Question

我想根据data中的项目过滤drop。

data = [
    ['Basket', 'NBA ET', 'Kobe'],
    ['Basket', 'NCAA', 'Shaq'],
    ['Basket', 'ENG', 'Shaq'],
]
drop = ['NBA', 'NCAA']

因为我希望列出NBA ET的列表也被忽略了，所以它必须超越：

filtered = [d for d in data if d[1] not in drop]   # assume d[1] will hold

我需要的是：

# pseudocode
filtered = [d for d in data if _ not in d[1] for _ in drop]

但我永远记不起语法。

对于记录，filtered应该会产生[['Basket', 'ENG', 'Shaq']]

Answer 1

您可以使用any()并在空白处拆分字符串：

filtered = [d for d in data if not any(dropped in d[1].split() for dropped in drop)]

如果您将drop设为设置，只需测试一个十字路口：

drop = set(drop)

filtered = [d for d in data if not drop.intersection(d[1].split())]

后者应该更大，drop变得更好。

演示：

>>> data = [
...     ['Basket', 'NBA ET', 'Kobe'],
...     ['Basket', 'NCAA', 'Shaq'],
...     ['Basket', 'ENG', 'Shaq'],
... ]
>>> drop = ['NBA', 'NCAA']
>>> [d for d in data if not any(dropped in d[1].split() for dropped in drop)]
[['Basket', 'ENG', 'Shaq']]
>>> drop = set(drop)
>>> [d for d in data if not drop.intersection(d[1].split())]
[['Basket', 'ENG', 'Shaq']]

Answer 2

[row for row in data if not any(league in row[1].split() for league in drop)]
# [['Basket', 'ENG', 'Shaq']]

根据其他列表过滤列表

2 个答案: