[英]How to replace indices values in a cell with the column value in a Pandas dataframe
使用replace
為Serie
:
df = df.replace(df['id'])
#or convert to dict (first solution)
#df = df.replace(df['id'].to_dict())
print (df)
id neighborl neighbor2 neighbor3
0 u1 u1 u4 u3
1 u2 u2 u3 u2
2 u3 u3 u1 u2
3 u4 u4 u1 u2
另一個解決方案:
cols = ['neighbor1', 'neighbor2', 'neighbor3']
df[cols] = df[cols].applymap(df['id'].to_dict().get)
print (df)
id neighbor1 neighbor2 neighbor3
0 u1 u1 u4 u3
1 u2 u2 u3 u2
2 u3 u3 u1 u2
3 u4 u4 u1 u2
如果要更多動態解決方案:
#select columns starting by neighbor
cols = df.filter(regex='^neighbor').columns
print (cols)
Index(['neighbor1', 'neighbor2', 'neighbor3'], dtype='object')
df[cols] = df[cols].replace(df['id'])
print (df)
id neighbor1 neighbor2 neighbor3
0 u1 u1 u4 u3
1 u2 u2 u3 u2
2 u3 u3 u1 u2
3 u4 u4 u1 u2
#create mask by columns names starting by neighbor
mask = df.columns.str.startswith('neighbor')
print (mask)
[False True True True]
df.loc[:, mask] = df.loc[:, mask].replace(df['id'])
print (df)
id neighbor1 neighbor2 neighbor3
0 u1 u1 u4 u3
1 u2 u2 u3 u2
2 u3 u3 u1 u2
3 u4 u4 u1 u2
采用
In [289]: cols = ['neighbor1', 'neighbor2', 'neighbor3']
In [290]: df[cols].replace(df.set_index('index')['id'].to_dict())
Out[290]:
neighbor1 neighbor2 neighbor3
0 u1 u4 u3
1 u2 u3 u2
2 u3 u1 u2
3 u4 u1 u2
In [291]: df[cols] = df[cols].replace(df.set_index('index')['id'].to_dict())
In [292]: df
Out[292]:
index id neighbor1 neighbor2 neighbor3
0 0 u1 u1 u4 u3
1 1 u2 u2 u3 u2
2 2 u3 u3 u1 u2
3 3 u4 u4 u1 u2
嘗試這個,
print df.replace(df['id'].to_dict())
輸入:
id neighbor1 neighbor2 neighbor3
index
0 u1 0 3 2
1 u2 1 2 1
2 u3 2 0 1
3 u4 3 0 1
輸出:
id neighbor1 neighbor2 neighbor3
index
0 u1 u1 u4 u3
1 u2 u2 u3 u2
2 u3 u3 u1 u2
3 u4 u4 u1 u2
df = pd.DataFrame([['u1', 0, 3, 2], ['u2', 1, 2, 1], ['u3', 2, 0, 1], ['u4', 3, 0, 1]], columns=['id', 'n1', 'n2', 'n3'])
print df
id n1 n2 n3
0 u1 0 3 2
1 u2 1 2 1
2 u3 2 0 1
3 u4 3 0 1
print df.assign(n1=df.n1.map(df.id),
n2=df.n2.map(df.id),
n3=df.n3.map(df.id))
id n1 n2 n3
0 u1 u1 u4 u3
1 u2 u2 u3 u2
2 u3 u3 u1 u2
3 u4 u4 u1 u2
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