从字典中的列表中查找元素，然后返回该键

Question

If I had someDictionary defined as

{'1': 33, '2': 44, '3': 55, '4': 10, '5': 66, '6': 60}

and someList defined as

[51, 52, 53, 54, 55, 56]

how can I find the key in someDictionary that matches an element in someList ? For now, I'm assuming there won't be more than one match.

I would think it would be something like this:

for k, v in someDictionary.items():
    if v == someList:
        myAnswer = k

Given the code above, myAnswer would be 3

The second part I need to do is, if someList does not contain an element that is in someDictionary , find the value in someDictionary larger than (but closest to) the last element in someList someList[-1]

In that case, myAnswer would be 6

Answer 1

It sounds like you want a bidict

>>> from bidict import bidict  # pip install bidict
>>> d = bidict({'1': 33, '2': 44, '3': 55, '4': 10, '5': 66, '6': 60})
>>> d['3'] 55
>>> d.inv[55] '3'

This allows O(1) lookups in either direction. Now, you can loop over someList and check if an element is in d.inv efficiently.

Answer 2

First part, Find the list of keys that have a dictionary value in list.

someDictionary = {'1': 33, '2': 44, '3': 55, '4': 10, '5': 66, '6': 60}
someList = [51, 52, 53, 54, 55, 56]

res = [key for key, value in someDictionary.items() if value in someList]

second part, if no result from first part, find the key with the closest larger value (continuing):

if not res:
    res = min ([(value, key) for key, value in someDictionary.items() if value > max(someList) ])[1]

Answer 3

myAnswer = ''
closest_to = someList[-1]
closest_diff = 0
for k in someDictionary:
    if someDictionary[k] in someList:
         myAnswer = k
         break; # stop the loop when exact match found
    diff = someDictionary[k] - closest_to
    if diff > 0 and diff < closest_diff:
         closest_diff = diff
         myAnswer = k # this would save key of larger but closest number to last in someList

Answer 4

First use list comprehension to gather all keys whose values are in lst , if there is no match then filter for keys whose value is larger than lst[-1] . After sort them based on the abs value of the difference of the keys values and the last item in lst and take the 0 index, closest item.

dct = {'1': 33, '2': 44, '3': 55, '4': 10, '5': 66, '6': 60}
lst = [51, 52, 53, 54, 55, 56]

my_answer = [int(i) for i in dct if dct[i] in lst]
if my_answer:
    print(*my_answer) # => 3
else:
    close = [i for i in dct if int(dct[i]) > max(lst)]
    closest = sorted(close, key=lambda x: abs(dct.get(x) - lst[-1]))[0]
    print(closest) # => 6

Answer 5

How can I find the key in someDictionary that matches an element in someList ?

A dictionary is used map keys to values. The reverse isn't possible in O(1) time. But you can iterate in O( n ) time until you can match a list element with a dictionary value.

You can use a simple for loop for this, either iterating your dictionary or list.

d = {'1': 33, '2': 44, '3': 55, '4': 10, '5': 66, '6': 60}
L = [51, 52, 53, 54, 55, 56]

def find_key_dict(d, L):
    L_set = set(L)  # convert to set for O(1) lookup
    for k, v in d.items():
        if v in L_set:
            return k

def find_key_list(d, L):
    d_rev = {v: k for k, v in d.items()}  # reverse dict for value -> key map
    for i in L:
        if i in d_rev:
            return d_rev[i]

find_key_dict(d, L)  # '3'
find_key_list(d, L)  # '3'

It's also possible to rewrite these functions as 1-line generator expressions with next , but this won't necessarily be more efficient.

The second part I need to do is, if someList does not contain an element that is in someDictionary , find the value in someDictionary larger than (but closest to) the last element in someList

You can write a similar function utilising a for... else... construct with min :

def find_key_dict(d, L):
    L_set = set(L)  # convert to set for O(1) lookup
    for k, v in d.items():
        if v in L_set:
            return k
    else:
        def min_func(x):
            diff = x[1] - L[-1]
            return diff <= 0, diff
        return min(d.items(), key=min_func)[0]

find_key_dict(d, L)  # '6'

Answer 6

For the first question, you are just using the wrong operator to check if the value is in someList

for k, v in someDictionary.items():
    if v in someList:
        myAnswer = k

About the second question you can extend the previous code in this way

for k, v in someDictionary.items():
     if v in someList:
         myAnswer = k
         break
else:
     myAnswer = None
     for k, v in someDictionary.items():
         if v > someList[-1] and (myAnswer is None or v < someDictionary[myAnswer]):
             myAnswer = k

Answer 7

If your dictionary and list are large, and especially if you have many lists to test against the same dictionary, it can be worth preparing your data in order to have a faster execution.

The worst operation in __init__ is sorting, which is O(n*log(n)). It allows us to use bisect to find the closest value in your last case in O(log(n)).

from bisect import bisect


class FindInDict:
    def __init__(self, someDictionary):
        self.dict_by_values = {val: key for key, val in someDictionary.items()}
        self.dict_values_set = set(sorted_values)
        self.sorted_values = sorted(someDictionary.values())

    def find(self, someList):
        common = set(someList).intersection(self.dict_values_set)
        if common:
            key = self.dict_by_values[common.pop()]
        else:
            closest_value = self.sorted_values[bisect(self.sorted_values, someList[-1])]
            key = self.dict_by_values[closest_value]
        return key



someDictionary = {'1': 33, '2': 44, '3': 55, '4': 10, '5': 66, '6': 60}    
finder = FindInDict(someDictionary)

finder.find([51, 52, 53, 54, 55, 56])
# 3

finder.find([51, 52, 53, 54, 56])  # no common value
# 6

Answer 8

Here's something that handles both parts of your question and should be easy to modify to handle any pesky "edge-cases" that might pop up:

someDictionary = {'1': 33, '2': 44, '3': 55, '4': 10, '5': 66, '6': 60}
someList = [51, 52, 53, 54, 55, 56]

myAnswer = [key for key in someDictionary if someDictionary[key] in someList]
if not myAnswer:
    diffs = {dict_key: dict_value-someList[-1]
                for dict_key, dict_value in someDictionary.items()
                    if dict_value-someList[-1] > 0}
    myAnswer = [min(diffs, key=diffs.get)]  # Key of value with minimum
                                            # (positive) difference
print(myAnswer)