Using np.where是比较快的。使用与您使用的类似的模式replace
:
df['col1'] = np.where(df['col1'] == 0, df['col2'], df['col1'])
df['col1'] = np.where(df['col1'] == 0, df['col3'], df['col1'])
然而,使用嵌套np.where
稍微快一点:
df['col1'] = np.where(df['col1'] == 0,
np.where(df['col2'] == 0, df['col3'], df['col2']),
df['col1'])
Timings
使用以下设置生成更大的示例 DataFrame 和计时函数:
df = pd.concat([df]*10**4, ignore_index=True)
def root_nested(df):
df['col1'] = np.where(df['col1'] == 0, np.where(df['col2'] == 0, df['col3'], df['col2']), df['col1'])
return df
def root_split(df):
df['col1'] = np.where(df['col1'] == 0, df['col2'], df['col1'])
df['col1'] = np.where(df['col1'] == 0, df['col3'], df['col1'])
return df
def pir2(df):
df['col1'] = df.where(df.ne(0), np.nan).bfill(axis=1).col1.fillna(0)
return df
def pir2_2(df):
slc = (df.values != 0).argmax(axis=1)
return df.values[np.arange(slc.shape[0]), slc]
def andrew(df):
df.col1[df.col1 == 0] = df.col2
df.col1[df.col1 == 0] = df.col3
return df
def pablo(df):
df['col1'] = df['col1'].replace(0,df['col2'])
df['col1'] = df['col1'].replace(0,df['col3'])
return df
我得到以下时间安排:
%timeit root_nested(df.copy())
100 loops, best of 3: 2.25 ms per loop
%timeit root_split(df.copy())
100 loops, best of 3: 2.62 ms per loop
%timeit pir2(df.copy())
100 loops, best of 3: 6.25 ms per loop
%timeit pir2_2(df.copy())
1 loop, best of 3: 2.4 ms per loop
%timeit andrew(df.copy())
100 loops, best of 3: 8.55 ms per loop
我尝试对你的方法进行计时,但它已经运行了几分钟而没有完成。作为比较,仅在 6 行示例 DataFrame(不是上面测试的更大的 DataFrame)上对方法进行计时需要 12.8 毫秒。