Question

目的：如果并且当COUNTRIES的键与Country_gdps的值匹配时，比较两个字典并返回第三个字典，其键为country_gdps和值COUNTRIES，否则删除所有键来自country_gdps的值对。

给出词典摘要，此列表包含许多国家/地区，很长，仅以示例为例

country_gdps : {"Arab World": 2504702625568, "Caribbean small states": 66707362091,"Italy": 1858913163927, "Jamaica": 14056908749} 

COUNTRIES ={'ad': 'Andorra', 'ae': 'United Arab Emirates', 'af','it': 'Italy','jp': 'Japan'}

我尝试定义的函数，该函数仅适用于最后一个键-值对

def compare(dict1,dict2):
    for key1,value1 in dict1.items():
    for key2 ,value2 in dict2.items():
    if key1 == dict2[key2]:
    key1 = key2
    dict3 = {key1: value1}
    return dict3 

print(compare(country_gdps,COUNTRIES))

我想要一本新字典，当dict1的键与dict2的值匹配且不应包含dict1的键值对不匹配时，其内容应类似于result = {''it'：1858913163927}例如{“阿拉伯世界”：2504702625568，“加勒比小州”：66707362091}在新词典中不应该存在。

Answer 1

假设您要将country_gdps的密钥与COUNTRY的值进行匹配，则代码如下：

   country_gdps ={"Arab World": 2504702625568, "Caribbean small states": 66707362091,"Italy": 1858913163927, "Jamaica": 14056908749} 

COUNTRIES ={'ad': 'Andorra', 'ae': 'United Arab Emirates','it': 'Italy','jp': 'Japan'}


dict1={}
for i in country_gdps:
    for j in COUNTRIES :
        if i == COUNTRIES[j]:
            dict1[j]=country_gdps[i]

print(dict1)

输出：

{'it': 1858913163927}

这是您想要的还是其他要求，请告诉我

Answer 2

纠正您的COUNTRIES词典中的错误。

country_gdps = {"Arab World": 2504702625568, "Caribbean small states": 66707362091,"Italy": 1858913163927, "Jamaica": 14056908749} 
COUNTRIES ={'ad': 'Andorra', 'ae': 'United Arab Emirates','it': 'Italy','jp': 'Japan'}

def compare(d1,d2,d):
    for i in d1.values():
        for j in d2.keys():
            if i == j:
                d[i] = j
                return d
    else:
        country_gdps = {}
        return country_gdps
print(compare(COUNTRIES,country_gdps,{}))

输出将为

{'Italy': 'Italy'}

欢迎其他任何疑问。

比较两个字典并返回新字典

2 个答案: