Question

我正在尝试在Python中找到正确的方法来完成以下任务（不能按照书面形式工作）：

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不幸的是（至少对我而言），Python中不存在++运算符或语句，作为将变量递增1的方法，但使用

myList = ["a", "b", "c"]
myCounter = 5

for item in myList:
  print("""Really long text 
in which I need to put the next iteration of myCounter (""", myCounter++, """) followed 
by a lot more text with many line breaks
followed by the next iteration of myCounter (""", myCounter++, """) followed by even
more long text until finally we get to the next
iteration of the for loop.""", sep='')

当我想要打印变量并同时递增变量时，

在它的位置似乎不起作用。我希望它首次通过for循环打印5和6，然后在下一次打印时打印7和8，然后在最后一次打印时打印9和10。应该如何在Python 3中完成？

Answer 1

我可能会考虑使用itertools.count：

import itertools

myCounter = itertools.count(5)
for item in myList:
    print("la la la.", next(myCounter), "foo foo foo", next(myCounter))

如果您希望避免导入，您可以很容易地编写自己的生成器来执行此类操作：

def counter(val=0):
    while True:
        yield val
        val += 1

Answer 2

我只是在print语句中使用myCounter + 1和myCounter + 2，然后在它之外将myCounter增加2.示例 -

myList = ["a", "b", "c"]
myCounter = 5

for item in myList:
  print("""Really long text 
in which I need to put the next iteration of myCounter (""", myCounter + 1, """) followed 
by a lot more text with many line breaks
followed by the next iteration of myCounter (""", myCounter + 2, """) followed by even
more long text until finally we get to the next
iteration of the for loop.""", sep='')
  myCounter += 2

pythonista如何在Python 3中编写与增量运算符等效的代码？

2 个答案: