Question

我正在尝试从pandas数据框中检索一行，其中单元格值是一个列表。我已经尝试过isin，但看起来它正在执行OR运算，而不是AND运算。

>>> import pandas as pd
>>> df = pd.DataFrame([['100', 'RB','stacked'], [['101','102'], 'CC','tagged'], ['102', 'S+C','tagged']],
    columns=['vlan_id', 'mode' ,    'tag_mode'],index=['dinesh','vj','mani'])

>>> df
           vlan_id  mode  tag_mode
dinesh         100   RB  stacked
vj      [101, 102]   CC   tagged
mani           102  S+C   tagged

>>> df.loc[df['vlan_id'] == '102']; # Fetching string value match
      vlan_id mode tag_mode
mani     102  S+C   tagged

>>> df.loc[df['vlan_id'].isin(['100','102'])]; # Fetching if contains either 100 or 102

       vlan_id mode tag_mode
dinesh     100   RB  stacked
mani       102  S+C   tagged
>>> df.loc[df['vlan_id'] == ['101','102']]; # Fails ? 
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Python27\lib\site-packages\pandas\core\ops.py", line 1283, in wrapper
    res = na_op(values, other)
  File "C:\Python27\lib\site-packages\pandas\core\ops.py", line 1143, in na_op
    result = _comp_method_OBJECT_ARRAY(op, x, y)
  File "C:\Python27\lib\site-packages\pandas\core\ops.py", line 1120, in _comp_method_OBJECT_ARRAY
    result = libops.vec_compare(x, y, op)
  File "pandas\_libs\ops.pyx", line 128, in pandas._libs.ops.vec_compare
ValueError: Arrays were different lengths: 3 vs 2

我可以将值获取到列表中并进行比较。相反，有什么方法可以使用.loc本身对列表值进行检查吗？

Answer 1

要查找列表，您可以遍历vlan_id的值并使用np.array_equal比较每个值：

df.loc[[np.array_equal(x, ['101','102']) for x in df.vlan_id.values]]


     vlan_id    mode    tag_mode
vj  [101, 102]  CC       tagged

尽管如此，建议避免将列表用作数据框中的单元格值。

DataFrame.loc可以使用标签列表或布尔数组来访问行和列。上面的列表理解构造一个布尔数组。

Answer 2

我不确定这是否是最好的方法，或者是否有好的方法，因为据我所知pandas并不是真的支持将lists存储在Series中。还是：

l = ['101', '102']

df.loc[pd.concat([df['vlan_id'].str[i] == l[i] for i in range(len(l))], axis=1).all(axis=1)]

输出：

       vlan_id mode tag_mode
vj  [101, 102]   CC   tagged

Answer 3

另一种解决方法是转换vlan_id列，以便可以将其查询为字符串。您可以通过将vlan_id列表值加入逗号分隔的字符串中来实现。

df['proxy'] = df['vlan_id'].apply(lambda x: ','.join(x) if type(x) is list else ','.join([x]) )

l = ','.join(['101', '102'])
print(df.loc[df['proxy'] == l])

根据列表中的单元格值检索数据框行

3 个答案: