我正在写一个函数,我需要分析一个字典并将其中的一个键及其所有值拉出并返回。从字典中返回特定的键和它的值
例词典:
{'P':[("Eight",1460, 225.0, 200.0, "fresco","Netherlands"),
("Six",1465,81.0, 127.1, "tempera", "Netherlands")],
'V':[("Four",1661, 148.0, 257.0,"oil paint", "Austria"),
("Two",1630, 91.0, 77.0, "oil paint","USA")],
'K':[("Five",1922,63.8,48.1,"watercolor","USA"),
("Seven",1950,61.0,61.0,"acrylic paint","USA"),
("Two",1965,81.3,100.3,"oil paint","United Kingdom")],
'C':[("Ten",1496,365.0,389.0,"tempera","Italy")],
'U':[("Nine",1203,182.0, 957.0,"egg tempera","Italy"),
("Twelve",1200,76.2,101.6,"egg tempera","France")]}
所以,如果我的函数被调用find_the_key,我一直在寻找“C”和它的值,那么就需要返回
find_the_key(dictionary2(),['C'])
return {'C': [('Ten', 1496, 365.0, 389.0, 'tempera','Italy')]}
如果需要找到'C'和'V'会返回
find_the_key(dictionary2(),['C','V'])
return {'C': [('Ten', 1496, 365.0, 389.0, 'tempera', 'Italy')], 'V': [('Four',
1661, 148.0, 257.0, 'oil paint', 'Austria'), ('Two', 1630, 91.0, 77.0, 'oil
paint', 'USA')]}
我已经做了一些关于从字典和我现在的代码可能很接近。
def find_the_key(dictionary,thekey):
for thekey in dictionary.keys():
if dictionary[thekey] == targetkey:
return key, targetkey
目前我得到的错误targetkey没有定义,虽然我认为我的逻辑是在正确的轨道上。有没有人有任何想法如何做到这一点?
如果密钥不存在于字典中,您想怎么做?只要忽略它,引发错误,或返回默认值? –
@ PM2Ring让 - 弗朗索瓦法布尔的解决方案做的是正确的事情,我认为它只是忽略它 – warrior4223
我相信这两个答案都可以处理问题,甚至不知道它:) –