从字典中的列表中查找元素，然后返回该键

Question

如果我将someDictionary定义为

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

和someList定义为

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

如何在someDictionary中找到与someDictionary中的元素匹配的someList ？ 现在，我假设不会有超过一场比赛。

我认为应该是这样的：

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

鉴于以上代码， myAnswer为3

第二部分我需要做的是，如果someList不包含在一个元素someDictionary ，找到在价值someDictionary大于（但最接近）在someList的最后一个元素someList[-1]

在这种情况下，myAnswer为6

Answer 1

听起来你想要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'

这允许在任一方向上进行O（1）查找。 现在，你可以遍历someList和检查元素是否是in d.inv有效。

Answer 2

第一部分，找到在列表中具有字典值的键的列表。

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]

第二部分，如果没有从第一部分得到的结果，则找到具有最接近的较大值的键（续）：

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

首先使用列表lst收集所有值在lst键，如果没有匹配项，则过滤值大于lst[-1] 。 根据键值和lst最后一项之差的abs值对它们进行排序后，取0索引，最接近的项。

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

如何在someDictionary中找到与someDictionary中的元素匹配的someList ？

字典用于将键映射到值。 在O（1）时间内不可能相反。 但是您可以在O（ n ）时间内进行迭代，直到可以将列表元素与字典值匹配为止。

为此，您可以使用简单的for循环，迭代字典或列表。

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'

也可以使用next将这些函数重写为1行生成器表达式，但这不一定会更有效。

第二部分我需要做的是，如果someList不包含在一个元素someDictionary ，在发现价值someDictionary大于（但最接近）中最后一个元素someList

您可以使用for... else...构造带有min的类似函数：

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

对于第一个问题，您只是使用错误的运算符来检查值是否在someList

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

关于第二个问题，您可以通过这种方式扩展先前的代码

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

如果您的字典和列表很大，尤其是如果您有很多列表要针对同一个字典进行测试，则值得准备数据以便更快地执行。

__init__最糟糕的操作是排序，即O（n * log（n））。 它使我们可以使用bisect在O（log（n））中最后一种情况下找到最接近的值。

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

这可以解决您的问题的两个部分，并且应该易于修改以处理可能弹出的任何讨厌的“边缘情况”：

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)