Question

3) Top-selling products (30 points) Use Python programming to find most popular single products and co-purchased products from the large transaction data: retail.csv. Each row in the file is one purchase transaction from a customer, including a set of product ids separated by commas. The first column is transaction ID, column 2-3 are the products ID purchased in this transaction. It is worth mentioning that if the value in third column is zero, it means this customer only purchased one product (the one in second column). Note: Co-purchased products is defined as a pair of products purchased in the same transaction. For example a row is: "2 24 35". Then 24 and 35 is a pair of co- purchased products IDS, To find co-purchased product in each transaction, you might use a nested loop. Write top 10 single products and top 10 co-purchased product pairs into a new file: output.txt

Answer 1

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sales.txt

purchase.py output.txt sales.txt X O . pyth > 1 2 sales.txt 12 54 45 11 1 0 15 54 45 88 1 0 87 34 44 89 34 45 85 34 44 90 2 0 91 20 92 44 0 0 A15 Ln 10, Col & Spaces: 4 UTF-8 LF Plain Text 3 0 0

output.txt:

sales.txt purchase.py output.txt X pyth > output.txt 1 Top 10 Purchased items: 2 1 2 4 5 6 44 Top 10 Co-Purchased items: 54 45 34 44 34 45 8 9 0 A15 Ln 9, Col1 Spaces: 4 UTF-8 LF Plain Text 3 0 0

code(use this for proper indentation)

a of Good purchase.py X output.txt sales.txt pyth > purchase.py > ... 1 filename=input("Enter the name of the file: "); purchased={} co_purchased={} file=open(filename) for line in file.readlines (: data=line.split() if data[2]=='0': if int(data[1]) in purchased: purchased [data[1]]+=1 else: purchased(data[1]]=1 else: t=tuple(data[1:3]). if t in co_purchased : co_purchased[t]+=1 else: co_purchased[t]=1 sorted purchased=sorted (purchased. items(), key=lambda x:x[1], reverse=True) sorted co purchased=sorted(co purchased items(), key=lambda x:x[1], reverse=True) 23 24 25 26 27 28 output file=open("output.txt", "W"); output file.write("Top 10 Purchased items:\n") for i in sorted_purchased: output file.write(i[0]+"\n"); output_file.write("Top 10 Co-Purchased items:\n") for i in sorted co purchased : output_file.write(i[0][0]+" "+i[0][1]+"\n"); 0 A1 % Ln 27, Col1 Spaces: 4 UTF-8 LF Python 0 5

CODE:

#take input, name of the file

filename=input("Enter the name of the file: ")

#open the file

file=open(filename)

#2 dictionaries to hold purchased and copurchased items and their counts

purchased={}

co_purchased={}

#file.readlines will return a list of all the lines from the file

#for every line in the file

for line in file.readlines():

#split the line, get transaction id at data[0], item1 id at data[1] and item2 id at data[2]

data=line.split()

#if item2 does not exist, its a single item purchase

if data[2]=='0':

#if item exists in purchased dictionary,increase it's count

if data[1] in purchased:

purchased[data[1]]+=1

#otherwise add it to purchase dictionary with count 1

else:

purchased[data[1]]=1

#else if item2 exists

else:

#form a tuple containing item1 and item2 together

#this is because a dictionary cannot have a list as key, only tuples,numbers,string,etc can be keys

t=tuple(data[1:3])

#if t is in copurchased history,increase it's count

if t in co_purchased:

co_purchased[t]+=1

#otherwise add it to copurchase dictionary with count 1

else:

co_purchased[t]=1

#sort the 2 dictionaries based on count

#this is a standard method to sort a dictionary by value

sorted_purchased=sorted(purchased.items(),key=lambda x:x[1],reverse=True)

sorted_co_purchased=sorted(co_purchased.items(),key=lambda x:x[1],reverse=True)

#open output.txt in write mode

output_file=open("output.txt","w")

#write a heading

output_file.write("Top 10 Purchased items:\n")

#for every item in purchased

for i in sorted_purchased:

#write it to file

output_file.write(i[0]+"\n")

#write one more heading

output_file.write("Top 10 Co-Purchased items:\n")

#for every set of co-items in copurchased

for i in sorted_co_purchased:

#write them into file

output_file.write(i[0][0]+" "+i[0][1]+"\n")