user2775128 - 1 year ago 67

Python Question

I have two dataframe like this

`df1 = pd.DataFrame({'A': ['1', '2', '3', '4','5'],`

'B': ['1', '1', '1', '1','1'],

'C': ['A', 'A1', 'A2', 'A3','A4'],

'D': ['B0', 'B1', 'B2', 'B3','B4'],

'E': ['A', 'A', 'S', 'S','S']})

df2 = pd.DataFrame({'A': ['1', '6', '9', '4'],

'C': ['c', 'c1', 'c2', 'c3'],

'D': ['d1', 'd1', 'd2', 'd3']})

and I want to update df1's C,D columns by df2 when they get same column values in A (if df1['A']==df2['A'] then df1['C']=df2['C'] and df1['D']=df2['D'])

the answer should be like this

`A B C D E`

0 1 1 c d1 A

1 2 1 A1 B1 A

2 3 1 A2 B2 S

3 4 1 c3 d3 S

4 5 1 A4 B4 S

I tried

`df1.update(df2)`

`>df1.update(df2)`

> A B C D E

0 1 1 c d1 A

1 6 1 c1 d1 A

2 9 1 c2 d2 S

3 4 1 c3 d3 S

4 5 1 A4 B4 S

and I tried

`pd.merge(df1, df2,how='inner' ,on=['A'])`

`A B C_x D_x E C_y D_y`

0 1 1 A B0 A c d1

1 4 1 A3 B3 S c3 d3

can anyone give me some suggestion?

Thank you

Answer Source

I think this will be more space efficient:

This may be more efficient:

```
In [22]: df1,df2 = df1.align(df2,join='left',axis=0)
In [23]: df1
Out[23]:
A B C D E
0 1 1 A B0 A
1 2 1 A1 B1 A
2 3 1 A2 B2 S
3 4 1 A3 B3 S
4 5 1 A4 B4 S
In [24]: df2
Out[24]:
A C D
0 1 c d1
1 6 c1 d1
2 9 c2 d2
3 4 c3 d3
4 NaN NaN NaN
```

Now you can do find a boolean array where the columns are equal, and use `loc`

based assignment to modify `df1`

inplace without needed the extra columns:

```
In [26]: equal_rows = df1.A == df2.A
In [27]: df1.loc[equal_rows]
Out[27]:
A B C D E
0 1 1 A B0 A
3 4 1 A3 B3 S
In [28]: df1.loc[equal_rows,['C','D']] = df2.loc[equal_rows,['C','D']]
In [29]: df1
Out[29]:
A B C D E
0 1 1 c d1 A
1 2 1 A1 B1 A
2 3 1 A2 B2 S
3 4 1 c3 d3 S
4 5 1 A4 B4 S
```

And if you really need df2 as it was originally:

```
In [30]: df2.dropna(how='all',axis=0, inplace=True)
In [31]: df2
Out[31]:
A C D
0 1 c d1
1 6 c1 d1
2 9 c2 d2
3 4 c3 d3
```

Here is a clunky way that is not space efficient:

```
In [13]: merged = pd.merge(df1,df2,how='left', on=['A'])
In [14]: merged
Out[14]:
A B C_x D_x E C_y D_y
0 1 1 A B0 A c d1
1 2 1 A1 B1 A NaN NaN
2 3 1 A2 B2 S NaN NaN
3 4 1 A3 B3 S c3 d3
4 5 1 A4 B4 S NaN NaN
In [15]: merged.fillna({'C_y':df1.C,'D_y':df1.D},inplace=True)
Out[15]:
A B C_x D_x E C_y D_y
0 1 1 A B0 A c d1
1 2 1 A1 B1 A A1 B1
2 3 1 A2 B2 S A2 B2
3 4 1 A3 B3 S c3 d3
4 5 1 A4 B4 S A4 B4
In [16]: merged.drop(['C_x','D_x'],axis=1,inplace=True)
In [17]: merged
Out[17]:
A B E C_y D_y
0 1 1 A c d1
1 2 1 A A1 B1
2 3 1 S A2 B2
3 4 1 S c3 d3
4 5 1 S A4 B4
```

And if you want the original names:

```
In [20]: merged.rename(columns={"C_y":'C','D_y':'D'},inplace=True)
In [21]: merged
Out[21]:
A B E C D
0 1 1 A c d1
1 2 1 A A1 B1
2 3 1 S A2 B2
3 4 1 S c3 d3
4 5 1 S A4 B4
```