python计数在列表中重复

我有以下清单：

['Boston Americans', 'New York Giants', 'Chicago White Sox', 'Chicago Cubs', 'Chicago Cubs', 'Pittsburgh Pirates', 'Philadelphia Athletics', 'Philadelphia Athletics', 'Boston Red Sox', 'Philadelphia Athletics', 'Boston Braves', 'Boston Red Sox', 'Boston Red Sox', 'Chicago White Sox', 'Boston Red Sox', 'Cincinnati Reds', 'Cleveland Indians', 'New York Giants', 'New York Giants', 'New York Yankees', 'Washington Senators', 'Pittsburgh Pirates', 'St. Louis Cardinals', 'New York Yankees', 'New York Yankees', 'Philadelphia Athletics', 'Philadelphia Athletics', 'St. Louis Cardinals', 'New York Yankees', 'New York Giants', 'St. Louis Cardinals', 'Detroit Tigers', 'New York Yankees', 'New York Yankees', 'New York Yankees', 'New York Yankees', 'Cincinnati Reds', 'New York Yankees', 'St. Louis Cardinals', 'New York Yankees', 'St. Louis Cardinals', 'Detroit Tigers', 'St. Louis Cardinals', 'New York Yankees', 'Cleveland Indians', 'New York Yankees', 'New York Yankees']

如何在不使用计数、追加或设置方法或导入的情况下从此列表中删除重复项

或者我真正想要的是：我怎样才能将列表打印成这样：

Boston Americans 5
New York Giants 2
team_name  number_of_duplicates
team_name  number_of_duplicates
team_name  number_of_duplicates

---

你可以列新的清单，例如

l = ['Boston Americans', 'New York Giants', 'Chicago White Sox', 'Chicago Cubs', 'Chicago Cubs', 'Pittsburgh Pirates', 'Philadelphia Athletics', 'Philadelphia Athletics', 'Boston Red Sox', 'Philadelphia Athletics', 'Boston Braves', 'Boston Red Sox', 'Boston Red Sox', 'Chicago White Sox', 'Boston Red Sox', 'Cincinnati Reds', 'Cleveland Indians', 'New York Giants', 'New York Giants', 'New York Yankees', 'Washington Senators', 'Pittsburgh Pirates', 'St. Louis Cardinals', 'New York Yankees', 'New York Yankees', 'Philadelphia Athletics', 'Philadelphia Athletics', 'St. Louis Cardinals', 'New York Yankees', 'New York Giants', 'St. Louis Cardinals', 'Detroit Tigers', 'New York Yankees', 'New York Yankees', 'New York Yankees', 'New York Yankees', 'Cincinnati Reds', 'New York Yankees', 'St. Louis Cardinals', 'New York Yankees', 'St. Louis Cardinals', 'Detroit Tigers', 'St. Louis Cardinals', 'New York Yankees', 'Cleveland Indians', 'New York Yankees', 'New York Yankees']
l2 = []
for v in l:
    if v not in l2:
        l2 = l2 + [v]

print(l2)

给出：

['Boston Americans', 'New York Giants', 'Chicago White Sox', 'Chicago Cubs', 'Pittsburgh Pirates', 'Philadelphia Athletics', 'Boston Red Sox', 'Boston Braves', 'Cincinnati Reds', 'Cleveland Indians', 'New York Yankees', 'Washington Senators', 'St. Louis Cardinals', 'Detroit Tigers']

不使用

列表。计数

：

for team in [ele for ind, ele in enumerate(l,1) if ele not in l[ind:]]:
    count = 0
    for ele in l:
        if team == ele:
            count += 1
    print("{} {}".format(team,count))
    count = 0

Boston Americans 1
Chicago Cubs 2
Boston Braves 1
Chicago White Sox 2
Boston Red Sox 4
Washington Senators 1
Pittsburgh Pirates 2
Philadelphia Athletics 5
New York Giants 4
Cincinnati Reds 2
Detroit Tigers 2
St. Louis Cardinals 6
Cleveland Indians 2
New York Yankees 13

您没有说您是否可以使用dict：

d = {}

for team in l:
    # if we have not seen team before, create k/v pairing
    # setting value to 0, if team already in dict this does nothing
    d.setdefault(team,0)
    # increase the count for the team
    d[team] += 1
for team, count in d.items():
    print("{} {}".format(team,count))

Chicago White Sox 2
New York Giants 4
Cincinnati Reds 2
Boston Red Sox 4
New York Yankees 13
Philadelphia Athletics 5
Pittsburgh Pirates 2
St. Louis Cardinals 6
Washington Senators 1
Boston Braves 1
Boston Americans 1
Cleveland Indians 2
Detroit Tigers 2
Chicago Cubs 2

结果:

