编写代码以打印过去一小时内访问过的所有唯一客户
我的尝试:
import datetime
def find_repeated_customer():
file_obj = open(" my file path","r")
customer_last_visit = {}
repeat_customer = set()
while line in file_obj:
timestamp,customer_id,page_id = line.split(" : ")
last_visit = customer_last_vist.get(customer_id,None)
if not last_visit:
customer_last_visit[customer_id] = last_visit
else:
# assuming time stamp looks like 2016-10-29 01:03:26.947000
year,month,date = timestamp.split(" ")[0].split("-")
current_visit = datetime.date(year,month,date)
day_diff = current_visit - last_visit
if day_diff >=1:
repeat_customer.add(customer_id)
customer_last_visit[customer_id] = current_visit
我完全失败了,以获得我想要的输出。通过这样做,我能够在最后一天获得重复的客户,但如何获得独特的用户?
答案 0 :(得分:1)
你无法一次性完成这种操作。你必须经过一次线路来获得客户,然后才能检查谁来过一次。在另一个过程中,您检查当前客户是否在一次客户列表中并与他做了一些事情。