Question

我正在尝试运行以下代码。它适用于较小的数据量，但对于较大的数据量，几乎要花一天的时间。

任何可以帮助优化代码或可以告诉我该方法的人。我们可以使用apply lambda解决问题吗？

for index in df.index:

        for i in df.index:

            if ((df.loc[index,"cityId"]==df.loc[i,"cityId"]) & (df.loc[index,"landingPagePath"]==df.loc[i,"landingPagePath"]) & 
                (df.loc[index,"exitPagePath"]==df.loc[i,"exitPagePath"]) &
                (df.loc[index,"campaign"]==df.loc[i,"campaign"]) &
                (df.loc[index,"pagePath"]==df.loc[i,"previousPagePath"]) &
                ((df.loc[index,"dateHourMinute"]+timedelta(minutes=math.floor(df.loc[index,"timeOnPage"]/60))==df.loc[i,"dateHourMinute"]) |
                        (df.loc[index,"dateHourMinute"]==df.loc[i,"dateHourMinute"]) |
                        ((df.loc[index,"dateHourMinute"]+timedelta(minutes=math.floor(df.loc[index,"timeOnPage"]/60))+timedelta(minutes=1))==df.loc[i,"dateHourMinute"]))
                ):
                    if(df.loc[i,"sess"]==0):


                        df.loc[i,'sess']=df.loc[index,'sess']

                    elif(df.loc[index,"sess"]>df.loc[i,"sess"] ):

                        df.loc[index,'sess']=df.loc[i,'sess']

                    elif(df.loc[index,"sess"]==0):
                        df.loc[index,'sess']=df.loc[i,'sess'] 

                    elif(df.loc[index,"sess"]<df.loc[i,"sess"] ):
                        x=df.loc[i,"sess"]
                        for q in df.index:
                            if(df.loc[q,"sess"]==x):
                                df.loc[q,"sess"]=df.loc[index,'sess']






            else:
                if (df.loc[index,"sess"]==0):

                    df.loc[index,'sess'] = max(df["sess"])+1

Answer 1

看起来您正在尝试手动进行数据库“连接”，Pandas将此功能公开为merge，使用它可以大大解决您的问题

我在遍及所有分支机构时遇到了麻烦，但是如果您使用merge，然后也许要进行一些后期处理/过滤以获得最终答案，那么您应该能够获得大部分途径< / p>

如何优化此代码的性能？

1 个答案: