Identifying Top Users and their Corresponding Purchases On a Website
Problem Description:
Click here to get the problem description.
Version of Python:
Solutions:
Method1:
sum=[]
dict={}
def id_best_users(*args):
# your code here
for arg in args:
sum.extend(arg)
#print(sum)
for item in sum:
dict[item]=dict.get(item,0)+1
#print("dict: ",dict)
#print("###################################################################################################################")
m2={}
for k,v in dict.items():
if v in m2.keys():
m2[v].append(k)
else:
m2[v] = [k]
#print(m2)
#print("###################################################################################################################")
list2=[]
for k,v in m2.items():
list1=[]
list1.append(k)
list1.append(v)
#print(list1)
list2.append(list1)
#print(list2)
#print("###################################################################################################################")
list2.sort(reverse=True)
#print(list2)
return list2
Grammars(Method1):
- python 統計list中各個元素出現的次數
dict[item]=dict.get(item,0)+1Click here to learn how it work. - python字典使用法 統計具有相同value的key
for k,v in dict.items():Click here to learn how it work. - 判斷python字典中key是否存在的
if v in m2.keys():Click here to learn how it work. - Python list排序的兩種方法及實例講解
list2.sort(reverse=True)Click here to learn how it work.
Method2:
from collections import Counter
from itertools import chain
def id_best_users(*args):
best_users = set.intersection(*(set(a) for a in args))
cnt = Counter(chain(*args))
users = {}
for k, v in cnt.iteritems():
if k in best_users:
users.setdefault(v, []).append(k)
return [[k, sorted(v)] for k, v in sorted(users.iteritems(), reverse=True)]
Method3:
def id_best_users(*args):
dct_m = set.intersection( *[ set( x for x in arg ) for arg in args ])
dct = { u:sum( arg.count(u) for arg in args) for u in dct_m }
dct_r = [ [x,sorted(u)] for x in set(dct.values()) for u in [[ y for y in dct.keys() if dct[y] == x ]] ]
return sorted(dct_r, reverse=True)
Method4:
from collections import Counter
def id_best_users(*months):
n = Counter([user for month in months for user in month])
c = {}
for user in set.intersection(*[set(month) for month in months]):
c[n[user]] = c.get(n[user], []) + [user]
return [[k, sorted(c[k])] for k in sorted([k for k in c], reverse=True)]
Method5:
def id_best_users(*args):
from collections import Counter
sum_counts = Counter(sum(args, []))
common_set = set.intersection(*(set(arg) for arg in args))
def is_common(user):
return user in common_set
common_users = [user for user, _ in sum_counts.most_common()
if is_common(user)]
from itertools import groupby
return [[count, sorted(users)] for count, users in
groupby(common_users, lambda user: sum_counts[user])]