python - 错误信息 : argument of type 'type' is not iterable

Question

我刚开始使用 Python，我有一个简短的问题。

import sys

# Function creates a list [1,3,5..99]
def createlist():
    list = []
    for i in range(1, 100, 2):
        list.append(i)
    return(list)

# Function asks for an odd integer between 1..100
def numberinsert():
    n = int(input("Please enter an odd number between 1 and 100: "))
    if n in list:
        return(n)
    else:
        sys.exit("Number does not match requirements!")

def main():
    createlist()
    numberinsert()

它给了我这个: TypeError: argument of type 'type' is not iterable
我究竟做错了什么？

Answer 1

您正在尝试访问函数 createlist 的局部变量来自另一个函数 numberinsert并且这是不可能的(局部变量只能在它们定义的范围内访问)。
我建议您按如下方式更改代码:

import sys

# Function creates a list [1,3,5..99]
def createlist():
    odd_list = []
    for i in range(1, 100, 2):
        odd_list.append(i)
    return(odd_list)

# Function asks for an odd integer between 1..100
def numberinsert():
    num_list = createlist()
    n = int(input("Please enter an odd number between 1 and 100: "))
    if n in num_list:
        return(n)
    else:
        sys.exit("Number does not match requirements!")

def main():
    numberinsert()

作为旁注，尽量避免使用内置函数的名称(如 list )命名变量。

您还可以使用 range(1, 100, 2) 以更紧凑的方式编写代码。直接在您的 numberinsert功能:

# Function asks for an odd integer between 1..100
def numberinsert():
    n = int(input("Please enter an odd number between 1 and 100: "))
    if n in range(1, 100, 2):
        return(n)
    else:
        sys.exit("Number does not match requirements!")

def main():
    numberinsert()