Question

我的pandas数据框如下所示：

date     |    location          | occurance <br>
------------------------------------------------------
somedate |united_kingdom_london | 5  
somedate |united_state_newyork  | 5

我希望它转变为

date     | country        | city    | occurance <br>
---------------------------------------------------
somedate | united kingdom | london  | 5  
---------------------------------------------------
somedate | united state   | newyork | 5

我是Python新手，经过一些研究，我编写了以下代码，但似乎无法提取国家和城市：

df.location= df.location.replace({'-': ' '}, regex=True)
df.location= df.location.replace({'_': ' '}, regex=True)

temp_location = df['location'].str.split(' ').tolist() 

location_data = pd.DataFrame(temp_location, columns=['country', 'city'])

感谢您的回复。

Answer 1

从这开始：

df = pd.DataFrame({'Date': ['somedate', 'somedate'],
                   'location': ['united_kingdom_london', 'united_state_newyork'],
                   'occurence': [5, 5]})

试试这个：

df['Country'] = df['location'].str.rpartition('_')[0].str.replace("_", " ")
df['City']    = df['location'].str.rpartition('_')[2]
df[['Date','Country', 'City', 'occurence']]

      Date        Country      City  occurence
0  somedate  united kingdom   london          5
1  somedate    united state  newyork          5

借用@MaxU的想法

df[['Country'," " , 'City']] = (df.location.str.replace('_',' ').str.rpartition(' ', expand= True ))
df[['Date','Country', 'City','occurence' ]]

      Date        Country      City  occurence
0  somedate  united kingdom   london          5
1  somedate    united state  newyork          5

Answer 2

str.rsplit的另一个解决方案，如果国家/地区没有_（只包含一个字），则效果很好：

import pandas as pd

df = pd.DataFrame({'date': {0: 'somedate', 1: 'somedate', 2: 'somedate'}, 
                   'location': {0: 'slovakia_bratislava', 
                                1: 'united_kingdom_london', 
                                2: 'united_state_newyork'}, 
                   'occurance <br>': {0: 5, 1: 5, 2: 5}})    
print (df)
       date               location  occurance <br>
0  somedate    slovakia_bratislava               5
1  somedate  united_kingdom_london               5
2  somedate   united_state_newyork               5

df[['country','city']] = df.location.str.replace('_', ' ').str.rsplit(n=1, expand=True)
#change ordering of columns, remove location column
cols = df.columns.tolist()
df = df[cols[:1] + cols[3:5] + cols[2:3]]
print (df)
       date         country        city  occurance <br>
0  somedate        slovakia  bratislava               5
1  somedate  united kingdom      london               5
2  somedate    united state     newyork               5

Answer 3

试试这个：

temp_location = {}
splits = df['location'].str.split(' ')
temp_location['country'] = splits[0:-1].tolist() 
temp_location['city'] = splits[-1].tolist() 

location_data = pd.DataFrame(temp_location)

如果你想要它回到原来的df：

df['country'] = splits[0:-1].tolist() 
df['city'] = splits[-1].tolist()

Answer 4

考虑使用rfind()

拆分列的字符串值

import pandas as pd

df = pd.DataFrame({'Date': ['somedate', 'somedate'],
                   'location': ['united_kingdom_london', 'united_state_newyork'],
                   'occurence': [5, 5]})

df['country'] = df['location'].apply(lambda x: x[0:x.rfind('_')])
df['city'] = df['location'].apply(lambda x: x[x.rfind('_')+1:])

df = df[['Date', 'country', 'city', 'occurence']]
print(df)

#        Date         country     city  occurence
# 0  somedate  united_kingdom   london          5
# 1  somedate    united_state  newyork          5

Answer 5

像这样的东西

import pandas as pd

df = pd.DataFrame({'Date': ['somedate', 'somedate'],
                   'location': ['united_kingdom_london', 'united_state_newyork'],
                   'occurence': [5, 5]})

df.location = df.location.str[::-1].str.replace("_", " ", 1).str[::-1]
newcols = df.location.str.split(" ")
newcols = pd.DataFrame(df.location.str.split(" ").tolist(),
                         columns=["country", "city"])
newcols.country = newcols.country.str.replace("_", " ")
df = pd.concat([df, newcols], axis=1)
df.drop("location", axis=1, inplace=True)
print(df)

         Date  occurence         country     city
  0  somedate          5  united kingdom   london
  1  somedate          5    united state  newyork

你可以在替换中使用正则表达式来处理更复杂的模式但是如果它只是在最后一个_之后的单词我发现更容易将str作为hack反转两次而不是摆弄正则表达式< / p>

Answer 6

我会使用.str.extract()方法：

In [107]: df
Out[107]:
       Date               location  occurence
0  somedate  united_kingdom_london          5
1  somedate   united_state_newyork          5
2  somedate         germany_munich          5

In [108]: df[['country','city']] = (df.location.str.replace('_',' ')
   .....:                             .str.extract(r'(.*)\s+([^\s]*)', expand=True))

In [109]: df
Out[109]:
       Date               location  occurence         country     city
0  somedate  united_kingdom_london          5  united kingdom   london
1  somedate   united_state_newyork          5    united state  newyork
2  somedate         germany_munich          5         germany   munich

In [110]: df = df.drop('location', 1)

In [111]: df
Out[111]:
       Date  occurence         country     city
0  somedate          5  united kingdom   london
1  somedate          5    united state  newyork
2  somedate          5         germany   munich

PS请注意，无法正确解析（区分）包含两个单词country + one-word city的行和包含一个单词country + two-words city的行（除非你有一个完整的列表国家，所以你检查这个列表）...

Panda的数据框将一列分成多列

6 个答案: