问题
I have a DataFrame:
A B
1: 0 1
2: 0 0
3: 1 1
4: 0 1
5: 1 0
I want to update each item column A of the DataFrame with values of column B if value from column A equals 0.
DataFrame I want to get:
A B
1: 1 1
2: 0 0
3: 1 1
4: 1 1
5: 1 0
I've already tried this code
df['A'] = df['B'].apply(lambda x: x if df['A'] == 0 else df['A'])
It raise an error :The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
回答1:
df['A'] = df.apply(lambda x: x['B'] if x['A']==0 else x['A'], axis=1)
Output
A B
1: 1 1
2: 0 0
3: 1 1
4: 1 1
5: 1 0
回答2:
Use where
In [348]: df.A = np.where(df.A.eq(0), df.B, df.A)
In [349]: df
Out[349]:
A B
1: 1 1
2: 0 0
3: 1 1
4: 1 1
5: 1 0
回答3:
You can perform this by using a mask:
df = pd.DataFrame()
df['A'] = [0,0,1,0,1]
df['B'] = [1,0,1,1,0]
mask = (df.A == 0)
df.loc[mask,'A'] = df.loc[mask,'B']
A B
0 1 1
1 0 0
2 1 1
3 1 1
4 1 0
EDIT: Ok this is actually a unefficient solution:
%timeit df.loc[mask,'A'] = df.loc[mask,'B']
%timeit df.apply(lambda x: x['B'] if x['A']==0 else x['A'], axis=1)
%timeit np.where(df.A.eq(0), df.B, df.A)
5.52 ms ± 556 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
1.27 ms ± 167 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
796 µs ± 89.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
So thanks to zero for this efficient solution with np.where!
来源：https://stackoverflow.com/questions/51787247/pandas-update-column-values-from-another-column-if-criteria