[英]Find an element from a list in a dictionary and return that key
如果我將someDictionary
定義為
{'1': 33, '2': 44, '3': 55, '4': 10, '5': 66, '6': 60}
和someList
定義為
[51, 52, 53, 54, 55, 56]
如何在someDictionary
中找到與someDictionary
中的元素匹配的someList
? 現在,我假設不會有超過一場比賽。
我認為應該是這樣的:
for k, v in someDictionary.items():
if v == someList:
myAnswer = k
鑒於以上代碼, myAnswer
為3
第二部分我需要做的是,如果someList不包含在一個元素someDictionary
,找到在價值someDictionary
大於(但最接近)在someList的最后一個元素someList[-1]
在這種情況下,myAnswer為6
聽起來你想要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'
這允許在任一方向上進行O(1)查找。 現在,你可以遍歷someList
和檢查元素是否是in d.inv
有效。
第一部分,找到在列表中具有字典值的鍵的列表。
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,如果沒有從第一部分得到的結果,則找到具有最接近的較大值的鍵(續):
if not res:
res = min ([(value, key) for key, value in someDictionary.items() if value > max(someList) ])[1]
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
首先使用列表lst
收集所有值在lst
鍵,如果沒有匹配項,則過濾值大於lst[-1]
。 根據鍵值和lst
最后一項之差的abs值對它們進行排序后,取0
索引,最接近的項。
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
如何在
someDictionary
中找到與someDictionary
中的元素匹配的someList
?
字典用於將鍵映射到值。 在O(1)時間內不可能相反。 但是您可以在O( n )時間內進行迭代,直到可以將列表元素與字典值匹配為止。
為此,您可以使用簡單的for
循環,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
將這些函數重寫為1行生成器表達式,但這不一定會更有效。
第二部分我需要做的是,如果
someList
不包含在一個元素someDictionary
,在發現價值someDictionary
大於(但最接近)中最后一個元素someList
您可以使用for... else...
構造帶有min
的類似函數:
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'
對於第一個問題,您只是使用錯誤的運算符來檢查值是否在someList
for k, v in someDictionary.items():
if v in someList:
myAnswer = k
關於第二個問題,您可以通過這種方式擴展先前的代碼
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
如果您的字典和列表很大,尤其是如果您有很多列表要針對同一個字典進行測試,則值得准備數據以便更快地執行。
__init__
最糟糕的操作是排序,即O(n * log(n))。 它使我們可以使用bisect
在O(log(n))中最后一種情況下找到最接近的值。
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
這可以解決您的問題的兩個部分,並且應該易於修改以處理可能彈出的任何討厭的“邊緣情況”:
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)
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