Boston Americans : 1
New York Giants : 4
Chicago White Sox : 2
Chicago Cubs : 2
Pittsburgh Pirates : 2
Philadelphia Athletics : 5
Boston Red Sox : 4
Boston Braves : 1
Cincinnati Reds : 2
Cleveland Indians : 2
New York Yankees : 13
Washington Senators : 1
St. Louis Cardinals : 6
Detroit Tigers : 2

---

要计算列表中每个条目的数量，可以使用

集合

模块中的

计数器

类：

l =['Boston Americans', 'New York Giants', 'Chicago White Sox', 'Chicago Cubs', 'Chicago Cubs', 'Pittsburgh Pirates', 'Philadelphia Athletics', 'Philadelphia Athletics', 'Boston Red Sox', 'Philadelphia Athletics', 'Boston Braves', 'Boston Red Sox', 'Boston Red Sox', 'Chicago White Sox', 'Boston Red Sox', 'Cincinnati Reds', 'Cleveland Indians', 'New York Giants', 'New York Giants', 'New York Yankees', 'Washington Senators', 'Pittsburgh Pirates', 'St. Louis Cardinals', 'New York Yankees', 'New York Yankees', 'Philadelphia Athletics', 'Philadelphia Athletics', 'St. Louis Cardinals', 'New York Yankees', 'New York Giants', 'St. Louis Cardinals', 'Detroit Tigers', 'New York Yankees', 'New York Yankees', 'New York Yankees', 'New York Yankees', 'Cincinnati Reds', 'New York Yankees', 'St. Louis Cardinals', 'New York Yankees', 'St. Louis Cardinals', 'Detroit Tigers', 'St. Louis Cardinals', 'New York Yankees', 'Cleveland Indians', 'New York Yankees', 'New York Yankees']

from collections import Counter
c = Counter(l) 
print(c)

c

是一个

计数器

对象，它保存每个不同条目/键在列表中出现的次数。由于

计数器

是从

dict

派生出来的，所以您可以像访问任何其他字典一样访问它

Counter({'New York Yankees': 13, 'St. Louis Cardinals': 6, 'Philadelphia Athletics': 5, 'New York Giants': 4, 'Boston Red Sox': 4, 'Chicago White Sox': 2, 'Pittsburgh Pirates': 2, 'Detroit Tigers': 2, 'Cincinnati Reds': 2, 'Cleveland Indians': 2, 'Chicago Cubs': 2, 'Boston Americans': 1, 'Boston Braves': 1, 'Washington Senators': 1})

我使用了以下代码：

from collections import Counter
a=input().split()
print(a)
c=Counter(a) 
for i in c:
    print(str(i),"appears", c[i],"times")

它产生了这样的结果：

希望有帮助。

试试这个：

new_list =  ['a', 'b', 'a']
new_dict = {}
for i in new_list:
 new_dict[i]=new_list.count(i)       
print(new_dict)

Result - {'a': 2, 'b': 1}

---

您想删除还是只计算每次出现的次数？我想让它输入名称，然后该名称在列表中出现了多少次。就像我举的例子。只是它不使用count、append或set方法如果不使用特定函数的原因是一个赋值，那么您很可能可以从讲座中推断出要使用什么。通常，您会进行排序，然后从第一个到最后一个，每当上一个/当前不同时，您就会打开一个新的“组”并打印出前一个“组”的计数。问题不在于

如何在不使用计数、追加或set方法或导入的情况下从该列表中删除重复项？

@helloV No

append

。谢谢。我错过了。这里的否决票是怎么回事？这正是OP想要的

或者我真正想要的是：我怎样才能把这个列表像这样打印出来呢

好吧，至少在他添加了禁止导入的部分之前……你能解释一下你代码的以下部分吗

[ele for ind，ele in enumerate（l，1）如果ele不在l[ind:]

实际上，您是否介意解释一下您给出的字典示例是如何实现的？我想对它有更深入的了解。@ea87，dict只使用每个团队名称作为键，并在每次遇到新团队时将值设置为0，每个+=1只会增加dict中每个团队的计数，因此最后我们会得到每个团队的频率/计数，如果我们不局限于没有进口，我只会使用一个集合

Counter({'New York Yankees': 13, 'St. Louis Cardinals': 6, 'Philadelphia Athletics': 5, 'New York Giants': 4, 'Boston Red Sox': 4, 'Chicago White Sox': 2, 'Pittsburgh Pirates': 2, 'Detroit Tigers': 2, 'Cincinnati Reds': 2, 'Cleveland Indians': 2, 'Chicago Cubs': 2, 'Boston Americans': 1, 'Boston Braves': 1, 'Washington Senators': 1})

from collections import Counter
a=input().split()
print(a)
c=Counter(a) 
for i in c:
    print(str(i),"appears", c[i],"times")

new_list =  ['a', 'b', 'a']
new_dict = {}
for i in new_list:
 new_dict[i]=new_list.count(i)       
print(new_dict)

Result - {'a': 2, 'b': 1}