识别包含特定键值对的字典的“名称”的最快方法是什么？

Question

I'd like to identify the dictionary within the following list that contains the key-value pair 'Keya':'123a', which is ID1 in this case.

lst = {ID1:{'Keya':'123a','Keyb':456,'Keyc':789},ID2:{'Keya':'132a','Keyb':654,'Keyc':987},ID3:{'Keya':'5433a','Keyb':222,'Keyc':333},ID4:{'Keya':'444a','Keyb':777,'Keyc':666}}

It's safe to assume all dictionaries have the same key's, but have different values.

I currently have the following to identify which dictionary has the value '123a' for the key 'Keya', but is there a shorter and faster way?

    DictionaryNames = map(lambda Dict: str(Dict),lst)
    Dictionaries = [i[1] for i in lst.items()]
    Dictionaries = map(lambda Dict: str(Dict),Dictionaries) 

    Dict = filter(lambda item:'123a' in item,Dictionaries)
    val = DictionaryNames[Dictionaries.index(Dict[0])]
    return val

Answer 1

If you actually had a list of dictionaries, this would be:

next(d for d in list_o_dicts if d[key]==value)

Since you actually have a dictionary of dictionaries, and you want the key associated with the dictionary, it's:

next(k for k, d in dict_o_dicts.items() if d[key]==value)

This returns the first matching value. If you're absolutely sure there is exactly one, or if you don't care which you get if there are more than one, and if you're happy with a StopIteration exception if you were wrong and there isn't one, that's exactly what you want.

If you need all matching values, just do the same with a list comprehension:

[k for k, d in dict_o_dicts.items() if d[key]==value]

That list can of course have 0, 1, or 17 values.

Answer 2

You can just do [name for name, d in lst.iteritems() if d['Keya']=='123a'] to get a list of all the dictionaries in lst that have that value for that key. If you know there is only one, you can get it with [name for name, d in lst.iteritems() if d['Keya']=='123a'][0] . (As Andy mentions in a comment, your name lst is misleading, since lst is actually a dictionary of dictionaries, not a list.)

Answer 3

Å different way to do this is to use the appropriate data structure: Keep a "reverse map" of key-value pairs to names. If your dictionary of dictionaries is static after being built, you can build the reverse dictionary like this:

revdict = {(key, value): name
           for name, subdict in dictodicts.items()
           for key, value in subdict.items()}

If not, you just need to add revdict[key, value] = name for each d[name][key] = value statement and build them up in parallel.

Either way, to find the name of the dict that maps key to value , it's just:

revdict[key, value]

For (a whole lot) more information (than you actually want), and some sample code for wrapping things up in different ways… I dug up an unfinished blog post, considered editing it, and decided to not bother and just clicked Publish instead, so: Reverse dictionary lookup and more, on beyond z .

Answer 4

Since you want the fastest, you should short-cut your search as soon as you find the data you are after. Iterating through the whole list is not necessary, nor is producing any temporary dictionary:

for key,data in lst.iteritems():
    if data['Keya']=='132a':
        return key   #or break is not in a function