python在列表中计数重复答案

【问题标题】：python count duplicate in listpython在列表中计数重复
【发布时间】：2014-12-03 23:51:52
【问题描述】：

我有这个清单：

['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 count、append 或 set 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：我怎样才能把这个列表打印成这样：

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

【问题讨论】：

您要删除还是只计算每个出现的次数？
我希望它把名字放在列表中的多少次。就像我举的例子。只是它不使用count、append或set方法
如果不使用特定功能的原因是作业，您很可能可以从讲座中推断出要使用什么。通常，您将排序然后从第一个到最后一个，每当上一个/当前不同时，您打开一个新的“组”并打印出前一个“组”的计数。

标签： python count duplicates

【解决方案1】：

要计算列表中每个条目的数量，您可以使用collections 模块中的Counter 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 Counter 对象，它保存列表中每个不同条目/键的出现次数。由于Counter 派生自dict，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})

【讨论】：

这里的反对票是怎么回事？这正是OP所要求的？！ Or what i really want is: how can i turn that list to print out like this 好吧，至少在他添加无导入部分之前...

【解决方案2】：

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']

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

完全不使用list.count：

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

你没有说你是否可以使用字典：

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

【讨论】：

您能解释一下您的代码的以下部分吗？ [ele for ind, ele in enumerate(l,1) if ele not in l[ind:]]
实际上，您介意解释一下您给出的字典示例的具体方式吗？我想对它有更深入的了解。
@ea87, dict 只是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 只会增加我们 dict 中每个团队的计数所以最后我们得到每个团队的频率/计数，如果我们不限于没有导入，我只会使用 collections.Counter dict

【解决方案3】：

players = ['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']

players_details, players_name = [], []
for each_player in players:
    if not(each_player in players_name):
        players_name = players_name + [each_player]
        players_details = players_details + [[each_player, 1]]
    else:
        for index in range(len(players_details)):
            if players_details[index][0] == each_player:
                players_details[index][1] = players_details[index][1]+1

for each in players_details:
    print '{} : {}'.format(*each)

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

【讨论】：

【解决方案4】：

我使用了这个代码：

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： output of the code

希望对你有帮助。

【讨论】：

【解决方案5】：

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}

【讨论】：

【解决方案6】：

你可以创建新的列表，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)

给予：

['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']

【讨论】：

不是问题How can i remove duplicate from this list without using the count, append or the set method or imports?
@helloV 不再有append。谢谢。我错过了。

【解决方案7】：

list=['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']
list1=[]
list2=[]
for x in list:
    if not x in list1:
        list1.append(x)
    if x in list1:
        list2.append(x)
list2.sort()
for num,og in enumerate(list2,1):
    print (num,og)

【讨论】：