Python Pandas：如何分组并为组中的所有项目分配一个 id？
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Python Pandas: How can I group by and assign an id to all the items in a group?
提问by UserYmY
I have df:
我有 df：
domain orgid
csyunshu.com 108299
dshu.com 108299
bbbdshu.com 108299
cwakwakmrg.com 121303
ckonkatsunet.com 121303
I would like to add a new column with replaces domain column with numeric ids per orgid:
我想添加一个新列，用每个 orgid 的数字 ID 替换域列：
domain orgid domainid
csyunshu.com 108299 1
dshu.com 108299 2
bbbdshu.com 108299 3
cwakwakmrg.com 121303 1
ckonkatsunet.com 121303 2
I have already tried this line but it does not give the result I want:
我已经尝试过这条线，但它没有给出我想要的结果：
df.groupby('orgid').count['domain'].reset_index()
Can anybody help?
有人可以帮忙吗？
回答by EdChum
You can call rankon the groupbyobject and pass param method='first':
你可以调用rank的groupby对象，并通过PARAM method='first'：
In [61]:
df['domainId'] = df.groupby('orgid')['orgid'].rank(method='first')
df
Out[61]:
domain orgid domainId
0 csyunshu.com 108299 1
1 dshu.com 108299 2
2 bbbdshu.com 108299 3
3 cwakwakmrg.com 121303 1
4 ckonkatsunet.com 121303 2
If you want to overwrite the column you can do:
如果要覆盖该列，可以执行以下操作：
df['domain'] = df.groupby('orgid')['orgid'].rank(method='first')
回答by Shahnawaz Akhtar
You can use LabelEncoder from sklearn.preprocessing like :
您可以使用 sklearn.preprocessing 中的 LabelEncoder ，例如：
df["domain"] = LabelEncoder().fit_transform(df.domain)
