[英]How to assign value to a pandas dataframe, when subset by complex index and boolean based conditions?
I would like to replace values in a pandas dataframe, with a complex subsetting pattern. 我想用复杂的子集模式替换熊猫数据框中的值。
With the .loc accessor, I was only able to subset by chaining multiple conditions, because some of the conditions are index based. 使用.loc访问器,我只能通过链接多个条件来进行子集化,因为某些条件是基于索引的。 But it seems I can not assign values after such a chain of subsetting.
但是似乎在这样的子集链之后我无法分配值。 UPDATE: A further problem is caused by the duplicated indicies.
更新:另一个问题是由重复的索引引起的。 I have updated the example accordingly.
我已经相应地更新了示例。
import numpy as np
import pandas as pd
df = pd.DataFrame({'a': ['foo'] * 10 + ['bar'] * 10, 'b': range(20)}, index=pd.date_range('2019-01-01','2019-01-10').append(pd.date_range('2019-01-01','2019-01-10')))
df.loc[df['a'] == 'foo', 'b'].loc[pd.to_datetime(['2019-01-05','2019-01-09'])] = np.nan
df
Result: 结果:
a b
2019-01-01 foo 0
2019-01-02 foo 1
2019-01-03 foo 2
2019-01-04 foo 3
2019-01-05 foo 4
2019-01-06 foo 5
2019-01-07 foo 6
2019-01-08 foo 7
2019-01-09 foo 8
2019-01-10 foo 9
2019-01-01 bar 10
2019-01-02 bar 11
2019-01-03 bar 12
2019-01-04 bar 13
2019-01-05 bar 14
2019-01-06 bar 15
2019-01-07 bar 16
2019-01-08 bar 17
2019-01-09 bar 18
2019-01-10 bar 19
Expected: 预期:
a b
2019-01-01 foo 0
2019-01-02 foo 1
2019-01-03 foo 2
2019-01-04 foo 3
2019-01-05 foo NaN
2019-01-06 foo 5
2019-01-07 foo 6
2019-01-08 foo 7
2019-01-09 foo NaN
2019-01-10 foo 9
2019-01-01 bar 10
2019-01-02 bar 11
2019-01-03 bar 12
2019-01-04 bar 13
2019-01-05 bar 14
2019-01-06 bar 15
2019-01-07 bar 16
2019-01-08 bar 17
2019-01-09 bar 18
2019-01-10 bar 19
I have tried alternative approaches like: 我尝试了其他方法,例如:
df.loc[df['a'] == 'foo' and df.index.isin(['2019-01-05','2019-01-09']), 'b']
which drops: 下降:
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
Not even this works, as the isin returns an array without the date based indexing: 甚至都不行,因为isin返回一个没有基于日期的索引的数组:
df['a'] == 'foo' and pd.Series(df.index.isin(['2019-01-05','2019-01-09']))
您可以使用loc
分配的一个.loc
链来做将是不安全的
df.loc[df.index.isin(['2019-01-05','2019-01-09'])&df.a.eq('foo'),'b']=np.nan
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