Pandas DataFrame 列表列：删除特定值

Question

我有一个带有一列列表的 Pandas DataFrame。我想创建一个由这些相同列表组成的新列，减去一个特定元素：

[In]:
key1 = 'site channel fiscal_week'.split()
key2 = 'site dude fiscal_week'.split()
key3 = 'site eng fiscal_week'.split()

keys = pd.DataFrame({'key': [1,2,3],
                     'dims': [key1,key2,key3]})

keys

[Out]:
                         dims  key
[site, channel, fiscal_week]    1
[site, dude, fiscal_week]       2
[site, eng, fiscal_week]        3

这是我失败的方法：

keys['reduced_dims'] = keys['dims'].remove('fiscal_week')

我需要能够删除特定元素，而不是 pop() 从最后一个元素中删除。

期望的输出：

[Out]:
                        dims  key  reduced_dims
[site, channel, fiscal_week]    1  [site, channel]
[site, dude, fiscal_week]       2  [site, dude]
[site, eng, fiscal_week]        3  [site, eng]

Answer 1

keys['dims'] 是 pd.Series，而不是 list，并且没有 list.remove() 方法。您应该使用pd.Series.apply() 方法，该方法将某些函数应用于每行中的值。

keys['reduced_dims'] = keys['dims'].apply(
    lambda row: [val for val in row if val != 'fiscal_week']
)

keys['reduced_dims']

Out:
0    [site, channel]
1       [site, dude]
2        [site, eng]
Name: reduced_dims, dtype: object

而且你不能只使用list.remove() 函数而不是列表理解，

lambda row: [val for val in row if val != 'fiscal_week']

因为list.remove() 返回None 你会得到这样的系列：

keys['reduced_dims'] = keys['dims'].apply(lambda x: x.remove('fiscal_week'))
keys['reduced_dims']

Out:
0    None
1    None
2    None
Name: reduced_dims, dtype: object

Answer 2

你可以试试

def rem_fw(lst):
    lst.remove('fiscal_week')
    return lst

keys['reduced_dims'] = keys['dims'].apply(rem_fw)