Iterating over the keys of a python dictionary, when keys are integers, I get this error, "TypeError: argument of type 'int' is not iterable"

Question

我正在研究 employees 的 pay raise，特别是 ids。假设，我公司有 5 名员工。我已经在 employee_id_list 中展示了它们。我希望 python 接受我的意见，其中包括 employees 的特定 ids，我想提高工资，以及他们的 salary。然后，我从这些输入中创建 dictionary。现在我想遍历 employee_id_list，使其与 input ids 相匹配。如果匹配，我想接受相应的 value of key 即 salary 并提高工资。但是我收到一个错误。我已经搜索了 Whosebug 上的所有内容，但没有任何内容符合我的问题

employee_id_list = [27, 25, 98, 78, 66]
employee_dict = dict()
while True:
    x = input("Enter an key to continue and 'r' for result: ").lower()
    if x== 'r':
        break
    try:
        employee_id = int(input("Enter key the Employee id: ")) 
        salary = int(input(f"Enter the {employee_id}'s salary: "))
        employee_dict[employee_id] = salary
    except ValueError:
        print("Please Enter the Integers!")
        continue 
print(employee_dict)
for e_ids in employee_id_list:
    for key, value in employee_dict.items():
        if e_ids in employee_dict[key] :
            employee_dict[value] = 0.8*value + value
print(employee_dict)

我遇到了这个错误

TypeError: argument of type 'int' is not iterable

Answer 1

是这个：

if e_ids in employee_dict[key] :

employee_dict 是 string-integer 对的字典，您正在尝试检查 e_ids 是否在 employee_dict[key] 中，这是一个 int，不像列表那样可迭代，您可以在列表中检查元素是否包含在其中。

另外，你不是说employee_dict[key] = 0.8*value + value吗？

Answer 2

您混淆了 for 循环的 key 和 value 变量

试试这个：

employee_id_list = [27, 25, 98, 78, 66]
employee_dict = dict()
while True:
    x = input("Enter an key to continue and 'r' for result: ").lower()
    if x == 'r':
        break
    try:
        employee_id = int(input("Enter key the Employee id: "))
        salary = int(input(f"Enter the {employee_id}'s salary: "))
        employee_dict[employee_id] = salary
    except ValueError:
        print("Please Enter the Integers!")
        continue
for e_ids in employee_id_list:
    for key, value in employee_dict.items():
        if e_ids in employee_dict:
            employee_dict[key] = 0.8*value + value
print(employee_dict)

你必须写 employee_dict[key] = 0.8*value+value 而不是 employee_dict[value] = 0.8*value+value。

您也可以写 employee_dict[key] = value*1.8 而不是 employee_dict[key] = 0.8*value+value。

Answer 3

把@Konny 和@Sunderam 的正确思路结合起来，加上我自己的改变来检查if语句，答案是：

employee_id_list = [27, 25, 98, 78, 66]
    employee_dict = dict()

    while True:
        x = input("Enter an key to continue and 'r' for result: ").lower()
        if x== 'r':
            break
        try:
            employee_id = int(input("Enter key the Employee id: "))
            salary = int(input(f"Enter the {employee_id}'s salary: "))
            employee_dict[employee_id] = salary
        except ValueError:
            print("Please Enter the Integers!")
            continue

    print(employee_dict)
    for e_ids in employee_id_list:
        for key, value in employee_dict.items():
            if e_ids == key:    # This is my contribution
                employee_dict[key] = 1.8*value
    print(employee_dict)

Iterating over the keys of a python dictionary, when keys are integers, I get this error, "TypeError: argument of type 'int' is not iterable"

Iterating over the keys of a python dictionary, when keys are integers, I get this error, "TypeError: argument of type 'int' is not iterable"

python

django

list

dictionary

while-loop