]
assert new == [['abc', 'cdf', 'efgh', 'x', 'hijk'], ['xyz', 'qwerty', 'uiop', 'x', 'asdf']]

忽略嵌套列表中的某些元素

ignoring certain elements in a nested list

python

list-comprehension

nested-lists

web-scraping