如何使用字典理解从已知值中获取键

Question

假设我有一个可以在字典中找到的变量state = "MD"：

stateabb = {'Alabama': 'AL','Alaska': 'AK','Arizona': 'AZ','Arkansas': 'AR','California': 'CA','Colorado': 'CO','Connecticut': 'CT','Delaware': 'DE','DistrictOfColumbia': 'DC','Florida': 'FL',
'Georgia': 'GA','Hawaii': 'HI','Idaho': 'ID','Illinois': 'IL','Indiana': 'IN','Iowa': 'IA','Kansas': 'KS','Kentucky': 'KY','Louisiana': 'LA','Maine': 'ME','Maryland': 'MD',
'Massachusetts': 'MA','Michigan': 'MI','Minnesota': 'MN','Mississippi': 'MS','Missouri': 'MO','Montana': 'MT','Nebraska': 'NE','Nevada': 'NV','NewHampshire': 'NH',
'NewJersey': 'NJ','NewMexico': 'NM','NewYork': 'NY','NorthCarolina': 'NC','NorthDakota': 'ND','Ohio': 'OH','Oklahoma': 'OK','Oregon': 'OR','Pennsylvania': 'PA',
'RhodeIsland': 'RI','SouthCarolina': 'SC','SouthDakota': 'SD','Tennessee': 'TN','Texas': 'TX','Utah': 'UT','Vermont': 'VT','Virginia': 'VA','Washington': 'WA',
'WestVirginia': 'WV','Wisconsin': 'WI','Wyoming': 'WY'}

我想要一个快速函数，它使用字典理解来根据 value['state'] 获取匹配的 key['stateName'] 并将其分配给一个空变量：

stateName = ""
def long_name(stAbbrev):
    stateName = [key: key for v[stAbbrev]in stateabb.items()]
    return stateName
stateName = long_name(state)

注意：这不太正确；我在这里做错了什么？

Answer 1

抱歉，答案很简单。我只是稍微调整并添加一个 if 语句：

def long_name(stAbbrev):
stateName = [k for k, v in stateabb.items() if v == stAbbrev]
return stateName

Answer 2

您对自己问题的回答是正确的，但不是最优的。每次你想通过缩写获取州名称时，你都会迭代 state/abbr 对，直到找到预期的缩写（时间复杂度为 O(n)，其中 n 是州的数量）。因为你只有一个州的缩写，所以很容易反转字典：

>>> stateabb = {'Alabama': 'AL','Alaska': 'AK','Arizona': 'AZ','Arkansas': 'AR','California': 'CA','Colorado': 'CO','Connecticut': 'CT','Delaware': 'DE','DistrictOfColumbia': 'DC','Florida': 'FL', 'Georgia': 'GA','Hawaii': 'HI','Idaho': 'ID','Illinois': 'IL','Indiana': 'IN','Iowa': 'IA','Kansas': 'KS','Kentucky': 'KY','Louisiana': 'LA','Maine': 'ME','Maryland': 'MD', 'Massachusetts': 'MA','Michigan': 'MI','Minnesota': 'MN','Mississippi': 'MS','Missouri': 'MO','Montana': 'MT','Nebraska': 'NE','Nevada': 'NV','NewHampshire': 'NH', 'NewJersey': 'NJ','NewMexico': 'NM','NewYork': 'NY','NorthCarolina': 'NC','NorthDakota': 'ND','Ohio': 'OH','Oklahoma': 'OK','Oregon': 'OR','Pennsylvania': 'PA', 'RhodeIsland': 'RI','SouthCarolina': 'SC','SouthDakota': 'SD','Tennessee': 'TN','Texas': 'TX','Utah': 'UT','Vermont': 'VT','Virginia': 'VA','Washington': 'WA', 'WestVirginia': 'WV','Wisconsin': 'WI','Wyoming': 'WY'}
>>> state_name_by_abb = {v: k for k, v in stateabb.items()}
>>> state_name_by_abb
{'AL': 'Alabama', 'AK': 'Alaska', 'AZ': 'Arizona', 'AR': 'Arkansas', 'CA': 'California', 'CO': 'Colorado', 'CT': 'Connecticut', 'DE': 'Delaware', 'DC': 'DistrictOfColumbia', 'FL': 'Florida', 'GA': 'Georgia', 'HI': 'Hawaii', 'ID': 'Idaho', 'IL': 'Illinois', 'IN': 'Indiana', 'IA': 'Iowa', 'KS': 'Kansas', 'KY': 'Kentucky', 'LA': 'Louisiana', 'ME': 'Maine', 'MD': 'Maryland', 'MA': 'Massachusetts', 'MI': 'Michigan', 'MN': 'Minnesota', 'MS': 'Mississippi', 'MO': 'Missouri', 'MT': 'Montana', 'NE': 'Nebraska', 'NV': 'Nevada', 'NH': 'NewHampshire', 'NJ': 'NewJersey', 'NM': 'NewMexico', 'NY': 'NewYork', 'NC': 'NorthCarolina', 'ND': 'NorthDakota', 'OH': 'Ohio', 'OK': 'Oklahoma', 'OR': 'Oregon', 'PA': 'Pennsylvania', 'RI': 'RhodeIsland', 'SC': 'SouthCarolina', 'SD': 'SouthDakota', 'TN': 'Tennessee', 'TX': 'Texas', 'UT': 'Utah', 'VT': 'Vermont', 'VA': 'Virginia', 'WA': 'Washington', 'WV': 'WestVirginia', 'WI': 'Wisconsin', 'WY': 'Wyoming'}

然后在摊销的 O(1) 时间内通过缩写找到任何州名称：

>>> state_name_by_abb["HI"]
'Hawaii'

抛开时间复杂度的考虑，这种方法（对我来说）似乎更容易理解。

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