Python Pandas KeyError：'标签不在[索引]中'
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Python Pandas KeyError: 'the label is not in the [index]'
提问by J. Doe
I have a duplicates_to_fetchdata frame of index :
我有一个duplicates_to_fetchindex 数据框：
mail_domaine Values
0 @A.com [0, 2]
1 @B.com [1, 4]
And the following df_to_rearrangedata frame
以及以下df_to_rearrange数据框
movie_name df_pname
0 A [mr a]
1 B [mr b]
2 Aa [mr aa]
3 D [mr d]
4 Bb [mr Bb, mr Bbb]
5 E [mr e]
I would like to have the following transformed data frame
我想要以下转换后的数据框
movie_name df_pname
0 [A, Aa] [mr a, mr aa]
1 [B, Bb] [mr b, mr Bb, mr Bbb]
3 [D] [mr d]
5 [E] [mr e]
However... when I drop the lines, the algorithm stops because of missing index
但是......当我删除这些行时，算法会因为缺少索引而停止
I did like
我喜欢
for i in range(0,len(druplicates_to_fetch)):
mylist = duplicates_to_fetch.loc[i,"Values"]
index_to_fetch_on = mylist[0]
# rearrange mylist (which can have >2 values)
mylist = [myindex for myindex in mylist if myindex != index_to_fetch_on]
for j in mylist:
df_to_rearrange.loc[index_to_fetch, "df_pname"].append(df_to_rearrange.loc[j, "df_pname"])
df_to_rearrange.drop(df_to_rearrange.index[j], inplace=True)
The error is the following KeyError: 'the label [179] is not in the [index]'. Id there a more Pythonic way to do this ?
错误如下KeyError: 'the label [179] is not in the [index]'。有没有更 Pythonic 的方法来做到这一点？
回答by Zero
Here's another to way to do it. Use lookupfor groups, and aggthe columns to list
这是另一种方法。使用lookup团体，并agg在列list
In [78]: lookup = {v: i for i, x in enumerate(duplicates_to_fetch['Values']) for v in x}
In [79]: (df_to_rearrange.groupby(df_to_rearrange.index.to_series().map(lookup).fillna(''))
.agg({'movie_name': lambda x: [v for v in x],
'df_pname': lambda x: [a for v in x.values for a in v]})
.reset_index(drop=True))
Out[79]:
movie_name df_pname
0 [A, Aa] [mr a, mr aa]
1 [B, Bb] [mr b, mr Bb, mr Bbb]
2 [D] [mr d]
Details
细节
In [959]: duplicates_to_fetch
Out[959]:
mail_domaine Values
0 @A.com [0, 2]
1 @B.com [1, 4]
In [960]: df_to_rearrange
Out[960]:
movie_name df_pname
0 A [mr a]
1 B [mr b]
2 Aa [mr aa]
3 D [mr d]
4 Bb [mr Bb, mr Bbb]
In [958]: lookup
Out[958]: {0: 0, 1: 1, 2: 0, 4: 1}
回答by Expired Brain
.reindex() as an alternative of .loc
.reindex() 作为 .loc 的替代品
df_to_rearrange.reindex([index_to_fetch], ["df_pname"]).append(df_to_rearrange.reindex([j], ["df_pname"]))
回答by J. Doe
One solution is
一种解决方案是
df_to_rearrange.drop(df_to_rearrange.index[np.where((df_to_rearrange.index==j))], inplace=True)
