Question

我有两个数据框如下

df1     df2 
A       A   C
A1      A1  C1
A2      A2  C2
A3      A3  C3
A1      A4  C4
A2          
A3          
A4

列＆＃39; A＆＃39;的值在df2的C＆＃39;列中定义。我想在df1中为df1添加一个新列，其值来自df2列＆＃39; C＆＃39;

最终的df1应如下所示

df1
A   B
A1  C1
A2  C2
A3  C3
A1  C1
A2  C2
A3  C3
A4  C4

我可以遍历df2并将值添加到df1，但由于数据量很大，所以耗费时间。

    for index, row in df2.iterrows():
           df1.loc[df1.A.isin([row['A']]), 'B']= row['C']

有人可以帮助我理解如何在不绕过df2的情况下解决这个问题。

由于

Answer 1

您可以Series使用map：

df1['B'] = df1.A.map(df2.set_index('A')['C'])
print (df1)
    A   B
0  A1  C1
1  A2  C2
2  A3  C3
3  A1  C1
4  A2  C2
5  A3  C3
6  A4  C4

map与dict相同：

d = df2.set_index('A')['C'].to_dict()
print (d)
{'A4': 'C4', 'A3': 'C3', 'A2': 'C2', 'A1': 'C1'}

df1['B'] = df1.A.map(d)
print (df1)
    A   B
0  A1  C1
1  A2  C2
2  A3  C3
3  A1  C1
4  A2  C2
5  A3  C3
6  A4  C4

<强>计时：

len(df1)=7：

In [161]: %timeit merged = df1.merge(df2, on='A', how='left').rename(columns={'C':'B'})
1000 loops, best of 3: 1.73 ms per loop

In [162]: %timeit df1['B'] = df1.A.map(df2.set_index('A')['C'])
The slowest run took 4.44 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 873 µs per loop

len(df1)=70k：

In [164]: %timeit merged = df1.merge(df2, on='A', how='left').rename(columns={'C':'B'})
100 loops, best of 3: 12.8 ms per loop

In [165]: %timeit df1['B'] = df1.A.map(df2.set_index('A')['C'])
100 loops, best of 3: 6.05 ms per loop

Answer 2

IIUC你可以合并并重命名col

df1.merge(df2, on='A', how='left').rename(columns={'C':'B'})

In [103]:
df1 = pd.DataFrame({'A':['A1','A2','A3','A1','A2','A3','A4']})
df2 = pd.DataFrame({'A':['A1','A2','A3','A4'], 'C':['C1','C2','C4','C4']})
merged = df1.merge(df2, on='A', how='left').rename(columns={'C':'B'})
merged

Out[103]:
    A   B
0  A1  C1
1  A2  C2
2  A3  C4
3  A1  C1
4  A2  C2
5  A3  C4
6  A4  C4

Answer 3

基于searchsorted方法，以下是三种采用不同索引方案的方法 -

df1['B'] = df2.C[df2.A.searchsorted(df1.A)].values
df1['B'] = df2.C[df2.A.searchsorted(df1.A)].reset_index(drop=True)
df1['B'] = df2.C.values[df2.A.searchsorted(df1.A)]

比较Pandas数据帧并添加列

3 个答案: