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一起使用.loc和.replace()

[英]Using .loc and .replace() Together

How can I update a dataframe's columns values based on dictionary? 如何根据字典更新数据框的列值?

For example, I have df that looks like: df = pd.DataFrame({'B' : [100,101,102,103],'E' : pd.Categorical(["test","train","test","train"]), 'F' : [128,300,1205,2000]}) 例如,我的df如下所示: df = pd.DataFrame({'B' : [100,101,102,103],'E' : pd.Categorical(["test","train","test","train"]), 'F' : [128,300,1205,2000]})

Out[28]:

     B      E     F
0  100   test   128
1  101  train   300
2  102   test  1205
3  103  train  2000

dict = {300:301, 2000:2001}

df.loc[df['B'].isin([101,103])].replace(dict)

Out[31]: 
     B      E     F
1  101  train   301
3  103  train  2001

This gives the proper results but doing this inplace gives a Copy Warning and I need to update the original dataframe with this logic. 这样可以得出正确的结果,但是就地执行此操作则会发出“复制警告”,我需要使用此逻辑来更新原始数据框。

Also, doing a very inefficient groupby & apply combo works but clearly not optimal. 另外,执行非常低效的groupbyapply组合工作,但显然不是最佳选择。

How can I accomplish this? 我该怎么做?

You can assign the result back to the same positions of the data frame: 您可以将结果分配回数据框的相同位置:

d = {300:301, 2000:2001}  
mask = df.B.isin([101, 103])  
df.loc[mask] = df.loc[mask].replace(d)

df
#     B     E      F
#0  100 test    128
#1  101 train   301
#2  102 test    1205
#3  103 train   2001

Or you can use update : 或者您可以使用update

df.update(df.loc[df.B.isin([101, 103])].replace(d))

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