对于pandas操作来说,我有两个这样的数据帧：
import pandas as pd
df = pd.DataFrame({'name': ['a','a','b','b','c','c'], 'id':[1,2,1,2,1,2], 'val1':[0,0,0,0,0,0],'val2':[0,0,0,0,0,0],'val3':[0,0,0,0,0,0]})
id name val1 val2 val3
0 1 a 0 0 0
1 2 a 0 0 0
2 1 b 0 0 0
3 2 b 0 0 0
4 1 c 0 0 0
5 2 c 0 0 0
subdf = pd.DataFrame({'name': ['a','b','c'], 'id':[1,1,2],'val1':[0.3,0.4,0.7], 'val2':[4,5,4]}
id name val1 val2
0 1 a 0.3 4
1 1 b 0.4 5
2 2 c 0.7 4
我想获得输出：
id name val1 val2 val3
0 1 a 0.3 4 0
1 2 a 0.0 0 0
2 1 b 0.4 5 0
3 2 b 0.0 0 0
4 1 c 0.0 0 0
5 2 c 0.7 4 0
但我没有抓住替换的例子,只是添加了我看到的教程中的列/行！
解决方法:
这需要几个步骤,在匹配的列上留下merge,这将创建“x”和“y”,其中存在冲突：
In [25]:
merged = df.merge(subdf, on=['id', 'name'], how='left')
merged
Out[25]:
id name val1_x val2_x val3 val1_y val2_y
0 1 a 0 0 0 0.3 4
1 2 a 0 0 0 NaN NaN
2 1 b 0 0 0 0.4 5
3 2 b 0 0 0 NaN NaN
4 1 c 0 0 0 NaN NaN
5 2 c 0 0 0 0.7 4
In [26]:
# take the values that of interest from the clashes
merged['val1'] = np.max(merged[['val1_x', 'val1_y']], axis=1)
merged['val2'] = np.max(merged[['val2_x', 'val2_y']], axis=1)
merged
Out[26]:
id name val1_x val2_x val3 val1_y val2_y val1 val2
0 1 a 0 0 0 0.3 4 0.3 4
1 2 a 0 0 0 NaN NaN 0.0 0
2 1 b 0 0 0 0.4 5 0.4 5
3 2 b 0 0 0 NaN NaN 0.0 0
4 1 c 0 0 0 NaN NaN 0.0 0
5 2 c 0 0 0 0.7 4 0.7 4
In [27]:
# drop the additional columns
merged = merged.drop(labels=['val1_x', 'val1_y','val2_x', 'val2_y'], axis=1)
merged
Out[27]:
id name val3 val1 val2
0 1 a 0 0.3 4
1 2 a 0 0.0 0
2 1 b 0 0.4 5
3 2 b 0 0.0 0
4 1 c 0 0.0 0
5 2 c 0 0.7 4
另一种方法是在“id”和“name”上对df进行排序,然后调用update：
In [30]:
df = df.sort(columns=['id','name'])
subdf = subdf.sort(columns=['id','name'])
df.update(subdf)
df
Out[30]:
id name val1 val2 val3
0 1 a 0.3 4 0
2 2 c 0.7 4 0
4 1 c 0.0 0 0
1 1 b 0.4 5 0
3 2 b 0.0 0 0
5 2 c 0.0 0 0
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