Question

不确定我在做什么错。我以为zip方法可以检查一个值是否在多个列表中？我希望它执行的操作是检查该值是否在那些词典中的任何一个中，如果是，则打印其键，但如果不是，则仅打印一个字符串（“不在任何词典中”）。此方法出于某些原因会用实际字典打印其中的40个。

MLB_Teams = {1: 'New York Yankees', 2: 'Pittsburgh Pirates'}
NBA_Teams = {1: 'Houston Rockets', 2: 'Brooklyn Nets'}
NFL_Teams = {1: 'Philadelphia Eagles', 2: 'Detroit Lions'}

for (key,value), (key,value), (key, value) in zip(MLB_Teams.items(), NBA_Teams.items(), NFL_Teams.items()):
    reply = 'Houston Rockets'
    if reply == value:
        print(key)
    else:
        print('Not In Any Dictionary')

Answer 1

我认为您可以通过一种非常简单的方式做到这一点：

MLB_Teams = {1: 'New York Yankees', 2: 'Pittsburgh Pirates'}
NBA_Teams = {1: 'Houston Rockets', 2: 'Brooklyn Nets'}
NFL_Teams = {1: 'Philadelphia Eagles', 2: 'Detroit Lions'}

v = 'Philadelphia Eagles'

def find_in_dict(val, d):
    for k, v in d.items():
        if v == val:
            print(k)
            return True

for dd in (MLB_Teams, NBA_Teams, NFL_Teams):
    if find_in_dict(v, dd):
        break
else:
    print('Not In Any Dictionary')

Answer 2

问题在于您如何将变量名重用于键和值。添加打印语句以查看效果。

for (key,value), (key,value), (key, value) in zip(MLB_Teams.items(), NBA_Teams.items(), NFL_Teams.items()):
    print(value) #added
    reply = 'Houston Rockets'
    if reply == value:
        print(key)
    else:
        print('Not In Any Dictionary')
#output
Philadelphia Eagles
Not In Any Dictionary
Detroit Lions
Not In Any Dictionary

将变量键和值重新分配给元组中的最后一个条目。

如果稍后再处理包装，则可以使用zip。

MLB_Teams = {1: 'New York Yankees', 2: 'Pittsburgh Pirates'}
NBA_Teams = {1: 'Houston Rockets', 2: 'Brooklyn Nets'}
NFL_Teams = {1: 'Philadelphia Eagles', 2: 'Detroit Lions'}
reply = 'Houston Rockets'
for tups in zip(MLB_Teams.items(), NBA_Teams.items(), NFL_Teams.items()):
    if any(reply == val for key,val in tups):
        print(tups[0][0]) #key
    else:
        print('Not In Any Dictionary')

#output
1
Not In Any Dictionary

Python：如何迭代两个以上的词典？

2 个答案: