Question

我正在尝试使用此SO as guide对由DocID和字符串组成的DataFrame进行分组，而不是每个DocID具有1行且所有字符串值都用空格分隔的数据帧，我最终得到包含列值的列。

有人可以指出我的错误吗？

样本数据

StringDF.head()

    DocID                                   LessStopWords
0   dd9ae7c8-7e98-4539-ab81-24c4780a6756    judgment of the court chamber 
1   dd9ae7c8-7e98-4539-ab81-24c4780a6756    the request proceedings
2   dd9ae7c8-7e98-4539-ab81-24c4780a6756    legal context law
3   dd9ae7c8-7e98-4539-ab81-24c4780a6756    article 1 directive
4   dd9ae7c8-7e98-4539-ab81-24c4780a6756    the status taken

我的代码

DocsForTopicModel=StringDF.groupby(['DocID'],as_index=False).agg(lambda x : ' '.join(x))

我的输出

     DocID                                  LessStopWords
 0  010b158d-8c0b-49ad-9340-774893e4f62f    DocID LessStopWords
 1  02874037-416d-4b91-8e2d-1a288b8c3a7b    DocID LessStopWords
 2  05b9ea7b-b5f0-4757-854c-b303a295f606    DocID LessStopWords
 3  06f87756-4dbe-4199-a8e2-b504451e823a    DocID LessStopWords
 4  070bd4d1-6830-447e-9042-12c6def18822    DocID LessStopWords

我希望得到的输出

     DocID                                      LessStopWords
     0  010b158d-8c0b-49ad-9340-774893e4f62f    judgment of the court chamber the request proceedings legal context law article 1 directive
     1  02874037-416d-4b91-8e2d-1a288b8c3a7b    ...

Answer 1

您还可以使用.str.cat(sep=' ')（进行串联）：

>>> df.groupby('DocID')['LessStopWords'].apply(lambda ser: ser.str.cat(sep=' '))
DocID
dd9ae7c8-7e98-4539-ab81-24c4780a6756    judgment of the court chamber the request proc...
Name: LessStopWords, dtype: object

Working with Text Data中的更多示例。

更多示例：

>>> import string
>>> import uuid
>>> 
>>> import numpy as np
>>> import pandas as pd
>>> 
>>> uids = np.random.choice([uuid.uuid4() for _ in range(3)], size=10)
>>> words = np.random.choice(list(string.ascii_letters), size=10)
>>> 
>>> df = pd.DataFrame({'DocID': uids, 'LessStopWords': words})
>>> df
                                  DocID LessStopWords
0  8ec3faf7-a771-4e50-87d7-127a69d4d738             p
1  0befc0aa-9311-4154-bced-00a280c99cdd             q
2  8ec3faf7-a771-4e50-87d7-127a69d4d738             t
3  de1021d3-ce47-4f56-8e4d-47d389473dd6             j
4  0befc0aa-9311-4154-bced-00a280c99cdd             L
5  8ec3faf7-a771-4e50-87d7-127a69d4d738             t
6  de1021d3-ce47-4f56-8e4d-47d389473dd6             g
7  0befc0aa-9311-4154-bced-00a280c99cdd             D
8  0befc0aa-9311-4154-bced-00a280c99cdd             d
9  8ec3faf7-a771-4e50-87d7-127a69d4d738             J
>>> df.groupby('DocID')['LessStopWords'].apply(lambda ser: ser.str.cat(sep=' '))
DocID
0befc0aa-9311-4154-bced-00a280c99cdd    q L D d
8ec3faf7-a771-4e50-87d7-127a69d4d738    p t t J
de1021d3-ce47-4f56-8e4d-47d389473dd6        j g
Name: LessStopWords, dtype: object

Pandas GroupBy String正在联接列名而不是列值

1 个答案: