Question

所以我有一个列表，列出了人们每次玩游戏是否赢，输还是输：

scores = [["win","lose","win"],["win","win"],["draw","win"],["lose"]]

我想为每个子列表计数'win' / 'lose' / 'draw'

我可以使用字典吗？

例如dict= {[win:2, draw:0, lose:1],[win:2, draw:0, lose:0].....}

我尝试通过执行以下操作来计数并放置在列表中：

countlose=0
for sublist in scores:
    for item in sublist:
        for item in range(len(sublist)):
            if item=="lose":
                countlose+=1
print(countlose)

但这只是返回0

让我知道您将如何解决问题

Answer 1

您想要的结果不是有效的语法。您最有可能需要字典列表。

collections.Counter仅对可迭代值进行计数；除非您提供其他逻辑，否则它不计算外部提供的密钥。

在这种情况下，您可以将列表理解与空字典结合使用：

from collections import Counter

scores = [["win","lose","win"],["win","win"],["draw","win"],["lose"]]

empty = dict.fromkeys(('win', 'lose', 'draw'), 0)

res = [{**empty, **Counter(i)} for i in scores]

[{'draw': 0, 'lose': 1, 'win': 2},
 {'draw': 0, 'lose': 0, 'win': 2},
 {'draw': 1, 'lose': 0, 'win': 1},
 {'draw': 0, 'lose': 1, 'win': 0}]

Answer 2

我可以使用字典吗？

您可以使用collections.Counter（它是dict的子类）来计算可哈希对象：

>>> from collections import Counter
>>> scores = [['win', 'lose', 'win'], ['win', 'win'], ['draw', 'win'], ['lose']]
>>> counts = [Counter(score) for score in scores]
>>> counts
[Counter({'win': 2, 'lose': 1}), Counter({'win': 2}), Counter({'draw': 1, 'win': 1}), Counter({'lose': 1})]

要为丢失的键添加零计数，可以使用其他循环：

>>> for c in counts:
...     for k in ('win', 'lose', 'draw'):
...         c[k] = c.get(k, 0)
... 
>>> counts
[Counter({'win': 2, 'lose': 1, 'draw': 0}), Counter({'win': 2, 'lose': 0, 'draw': 0}), Counter({'draw': 1, 'win': 1, 'lose': 0}), Counter({'lose': 1, 'win': 0, 'draw': 0})]

或者，您可以用collections.defaultdict包装计数器：

>>> counts = [defaultdict(int, Counter(score)) for score in scores]
>>> counts
[defaultdict(<class 'int'>, {'win': 2, 'lose': 1}), defaultdict(<class 'int'>, {'win': 2}), defaultdict(<class 'int'>, {'draw': 1, 'win': 1}), defaultdict(<class 'int'>, {'lose': 1})]
>>> counts[0]['draw']
0

Answer 3

您可以为给定列表中的每个sublist应用一个列表理解。

还要声明您自己的counter函数，该函数计算win，lose或draw中一项的外观。 / p>

scores = [["win","lose","win"],["win","win"],["draw","win"],["lose"]]
def get_number(sublist):
  counter = {'win': 0, 'draw' : 0, 'lose': 0}
  for item in sublist:
    counter[item] += 1
  return counter

result = [get_number(sublist) for sublist in scores]

输出

[{'win': 2, 'draw': 0, 'lose': 1}, 
 {'win': 2, 'draw': 0, 'lose': 0}, 
 {'win': 1, 'draw': 1, 'lose': 0}, 
 {'win': 0, 'draw': 0, 'lose': 1}]

计算项目在每个嵌套列表中出现的次数

3 个答案: