基于现有列值和条件列表的pandas数据框中的新列

时间：2019-10-18 13:00:55

标签： python pandas dataframe

下面的链接： New column in pandas dataframe based on existing column values

我有一个数据框，其中有一个名为“国家”的列，其中列出了世界上几个国家。我需要用“欧洲”这样的区域说明符来创建另一列。我有三个属于几个地区的国家/地区列表，因此，如果df ['Country']中的州与'Europe'列表中的州相匹配，则'Europe'说明符将插入到新列df ['Region']中

我的数据是： https://sendeyo.com/up/d/2acd2eb849

问题是当我使用上一个链接中表达的解决方案时，它们适用于示例数据帧，但不适用于我的数据库。 我的数据框是这样的：

Year    Country Population  GDP 
1870    Austria  4,520     8,419    
1870    Belgium  5,096     13,716   
1870    Denmark  1,888     3,782    
1870    Finland  1,754     1,999    
1870    France   38,440    72,100

我的列表：

Europa = ["Austria", "Belgium", "Denmark"]

RamasOccidentales = ["Australia","New Zealand","Canada","United States"]

Latinoamerica = ["Brazil","Chile","Uruguay"]

Asia = ["Indonesia","Japan","Sri Lanka"]

预期结果

Year    Country Population  GDP Region
1870    Austria 4,520   8,419   Europa 
1870    Belgium 5,096   13,716  Europa 
1870    Denmark 1,888   3,782   Europa 
1870    Finland 1,754   1,999   Europa 
1870    France  38,440  72,100  Europa

这是我尝试的代码：

def Continent(country):
    return "Europa" if country in Europa else "Unknown"

df['Region'] = df['Country'].apply(Continent)

谢谢。

3 个答案:

答案 0 :(得分：2)

将Series.map与从列表创建的字典一起使用：

Europa = ["Austria", "Belgium", "Denmark",'France','Finland']
RamasOccidentales = ["Australia","New Zealand","Canada","United States"]
Latinoamerica = ["Brazil","Chile","Uruguay"]
Asia = ["Indonesia","Japan","Sri Lanka"]

d = {'Europa':Europa,'RamasOccidentales':RamasOccidentales,
     'Latinoamerica':Latinoamerica,'Asia':Asia}

#swap key values in dict
#http://stackoverflow.com/a/31674731/2901002
d1 = {k: oldk for oldk, oldv in d.items() for k in oldv}

df['Region'] = df['Country'].map(d1)

print (df)
   Year  Country Population     GDP  Region
0  1870  Austria      4,520   8,419  Europa
1  1870  Belgium      5,096  13,716  Europa
2  1870  Denmark      1,888   3,782  Europa
3  1870  Finland      1,754   1,999  Europa
4  1870   France     38,440  72,100  Europa

print (d1)

{'Austria': 'Europa', 'Belgium': 'Europa', 'Denmark': 'Europa', 
 'France': 'Europa', 'Finland': 'Europa', 
 'Australia': 'RamasOccidentales', 
 'New Zealand': 'RamasOccidentales', 
 'Canada': 'RamasOccidentales', 
 'United States': 'RamasOccidentales', 
 'Brazil': 'Latinoamerica', 'Chile': 'Latinoamerica', 
 'Uruguay': 'Latinoamerica', 'Indonesia': 'Asia',
 'Japan': 'Asia', 'Sri Lanka': 'Asia'}

1万行的性能提高了2.58倍：

np.random.seed(2019)

Europa = ["Austria", "Belgium", "Denmark",'France','Finland']
RamasOccidentales = ["Australia","New Zealand","Canada","United States"]
Latinoamerica = ["Brazil","Chile","Uruguay"]
Asia = ["Indonesia","Japan","Sri Lanka"]

d = {'Europa':Europa,'RamasOccidentales':RamasOccidentales,
     'Latinoamerica':Latinoamerica,'Asia':Asia}

d1 = {k: oldk for oldk, oldv in d.items() for k in oldv}
df = pd.DataFrame({'Country': np.random.choice(list(d1.keys()), size=10000)})

