**第一个解决方案：**您可以使用map执行此操作

dic1 = {'2012-03-06':['1','4','5'],'2012-03-12':['7','3','10']}

result_dict = {key: sum(map(int, value)) for key, value in dic1.items()}
print(result_dict)

输出量：

{'2012-03-06': 10, '2012-03-12': 20}

**第二种解决方案：**并轻松转换为预期输出

dic1 = {'2012-03-06':['1','3','5'],'2012-03-12':['7','3','10']}
for key, value in dic1.items():
    print(f"{key}, {sum(map(int, value))}")

输出量：

2012-03-06, 10
2012-03-12, 20

它可以

dic1 = dict() 
dic1 = {'2012-03-06':['1','4','5'],'2012-03-12':['7','3','10']} 
for i in dic1: 
    print(i+','+ str(sum(map(int, dic1[i]))))

在python3中，map(func,iter)返回将给定函数应用于给定可迭代对象的每一项之后的结果的迭代器（列表、元组等