忽略嵌套列表中的某些元素

我会在你的任务中使用嵌套列表理解。

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)

请注意 new 列表外部列表压缩考虑每个子列表。内部列表理解考虑单个字符串。

过滤器出现在 if item not in droplist 已执行。

这个 如果项目不在下拉列表中 可以由您可以编码的任何条件替换。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]

如果要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']]