编写一个for循环，该循环遍历to_ten并打印出数字是偶数还是奇数

Question

to_ten=['1','2','3','4','5','6','7','8','9','10']

for i in range(10):

    if i %2==0:
        i='Number is even'
    else: i='Number is odd'
print(i)

创建一个名为to_ten的列表，其中包含从1到10的数字。编写一个for循环，该循环遍历to_ten并打印出数字是偶数还是奇数。请帮助我理解如何为1-10列表中的每个数字获取关于它是偶数还是奇数的语句。

Answer 1

to_ten=['1','2','3','4','5','6','7','8','9','10']

for i in to_ten:
    if int(i)%2 == 0:
        print(f'{i} is even')
    else:
        print(f'{i} is odd')

输出：

1 is odd
2 is even
3 is odd
4 is even
5 is odd
6 is even
7 is odd
8 is even
9 is odd
10 is even

Answer 2

​

你的代码几乎是正确的，

但是，您需要缩进print语句，以便程序在每次迭代中打印数字。

并且，range( num )将从0开始，并且num将不包括在迭代中。

[1,2,3,4,5,6,7,8,9,10]相当于python中的range(1,11)。

for i in range(1,11):
    if i % 2 == 0: # If the number divided by two leaves no remainder:
        print(f'{i} is even')
    else:
        print(f'{i} is odd')

输出：

1 is odd
2 is even
3 is odd
4 is even
5 is odd
6 is even
7 is odd
8 is even
9 is odd
10 is even

Answer 3

您的代码有三个问题:首先，您将range()替换为to_ten，因此实际上没有实现所述的问题；您将to_ten设置为str的list，而不是int的list；您的print(i)语句的顺序和缩进错误。@biplobbiswas的解决方案很好地解决了所有这些小问题。(+1)

我们可以通过使to_ten成为int的list来简化问题，并让Python为我们生成它，以避免错误。我们还可以使用i % 2的结果作为索引，而不是使用一系列if语句显式地测试它：

to_ten = [*range(1, 11)]  # a list containing the numbers from 1-10

for number in to_ten:
    print("{} is {}".format(number, ['even', 'odd'][number % 2]))

有很多方法可以解决这样的问题。