Pandas之Fillna填充缺失数据的方法
约定:
import pandas as pd
import numpy as np
from numpy import nan as NaN
填充缺失数据
fillna()是最主要的处理方式了。
df1=pd.DataFrame([[1,2,3],[NaN,NaN,2],[NaN,NaN,NaN],[8,8,NaN]])
df1
代码结果:
0 | 1 | 2 | |
---|---|---|---|
0 | 1.0 | 2.0 | 3.0 |
1 | NaN | NaN | 2.0 |
2 | NaN | NaN | NaN |
3 | 8.0 | 8.0 | NaN |
用常数填充:
df1.fillna(100)
代码结果:
0 | 1 | 2 | |
---|---|---|---|
0 | 1.0 | 2.0 | 3.0 |
1 | 100.0 | 100.0 | 2.0 |
2 | 100.0 | 100.0 | 100.0 |
3 | 8.0 | 8.0 | 100.0 |
通过字典填充不同的常数:
df1.fillna({0:10,1:20,2:30})
代码结果:
0 | 1 | 2 | |
---|---|---|---|
0 | 1.0 | 2.0 | 3.0 |
1 | 10.0 | 20.0 | 2.0 |
2 | 10.0 | 20.0 | 30.0 |
3 | 8.0 | 8.0 | 30.0 |
传入inplace=True直接修改原对象:
df1.fillna(0,inplace=True)
df1
代码结果:
0 | 1 | 2 | |
---|---|---|---|
0 | 1.0 | 2.0 | 3.0 |
1 | 0.0 | 0.0 | 2.0 |
2 | 0.0 | 0.0 | 0.0 |
3 | 8.0 | 8.0 | 0.0 |
传入method=” “改变插值方式:
df2=pd.DataFrame(np.random.randint(0,10,(5,5)))
df2.iloc[1:4,3]=NaN;df2.iloc[2:4,4]=NaN
df2
代码结果:
0 | 1 | 2 | 3 | 4 | |
---|---|---|---|---|---|
0 | 6 | 6 | 2 | 4.0 | 1.0 |
1 | 4 | 7 | 0 | NaN | 5.0 |
2 | 6 | 5 | 5 | NaN | NaN |
3 | 1 | 9 | 9 | NaN | NaN |
4 | 4 | 8 | 1 | 5.0 | 9.0 |
df2.fillna(method='ffill')#用前面的值来填充
代码结果:
0 | 1 | 2 | 3 | 4 | |
---|---|---|---|---|---|
0 | 6 | 6 | 2 | 4.0 | 1.0 |
1 | 4 | 7 | 0 | 4.0 | 5.0 |
2 | 6 | 5 | 5 | 4.0 | 5.0 |
3 | 1 | 9 | 9 | 4.0 | 5.0 |
4 | 4 | 8 | 1 | 5.0 | 9.0 |
传入limit=” “限制填充个数:
df2.fillna(method='bfill',limit=2)
代码结果:
0 | 1 | 2 | 3 | 4 | |
---|---|---|---|---|---|
0 | 6 | 6 | 2 | 4.0 | 1.0 |
1 | 4 | 7 | 0 | NaN | 5.0 |
2 | 6 | 5 | 5 | 5.0 | 9.0 |
3 | 1 | 9 | 9 | 5.0 | 9.0 |
4 | 4 | 8 | 1 | 5.0 | 9.0 |
传入axis=” “修改填充方向:
df2.fillna(method="ffill",limit=1,axis=1)
代码结果:
0 | 1 | 2 | 3 | 4 | |
---|---|---|---|---|---|
0 | 6.0 | 6.0 | 2.0 | 4.0 | 1.0 |
1 | 4.0 | 7.0 | 0.0 | 0.0 | 5.0 |
2 | 6.0 | 5.0 | 5.0 | 5.0 | NaN |
3 | 1.0 | 9.0 | 9.0 | 9.0 | NaN |
4 | 4.0 | 8.0 | 1.0 | 5.0 | 9.0 |
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