Jim Eisenberg
2020-02-01 00:54
如果df1和df2是您的两个数据框：
import numpy as np
def get_district(city):
dlist = list(df2.loc[df2['city_id']==city, 'district_id']) #get list of districts
p = list(df2.loc[df2['city_id']==city, 'probability']) #get corresponding odds
return np.random.choice(dlist, p=p) #give weighed random choice from list
并应用此：
df['district_id'] = df.city_id.apply(get_district)
在@JoeCondron的有用评论之后，另一种方法是：
def get_city_district(city,df1,df2):
l = len(df1[df1.city_id==city])
d = df2[df2['city_id']==city]
ds, p = list(d['district_id']),list(d['probability'])
df1.loc[df1.city_id==city,'district_id'] = np.random.choice(ds, size=l,p=p)
return df1
def f(df1,df2):
df1['district_id'] = None
for i in set(df1.city_id):
df1 = get_city_district(i,df1,df2)
return df1
经过测试的速度要快得多，但仅限少数几个城市。
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这将要求
get_district每一行，df而每一行又将df2不必要地进行切片。我们只需要获取每个唯一城市的权重一次。此外，您将两次生成相同的布尔键。