Question

df：

address        city
BlockOf13thSt  Treasure Isla
Lincoln        Presidio
Duboce Park    Unknown
Twin Peaks     Unknown
Bernal Heights NaN
Holly Courts   Unknown
Ocean Beach    NaN
Maiden Ln      NaN
Avenue N       NaN

输出

address city BlockOf13thSt Treasure Isla Lincoln Presidio Duboce Park San Francisco Twin Peaks San Francisco Bernal Heights San Francisco Holly Courts San Francisco Ocean Beach San Francisco Maiden Ln New York Avenue N New York

大熊猫中是否有任何语法，例如SQL（IN）？地址为IN（Duboce Park，Twin Peaks，Bernal Heights，Holly Courts / Ocean Beach）和“旧金山”和“纽约”的替换/ fillna

谢谢

Answer 1

熊猫df.fillna（）应该可以解决问题。阅读文档：https://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.fillna.html

Answer 2

首先使用所需条件过滤数据框，然后填充空值：

df[(df.address == 'Duboce Park') | (df.address == 'Twin Peaks') | (df.address =='Bernal Heights') | (df.address == 'Holly Courts') | (df.address == 'Ocean Beach')].fillna('San Francisco')

pandas中的管道|运算符类似于OR中的SQL分隔符。

因此，对于以上address，将NULL替换为San Francisco。这样做，其他地址也一样，并用New York替换NULL。

让我知道这是否有帮助。

Answer 3

import pandas as pd  
#Replace all those localities with 'San Francisco'. For this we use .isin() function
df.loc[df['address'].isin(pd.Series(['Duboce Park','Twin Peaks','Bernal Heights','Holly Courts','Ocean Beach'])),'city']='San Francisco'

#Replace all NaNs with 'New York' with fillna().
df = df.fillna('New York')

df
Out[47]: 
          address           city
0   BlockOf13thSt  Treasure Isla
1        Lincoln        Presidio
2     Duboce Park  San Francisco
3      Twin Peaks  San Francisco
4  Bernal Heights  San Francisco
5    Holly Courts  San Francisco
6     Ocean Beach  San Francisco
7       Maiden Ln       New York
8        Avenue N       New York

替换/填充某些特定行的na值

3 个答案: