pandas分组依据并分配组id，然后取消分组

Question

我有一个以下格式的大数据集：

id, socialmedia
1, facebook
2, facebook
3, google
4, google
5, google
6, twitter
7, google
8, twitter
9, snapchat
10, twitter
11, facebook

我想分组，然后分配一个group_id列，然后将其解组（展开）回单个记录。在

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我尝试了以下操作，但结果是“DataFrameGroupBy”对象不支持项分配。在

x['grpId'] = x.groupby('socialmedia')['socialmedia'].rank(method='dense').astype(int)

Answer 1

通过使用ngroup

df['grpId']=df.groupby(' socialmedia').ngroup().add(1)
df
Out[354]: 
    id  socialmedia  grpId
0    1     facebook      1
1    2     facebook      1
2    3       google      2
3    4       google      2
4    5       google      2
5    6      twitter      4
6    7       google      2
7    8      twitter      4
8    9     snapchat      3
9   10      twitter      4
10  11     facebook      1

---

或pd.factorize和{}

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---

df['grpId']=df[' socialmedia'].astype('category').cat.codes.add(1)
df
Out[356]: 
    id  socialmedia  grpId
0    1     facebook      1
1    2     facebook      1
2    3       google      2
3    4       google      2
4    5       google      2
5    6      twitter      4
6    7       google      2
7    8      twitter      4
8    9     snapchat      3
9   10      twitter      4
10  11     facebook      1

Answer 2

您可以使用sklearn.preprocessing.LabelEncoder方法：

In [79]: from sklearn.preprocessing import LabelEncoder

In [80]: le = LabelEncoder()

In [81]: df['groupId'] = le.fit_transform(df['socialmedia'])+1

In [82]: df
Out[82]:
    id socialmedia  groupId
0    1    facebook        1
1    2    facebook        1
2    3      google        2
3    4      google        2
4    5      google        2
5    6     twitter        4
6    7      google        2
7    8     twitter        4
8    9    snapchat        3
9   10     twitter        4
10  11    facebook        1

Answer 3

我们还可以创建一个字典并映射它：

import pandas as pd

df = pd.DataFrame(dict(id=range(1,5),social=["Facebook","Twitter","Facebook","Google"]))

d = dict((k,v) for v,k in enumerate(df['social'].unique(),1))
df['groupid'] = df['social'].map(m)

print(df)

退货

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或者像这样的一句话：

df['groupid'] = df['social'].map({k:v for v,k in enumerate(df['social'].unique(),1)})

---

计时：

%timeit df['grpId']=df.groupby('social').ngroup().add(1)
%timeit df['grpId']=pd.factorize(df['social'])[0]+1
%timeit df['grpId']=df['social'].astype('category').cat.codes.add(1)
%timeit df['groupid'] = df['social'].map(dict((k,v) for v,k in enumerate(df['social'].unique(),1)))

退货

100 loops, best of 3: 1.5 ms per loop   <- Wen1
1000 loops, best of 3: 493 µs per loop  <- Wen2
1000 loops, best of 3: 990 µs per loop  <- Wen3
1000 loops, best of 3: 802 µs per loop  <- Antonvbr