python-3.x - 如果所有星星都相同，则减少 df 中逗号分隔的字符串

Question

到目前为止，这是我的代码

import pandas as pd
from io import StringIO

data = StringIO("""
"name1","hej","7aa","a"
"name1","du","71al","a"
"name1","aj","74a","a"
"name1","oj","7aj","a"
"name2","fin","7ag","a"
"name2","katt","7a","a"
""")
df = pd.read_csv(data, header=0, names=["name","text2","text","as"])
df[['text2','text','as']] = df.groupby(['name']).transform(lambda 
x: ','.join(x))
df = df[['name','text','text2','as']].drop_duplicates()
df

让我大部分时间。

df
    name          text     text2     as
0  name1  71al,74a,7aj  du,aj,oj  a,a,a
3  name2        7ag,7a  fin,katt    a,a

我只需要一行来检查每个 cols ['text','text2','as'] 并且如果所有逗号分隔的元素都相同，则只返回第一个

所以我追求的结果是

df
    name          text     text2     as
0  name1  71al,74a,7aj  du,aj,oj    a
3  name2        7ag,7a  fin,katt    a

我试过用 split(',') 申请。无法让它工作。

我正在添加这篇文章的第一条评论。我未能正确描述我的问题

如果我的 df 是这样的:

df
    name          text     text2     as
0  name1  71al,74a,7aj  du,aj,oj  a,b,a
3  name2        7ag,7a  fin,katt    a,a

我需要将其修改为:

df
    name          text     text2     as
0  name1  71al,74a,7aj  du,aj,oj  a,b,a
3  name2        7ag,7a  fin,katt    a

不是:

df
    name          text     text2     as
0  name1  71al,74a,7aj  du,aj,oj  a,b
3  name2        7ag,7a  fin,katt    a

谢谢

这是我的最终解决方案:

data = StringIO("""
"name1","hej","7aa","a"
"name1","du","71al","b"
"name1","aj","74a","a"
"name1","oj","7aj","a"
"name2","fin","7ag","a"
"name2","katt","7a","a"
""")
df = pd.read_csv(data, header=0, names=["name","text2","text","as"])
df[['text2','text','as']] = df.groupby(['name']).transform(lambda x: ','.join(x))
df = df[['name','text','text2','as']].drop_duplicates()
for col in df.columns:
    df[col] = df[col].str.split(',').map(lambda x: ','.join(set(x) if len(set(x)) == 1 else x))
df

我不得不求助于迭代。我无法使用 agg 获得所需的结果。此外，如果有人可以向我解释 len(set(x)) == 1 在这里，那将不胜感激(由于逗号，它应该至少为 2 吗？)

Answer 1

尝试

df['as'] = df['as'].str.split(',').map(lambda x: ','.join(set(x) if len(set(x)) == 1 else x))