问题描述
我想将字典与函数内的精确模式进行比较。 我的最终目标是:
- 当两个字典具有相同的键和值时,将变量增加3。
- 如果值不相同,则将变量减-1。
- 如果不存在一对键/值,则为0
例如:
dict1 = {"EX1": "C","EX2": "D","EX4": "A","EX5": "A"}
dict2 = {"EX1": "C","EX2": "A","EX3": "A","EX5": "A"}
预期输出:
8
8,因为: -两个字典中的EX1键值相同(3)
-两个词典中的EX2键值都不相同(3-1 = 2)
-dict1中没有EX3键,因此没有任何操作(2)
-EX4键在两个字典中的值相同(2 + 3 = 5)
-EX5键在两个字典中具有相同的值(5 + 3 = 8)
我从互联网上获得了我不知道如何将其转换为函数的两个代码片段,我不知道这些代码片段是否可以帮助您
{k : dict1[k] for k in dict1 if k in rep_valid and dict1[k] == rep_valid[k]} #Get same items
{k : dict2[k] for k in set(dict2) - set(dict1)} #Get difference
解决方法
您可以使用列表理解来获得答案:
dict1 = {"EX1": "C","EX2": "D","EX4": "A","EX5": "A"}
dict2 = {"EX1": "C","EX2": "A","EX3": "A","EX5": "A"}
# (count matching keys)*3 - (count not matching keys)
ttl = len([k for k in dict1 if k in dict2 and dict1[k]==dict2[k]]) * 3 - len([k for k in dict1 if k in dict2 and dict1[k]!=dict2[k]])
print(ttl) # 8
,
您可以使用以下功能,该功能还支持嵌套字典:
def is_dict_equal(d1,d2):
# for each key in d1
for key in d1.keys():
# check if d2 also has the key
if key not in d2:
return False
value1 = d1.get[key]
value2 = d2.get[key]
if isinstance(value1,dict): # if the value is another dictionary call this function again
if not is_dict_equal(value1,value2):
return False
elif value1 != value2: # else just compare them with eachother
return False
# check for keys in d2 but not in d1
for key in d2.keys():
if key not in d1:
return False
return True
,
count = 0
for key in (set(dict1.keys()) & set(dict2.keys())):
if dict1.get(key) == dict2.get(key):
count += 3
else:
count -= 1
print(f"count = {count}")
#count = 8