Question

我的部分代码如下所示：

for i in [0,1]:
    ...
    print('{} used {} words that {} did not use.'.format(a[i], 50 , a[i+1]))

对于第一次迭代，我希望它能够做到，

print('{} used {} words that {} did not use.'.format(a[0], 50 , a[1]))

但是对于第二次迭代，我想要这个：

print('{} used {} words that {} did not use.'.format(a[1], 50 , a[0]))

如何做到这一点？

Answer 1

如果这正是您正在寻找的，您可以这样做：

for i in [0,1]:
    ...
    print('{} used {} words that {} did not use.'.format(a[i], 50 , a[(i+1)%2]))

Answer 2

您可以使用indice modulo 2（%2）：

a = ['first', 'second']

for idx in [0, 1]:
    print('{} used {} words that {} did not use.'.format(a[idx%2], 50 , a[(idx+1)%2]))

输出：

first used 50 words that second did not use.
second used 50 words that first did not use.

或者，如果只有两个项目：

可能更易于阅读和维护，如下所示：

a = ['first', 'second']
x, y = a
print('{} used {} words that {} did not use.'.format(x, 50 , y))
print('{} used {} words that {} did not use.'.format(y, 50 , x))

Answer 3

您可以使用modulous运算符%：

for i in [0,1]:
    ...
    print('{} used {} words that {} did not use.'.format(a[i % 2], 50 , a[(i + 1) % 2]))

在第一次迭代中，i = 0：

i % 2       == 0 % 2 == 0
(i + 1) % 2 == 1 % 2 == 1

在第二次迭代中，i = 1：

i % 2       == 1 % 2 == 1
(i + 1) % 2 == 2 % 2 == 0

请注意，此问题的特定实例的第一个i % 2 == i。

Answer 4

这是我的解决方案：

a = ['John','Doe']
amount = 50

# Use index to create strings to be formatted
s1 = '{0} used {2} words that {1} did not use.'
s2 = '{1} used {2} words that {0} did not use.'

print(s1.format(*a,amount))
print(s2.format(*a,amount))

返回：

John used 50 words that Doe did not use.
Doe used 50 words that John did not use.

或者：

# Use index to create strings to be formatted
s = '''\
{0} used {2} words that {1} did not use.
{1} used {2} words that {0} did not use.'''

print(s.format(*a,amount))

返回：

John used 50 words that Doe did not use.
Doe used 50 words that John did not use.

Python，每个循环交换变量

4 个答案:

输出：

或者，如果只有两个项目：