Question

我有一个数字列表和一个包含名称（键）和数字（值）的字典我需要迭代名为lotto的列表，然后根据字典中的名称值检查它们，如果我得到匹配，则从字典值中删除该数字。

继承人到目前为止我所拥有的只是打印出原始字典键和值

players = {'ray': [1,2,3,6,7,8,], 'al':[1,2,3,4,8,9,]}
lotto =[1,2,3,4,5,6,]

for i in players.values():  
    if i in lotto:
       players.values.remove(i)  

print (players)

任何帮助appriciated

Answer 1

您正以错误的方式访问dict值。看看以下内容：

players = {'ray': [1,2,3,6,7,8,], 'al':[1,2,3,4,8,9,]}
lotto =[1,2,3,4,5,6,]

for p, val in players.items():  
    for num in lotto:
        if num in val:
            players[p].remove(num)

print (players)  # {'al': [8, 9], 'ray': [7, 8]}

如果列表值中的项目顺序对您不重要，您可以使用以下更快的变体：

players = {'ray': [1,2,3,6,7,8], 'al':[1,2,3,4,8,9]}
lotto = [1,2,3,4,5,6]
lotto = set(lotto)

for p, val in players.items():
    players[p] = list(set(val) - lotto)

print (players)  # {'ray': [8, 7], 'al': [8, 9]}

也可以在下面的单行中压缩（它重新创建dict而不是修改它）：

players = {k: list(set(val) - lotto) for k, v in players.items()}

Answer 2

您的问题是您尝试从错误的列表中删除您的号码。 players.values是每个键值的列表，而不是每个键的个别值。请改用每个密钥的单个值：

players = {'ray': [1,2,3,6,7,8,], 'al':[1,2,3,4,8,9,]}
lotto = [1,2,3,4,5,6,]

for key, value in players.items():
    for i in lotto:
        if i in value:
            value.remove(i)

print(players)

然后可以通过使用词典理解缩短：

players = {'ray': [1,2,3,6,7,8,], 'al':[1,2,3,4,8,9,]}
lotto =[1,2,3,4,5,6,]

players = {k: list(set(v) - set(lotto)) for k, v in players.items()}
print(players) # {'ray': [8, 7], 'al': [8, 9]}

此外，如果您的计划中的列表顺序无关紧要，我建议您改为使用set。它们提供了有效的元素查找，并使这样的事情变得非常简单：

>>> players = {'ray': {1,2,3,6,7,8,}, 'al':{1,2,3,4,8,9,}}
>>> lotto = {1,2,3,4,5,6,}
>>> {k: v - lotto for k, v in players.items()}
{'ray': {8, 7}, 'al': {8, 9}}
>>>

Answer 3

你可以这样过滤：

players = {'ray': [1,2,3,6,7,8,], 'al':[1,2,3,4,8,9,]}

lotto =[1,2,3,4,5,6]

new_players = {a:[i for i in b if i not in lotto] for a, b in players.items()}

输出：

{'al': [8, 9], 'ray': [7, 8]}

python迭代列表并删除在字典中的列表中找到的任何重复项并删除重复项

3 个答案: