我试试看：

{k: [int(v[0]), int(v[1])] for k, v in some_dict.items()}

我的意见是只需要迭代dict项，然后迭代values列表：

for key, value in some_dict.items():
    for i in range(len(value)):
        value[i] = int(value[i])

我想这会有用的

new_dict = {k: [int(i) for i in v] for k, v in some_dict.items()}

• • 答案：**

some_dict = {k: [int(vi) for vi in v] for k, v in some_dict.items()}

• • 测试**

测试代码：

some_dict = {"B1": [-1.0, 3.0], "B2": [-2.0, 4.0], "B3": [-3.0, 5.0], "B4": [-5, -6]}
print('some_dict = {}'.format(some_dict))

some_dict = {k: [int(vi) for vi in v] for k, v in some_dict.items()}
print('some_dict = {}'.format(some_dict))

输出：

some_dict = {'B1': [-1.0, 3.0], 'B2': [-2.0, 4.0], 'B3': [-3.0, 5.0], 'B4': [-5, -6]}
some_dict = {'B1': [-1, 3], 'B2': [-2, 4], 'B3': [-3, 5], 'B4': [-5, -6]}