Pandas Dataframe：如何通过应用函数更新多列？
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Pandas Dataframe: How to update multiple columns by applying a function?
提问by John Smith
I have a Dataframe df like this:
我有一个像这样的 Dataframe df：
A B C D
2 1 O s h
4 2 P
7 3 Q
9 4 R h m
I have a function f to calculate C and D based on B for a row:
我有一个函数 f 来计算基于 B 的 C 和 D 为一行：
def f(p): #p is the value of column B for a row.
return p+'k', p+'n'
How can I populate the missing values for row 4&7 by applying the function f to the Dataframe?
如何通过将函数 f 应用于数据框来填充第 4 行和第 7 行的缺失值？
The expected outcome is like below:
预期结果如下：
A B C D
2 1 O s h
4 2 P Pk Pn
7 3 Q Qk Qn
9 4 R h m
The function f has to be used as the real function is very complicated. Also, the function only needs to be applied to the rows missing C and D
必须使用函数 f，因为实际函数非常复杂。此外，该函数只需要应用于缺少 C 和 D 的行
回答by Fabio Lamanna
Maybe there is a more elegant way, but I would do in this way:
也许有更优雅的方式，但我会这样做：
df['C'] = df['B'].apply(lambda x: f(x)[0])
df['D'] = df['B'].apply(lambda x: f(x)[1])
Applying the function to the columns and get the first and the second value of the outputs. It returns:
将函数应用于列并获得输出的第一个和第二个值。它返回：
A B C D
0 1 O Ok On
1 2 P Pk Pn
2 3 Q Qk Qn
3 4 R Rk Rn
EDIT:
编辑：
In a more concise way, thanks to this answer:
以更简洁的方式，感谢这个答案：
df[['C','D']] = df['B'].apply(lambda x: pd.Series([f(x)[0],f(x)[1]]))
回答by Zenith
I have a more easy way to do it.
我有一个更简单的方法来做到这一点。
If the table is not so big.
如果桌子不是那么大。
def f(row): #row is the value of row.
if row['C']=='':
row['C']=row['B']+'k'
if row['D']=='':
row['D']=row['B']+'n'
return row
df=df.apply(f,axis=1)
回答by Colonel Beauvel
If you want to use your function as such, here is a one liner:
如果你想使用你的函数为这样的，这里是一个班轮：
df.update(df.B.apply(lambda x: pd.Series(dict(zip(['C','D'],f(x))))), overwrite=False)
In [350]: df
Out[350]:
A B C D
2 1 O s h
4 2 P Pk Pn
7 3 Q Qk Qn
9 4 R h m
You can also do:
你也可以这样做：
df1 = df.copy()
df[['C','D']] = df.apply(lambda x: pd.Series([x['B'] + 'k', x['B'] + 'n']), axis=1)
df1.update(df, overwrite=False)
回答by Nader Hisham
simply by doing the following
只需执行以下操作
df.C.loc[df.C.isnull()] = df.B.loc[df.C.isnull()] + 'k'
df.D.loc[df.D.isnull()] = df.B.loc[df.D.isnull()] + 'n'
check this link indexing-view-versus-copyif you want to know why I've use loc
如果您想知道我为什么使用，请检查此链接indexing-view-versus-copyloc
