Question

我的数据框包含这些列

ID   Address1   Address1-State   Address1-City  Address2  Address2-State  Address2-City   Address    State      City
 1    6th street   MN              Mpls
 2                                               15th St      MI           Flint
 3                 MA              Boston          Essex St   NY            New York
 4    7 street SE  MN              Mpls           8th St      IL             Chicago

现在，我要以这样一种方式填充地址字段：如果地址1为空白，则填充地址2和地址2的州城市字段

在上述情况下，最终数据帧将是这样

 ID     Address   State    City
  1      6th street   MN    Mpls
  2      15th St      MI    Flint
  3      Essex St     NY     New York
  4     7 street SE  MN     Mpls

当前，我正在这样做

def fill_add(address1,address2):
if address1!='':
    address=address1

elif address1=='' and address2!='':
    address=address2
elif address1=='' and address2=='':
     address=''

return address


def fill_add_apply(df):
df['Address']=df.apply(lambda row:fill_add(row['Address1'],row['Address2']),axis=1)

我是否需要对所有其他列执行相同的操作？是否有更好的方法？

请澄清一下，在ID = 3中，地址，州，城市应为“ Essex St NY New York”，因为地址1为空白，因此应选择地址2和地址2的城市和州。简而言之，如果Address1为空白，则即使Address1-State和Address1-City不为空白，也应选择Address2，Address2-State和Address2-City。

Answer 1

首先修改您的列，然后使用groupby + first

df=df.replace('',np.nan)#prepare for first 

df.columns=df.columns.str.replace('\d+','')
df.columns=df.columns.str.split('-').str[-1]
newdf=df.groupby(level=0,axis=1).first()
newdf.loc[df.iloc[:,1].isnull(),:]=df.groupby(level=0,axis=1).last()
newdf
Out[40]: 
       Address      City  ID State
0  6th street       Mpls   1    MN
1      15th St     Flint   2    MI
2    Essexb St  New York   3    NY
3  7 street SE      Mpls   4    MN

Answer 2

import numpy as np

df=df.replace('',np.nan)

addr_1=['ID','Address1','Address1-State','Address1-City']
addr_2=['ID','Address2','Address2-State','Address2-City']

new_df=pd.DataFrame(df[addr_1].values.copy(),columns=['ID','Address','State','City'])

new_df.loc[new_df['Address'].isnull(),:]=df.loc[df['Address1'].isnull(),addr_2].values

#print(new_df)
    ID  Address     State   City
0   1   6th street  MN      Mpls
1   2   15th St     MI      Flint
2   3   Essex St    NY      New York
3   4   7 street SE MN      Mpls

Answer 3

（鉴于您没有重复的索引）

选择要用Adress1填充的索引：

Address1_index = df.loc[!df.Address1.empty() and !df.Address1-State.empty() and !df.Address1-City.empty()].index

然后将Address1数据放入所需的列：

df.loc[Adress1_index, ["Adress", "State", "City"]] = df.loc[Adress1_index, ["Adress1", "Adress1-State", "Adress1-City"]]

现在选择要用地址2填充的索引：

Address2_index = df.loc[df.Adress1.empty() or df.Adress1-State.empty() or df.Adress1-City.empty()].index

然后还填写以下内容：

df.loc[Adress2_index, ["Adress", "State", "City"]] = df.loc[Adress2_index, ["Adress2", "Adress2-State", "Adress2-City"]]

删除不需要的列：

df.drop(["Address1", "Adress1-State", "Adress1-City", "Address2", "Adress2-State", "Adress2-City"], axis = 1, inplace = True)

根据其他列中的值填充熊猫列

3 个答案: