将列表值与字典中的项目进行比较

Question

Let's say I have a dictionary of all states abbreviations as a key, and the long names as the value:

statesDict = {'AK': 'Alaska', 'AL': 'Alabama', 'AR': 'Arkansas',...}

I also have a list of pre-selected state abbreviations:

statesAbbrv = ['AL', 'CA', 'CO', 'DE']

Based on the items in the cleanStates list, I want to chose only the values (the long names) for the state abbreviations keys in the stateNames dictionary and place them in a new list; stateNames = [] So, the results of the comparison will look like this:

stateNames = ['Alabama', 'California', 'Colorado', 'Deleware']

I was thinking the following, but it's not working. What am I doing wrong here?

stateNames = []
for i in statesAbbrv:
    for k, v in statesDict:
        if stateDict[k] == stateAbbrv[i]:
            stateNames.append(stateDict(k))
print stateNames

Answer 1

Functionally ¹ , you can use map ² with dict.get :

stateNames = list(map(statesDict.get, statesAbbrv))

If no match is found for an abbreviation, this will give None . A stricter version which will yield KeyError :

stateNames = list(map(statesDict.__getitem__, statesAbbrv))

The latter is akin to the list comprehension, as [] is syntax to access __getitem__ :

stateNames = [statesDict[i] for i in statesAbbrv]

If you wish to supply a fallback if a key is not found, use a list comprehension with dict.get :

stateNames = [statesDict.get(i, 'Fallback State') for i in statesAbbrv]

---

¹ Functional programming is a style of programming that treats computation as the evaluation of mathematical functions. See also Functional programming vs Object Oriented programming .

² In Python 2.x, explicit list conversion is not necessary since map returns a list . In Python 3, map returns an iterable, which will need to be exhausted via list .

Answer 2

This is a typical job for a list comprehension:

stateNames = [statesDict.get(state, state) for state in statesAbbrv]
print(stateNames)
#['Alabama', 'CA', 'CO', 'DE']

Note that if a state abbreviation for some reason is not in the dictionary, it will be used as the state name itself.

Answer 3

stateNames = [stateDict[k] for k in statesAbbrv if k in stateDict]

使用list comprehensions 。

Answer 4

遍历列表statesAbbrv和查找值statesDict

statesNames = [statesDict[n] for n in statesAbbrv]

Answer 5

stateNames = []
for i statesAbbrv:
    for k, v in statesDict.iteritems():
        if stateDict[k] == stateAbbrv[i]:
            stateNames.append(stateDict(k))
print stateNames

You need to add iteritems() to make the dictionary iterable on python 2.7.

Answer 6

Serious typo errors in your question. Well, here is the answer (Just after removing errors in your own code):

statesDict = {'AK': 'Alaska', 'AL': 'Alabama', 'AR': 'Arkansas'}
statesAbbrv = ['AL', 'AK']

stateNames = []
for i in statesAbbrv:
    for k in statesDict.keys():
        if k == i:
            stateNames.append(statesDict[k])
print (stateNames)

Answer 7

stateNames = [statesDict.get(i) for i in statesAbbrv]