pandas Panda 的数据框将一列拆分为多列
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Panda's dataframe split a column into multiple columns
提问by Emdadul
I have a pandas dataframe looks like as below:
我有一个Pandas数据框，如下所示：
date | location | occurance <br>
------------------------------------------------------
somedate |united_kingdom_london | 5
somedate |united_state_newyork | 5
I want it to transform into
我想让它变成
date | country | city | occurance <br>
---------------------------------------------------
somedate | united kingdom | london | 5
---------------------------------------------------
somedate | united state | newyork | 5
I am new to Python and after some research I have written following code, but seems to unable to extract country and city:
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'])
I appreciate your response.
我很感激你的回应。
回答by Merlin
Starting with this:
从这个开始：
df = pd.DataFrame({'Date': ['somedate', 'somedate'],
'location': ['united_kingdom_london', 'united_state_newyork'],
'occurence': [5, 5]})
Try this:
尝试这个：
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
Borrowing idea from @MaxU
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
回答by Kartik
Try this:
尝试这个：
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)
If you want it back in the original df:
如果你想要它回到原来的 df：
df['country'] = splits[0:-1].tolist()
df['city'] = splits[-1].tolist()
回答by Parfait
Consider splitting the column's string value using rfind()
考虑使用拆分列的字符串值 rfind()
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
回答by mgilbert
Something like this works
像这样的工作
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
You could use regex in the replace for a more complicated pattern but if it's just the word after the last _I find it easier to just reverse the str twice as a hack rather than fiddling around with regular expressions
您可以在替换中使用正则表达式来获得更复杂的模式，但如果它只是最后一个之后的单词，_我发现将 str 反转两次作为一种黑客攻击更容易，而不是摆弄正则表达式
回答by MaxU
I would use .str.extract()method:
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 please be aware that it's not possible to parse properly (to distinguish) between rows containing two-words country + one-word city and rows containing one-word country + two-words city (unless you have a full list of countries so you check it against this list)...
PS请注意，无法正确解析（区分）包含两个词国家+一个词城市的行和包含一个词国家+两个词城市的行（除非您有完整的国家/地区列表，因此您请对照此列表进行检查）...
