Pandas数据帧:按行和行替换值的性能是否存在差异?

时间:2016-06-30 09:27:11

标签: python performance pandas indexing dataframe

我有以下pandas DataFrame:

import pandas as pd
df = pd.read_csv('filename.csv')

print(df)

     dog      A         B           C
0     dog1    0.787575  0.159330    0.053095
1     dog10   0.770698  0.169487    0.059815
2     dog11   0.792689  0.152043    0.055268
3     dog12   0.785066  0.160361    0.054573
4     dog13   0.795455  0.150464    0.054081
5     dog14   0.794873  0.150700    0.054426
..    ....
8     dog19   0.811585  0.140207    0.048208
9     dog2    0.797202  0.152033    0.050765
10    dog20   0.801607  0.145137    0.053256
11    dog21   0.792689  0.152043    0.055268
    ....

我知道行0和列df['dog'],第一个值需要显式更改。而不是dog1,它应该是dog11

用户可以通过哪些方式更改此值?他们之间的表现有很大差异吗?

1 个答案:

答案 0 :(得分:1)

如果希望按列名称选择,则可以使用ixloc

df.ix[0, 'dog'] = 'dog11'
df.loc[0, 'dog'] = 'dog11'

最快是iat,如果您想按位置选择:dog位置(例如,在此示例第一列 - 0中):

df.iat[0, 0] = 'dog11'

<强>计时

In [144]: %%timeit
     ...: df.ix[0, 'dog'] = 'dog11'
The slowest run took 5.37 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 308 µs per loop

In [146]: %%timeit
     ...: df.loc[0, 'dog'] = 'dog11'
     ...: 
The slowest run took 5.03 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 305 µs per loop

In [145]: %%timeit
     ...: df.iat[0, 0] = 'dog11'
The slowest run took 53.64 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 8.18 µs per loop

In [151]: %%timeit
     ...: df.iloc[0, 0] = 'dog11'
     ...: 
The slowest run took 5.69 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 392 µs per loop