In [280]: %%timeit
     ...: d1 = {k: oldk for oldk, oldv in d.items() for k in oldv}
     ...: 
     ...: df['Region'] = df['Country'].map(d1)
     ...: 
3.04 ms ± 43.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [281]: %%timeit
     ...: classification_countries={'Europa':Europa,
     ...:                           'RamasOccidentales':RamasOccidentales,
     ...:                           'Latinoamerica':Latinoamerica ,
     ...:                           'Asia':Asia}
     ...: 
     ...: cond=[df['Country'].isin(classification_countries[key]) for key in classification_countries]
     ...: values=[ key for key in classification_countries]
     ...: 
     ...: df['Region']=np.select(cond,values)
     ...: 
7.86 ms ± 56.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [282]: %%timeit
     ...: cond=[df['Country'].isin(Europa),df['Country'].isin(RamasOccidentales),df['Country'].isin(Latinoamerica),df['Country'].isin(Asia)]
     ...: values=['Europa','RamasOccidentales','Latinoamerica','Asia']
     ...: df['Region']=np.select(cond,values)
     ...: 
7.96 ms ± 281 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [293]: %%timeit
     ...: classification_countries={'Europa':Europa,
     ...:                           'RamasOccidentales':RamasOccidentales,
     ...:                           'Latinoamerica':Latinoamerica ,
     ...:                           'Asia':Asia}
     ...: 
     ...: dict_cond_values= {key:df['Country'].isin(classification_countries[key]) for key in classification_countries}
     ...: 
     ...: 
     ...: df['Region']=np.select(dict_cond_values.values(),dict_cond_values.keys())
     ...: 
8.54 ms ± 1.31 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)

答案 1 :(得分：1)

使用np.select + Series.isin：

Europa = ["Austria", "Belgium", "Denmark",'France','Finland']

RamasOccidentales = ["Australia","New Zealand","Canada","United States"]

Latinoamerica = ["Brazil","Chile","Uruguay"]

Asia = ["Indonesia","Japan","Sri Lanka"]


#using np.select
cond=[df['Country'].isin(Europa),df['Country'].isin(RamasOccidentales),df['Country'].isin(Latinoamerica),df['Country'].isin(Asia)]
values=['Europa','RamasOccidentales','Latinoamerica','Asia']
df['Region']=np.select(cond,values)

print(df)

   Year  Country Population     GDP  Region
0  1870  Austria      4,520   8,419  Europa
1  1870  Belgium      5,096  13,716  Europa
2  1870  Denmark      1,888   3,782  Europa
3  1870  Finland      1,754   1,999  Europa
4  1870   France     38,440  72,100  Europa

此外，您还可以使用字典创建cond和值的列表。速度更快：

classification_countries={'Europa':Europa,
                          'RamasOccidentales':RamasOccidentales,
                          'Latinoamerica':Latinoamerica ,
                          'Asia':Asia}

dict_cond_values= {key:df['Country'].isin(classification_countries[key]) for key in classification_countries}


df['Region']=np.select(dict_cond_values.values(),dict_cond_values.keys())
print(df)
   Year  Country Population     GDP  Region
0  1870  Austria      4,520   8,419  Europa
1  1870  Belgium      5,096  13,716  Europa
2  1870  Denmark      1,888   3,782  Europa
3  1870  Finland      1,754   1,999  Europa
4  1870   France     38,440  72,100  Europa

或

classification_countries={'Europa':Europa,
                          'RamasOccidentales':RamasOccidentales,
                          'Latinoamerica':Latinoamerica ,
                          'Asia':Asia}

cond=[df['Country'].isin(classification_countries[key]) for key in classification_countries]
values=[ key for key in classification_countries]

df['Region']=np.select(cond,values)
print(df)

   Year  Country Population     GDP  Region
0  1870  Austria      4,520   8,419  Europa
1  1870  Belgium      5,096  13,716  Europa
2  1870  Denmark      1,888   3,782  Europa
3  1870  Finland      1,754   1,999  Europa
4  1870   France     38,440  72,100  Europa
  

对比创建字典后直到执行打印（df）时jezrael测量的解决方案 

%%timeit
dict_cond_values= {key:df['Country'].isin(classification_countries[key]) for key in classification_countries}   
df['Region']=np.select(dict_cond_values.values(),dict_cond_values.keys())
print(df)
#5.06 ms ± 215 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%%timeit
cond=[df['Country'].isin(classification_countries[key]) for key in classification_countries]
values=[ key for key in classification_countries]

df['Region']=np.select(cond,values)
print(df)
#5.18 ms ± 652 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

@jezrael 

%%timeit

d1 = {k: oldk for oldk, oldv in d.items() for k in oldv}

df['Region'] = df['Country'].map(d1)

print (df)
#7.88 ms ± 824 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

答案 2 :(得分：0)

一种非常相似但替代的方法是使用基于字典的查找来确定国家/地区。在此实现中，您将创建一个单个字典，其中国家作为键，而其相应的区域作为配对值。

region_map = {
    'Austria': 'Europa',
    'Brazil': 'Latinoamerica',
    'Japan': 'Asia'  # so on and so forth
}
df['Region'] = df['Country'].apply(lambda c: region_map.get(c, 'Unknown'))

如果不存在键值对，则会从您的词典地图中生成相应的国家或字符串“未知”。

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