如何匹配python中运行时间过长的所有键值对 [英] how can I match all the key value pair in python which running too long

问题描述

用户-项目亲和度和推荐:
我正在创建一个表格，其中显示购买此商品的客户也购买了算法"
输入数据集

User-item affinity and recommendations :
I am creating a table which suggests "customers who bought this item also bought algorithm "
Input dataset

productId   userId
Prod1        a
Prod1        b
Prod1        c
Prod1        d
prod2        b
prod2        c
prod2        a
prod2        b
prod3        c
prod3        a
prod3        d
prod3        c
prod4        a
prod4        b
prod4        d
prod4        a
prod5        d
prod5        a

需要输出

Product1    Product2    score
Prod1       prod3
Prod1       prod4
Prod1       prod5
prod2       Prod1
prod2       prod3
prod2       prod4
prod2       prod5
prod3       Prod1
prod3       prod2

Using code : 
#Get list of unique items
itemList=list(set(main["productId"].tolist()))

#Get count of users
userCount=len(set(main["productId"].tolist()))

#Create an empty data frame to store item affinity scores for items.
itemAffinity= pd.DataFrame(columns=('item1', 'item2', 'score'))
rowCount=0

#For each item in the list, compare with other items.
for ind1 in range(len(itemList)):

    #Get list of users who bought this item 1.
    item1Users = main[main.productId==itemList[ind1]]["userId"].tolist()
    #print("Item 1 ", item1Users)

    #Get item 2 - items that are not item 1 or those that are not analyzed already.
    for ind2 in range(ind1, len(itemList)):

        if ( ind1 == ind2):
            continue

        #Get list of users who bought item 2
        item2Users=main[main.productId==itemList[ind2]]["userId"].tolist()
        #print("Item 2",item2Users)

        #Find score. Find the common list of users and divide it by the total users.
        commonUsers= len(set(item1Users).intersection(set(item2Users)))
        score=commonUsers / userCount

        #Add a score for item 1, item 2
        itemAffinity.loc[rowCount] = [itemList[ind1],itemList[ind2],score]
        rowCount +=1
        #Add a score for item2, item 1. The same score would apply irrespective of the sequence.
        itemAffinity.loc[rowCount] = [itemList[ind2],itemList[ind1],score]
        rowCount +=1

#Check final result
itemAffinity

代码在示例数据集上运行得非常好，但是
代码在包含 100,000 行的数据集中运行时间太长.请帮我优化代码.

推荐答案

这里的关键是创建一个 productId 的笛卡尔积.见下面的代码，

The key here is to create a cartesian product of productId. See code below,

result=(main.drop_duplicates(['productId','userId'])
            .assign(cartesian_key=1)
            .pipe(lambda x:x.merge(x,on='cartesian_key'))
            .drop('cartesian_key',axis=1)
            .loc[lambda x:(x.productId_x!=x.productId_y) & (x.userId_x==x.userId_y)]
            .groupby(['productId_x','productId_y']).size()
            .div(data['userId'].nunique()))

result

Prod1   prod2   0.75
Prod1   prod3   0.75
Prod1   prod4   0.75
Prod1   prod5   0.5
prod2   Prod1   0.75
prod2   prod3   0.5
prod2   prod4   0.5
prod2   prod5   0.25
prod3   Prod1   0.75
prod3   prod2   0.5
prod3   prod4   0.5
prod3   prod5   0.5
prod4   Prod1   0.75
prod4   prod2   0.5
prod4   prod3   0.5
prod4   prod5   0.5
prod5   Prod1   0.5
prod5   prod2   0.25
prod5   prod3   0.5
prod5   prod4   0.5

方法二

result = (df.groupby(['productId','userId']).size()
            .clip(upper=1)
            .unstack()
            .assign(key=1)
            .reset_index()
            .pipe(lambda x:x.merge(x,on='key'))
            .drop('key',axis=1)
            .loc[lambda x:(x.productId_x!=x.productId_y)]
            .set_index(['productId_x','productId_y'])
            .pipe(lambda x:x.set_axis(x.columns.str.split('_',expand=True),axis=1,inplace=False))
            .swaplevel(axis=1)
            .pipe(lambda x:(x['x']+x['y']))
            .fillna(0)
            .div(2) 
            .mean(axis=1))

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