如何在不中断循环的情况下返回值? [英] How to return values without breaking the loop?
问题描述
我想知道如何在不破坏Python循环的情况下返回值.
I want to know how to return values without breaking a loop in Python.
以下是示例:
def myfunction():
list = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
print(list)
total = 0
for i in list:
if total < 6:
return i #get first element then it breaks
total += 1
else:
break
myfunction()
return仅会得到第一个答案，然后离开循环，我不希望那样，我想返回多个元素直到循环结束.
return will only get the first answer then leave the loop, I don't want that, I want to return multiple elements till the end of that loop.
如何解决这个问题，有什么解决办法吗?
How can resolve this, is there any solution?
推荐答案
您可以创建生成器这样，您就可以从生成器中获取yield值(使用yield语句后，您的函数将成为生成器).
You can create a generator for that, so you could yield values from your generator (your function would become a generator after using the yield statement).
请参阅以下主题，以更好地了解如何使用它:
See the topics below to get a better idea of how to work with it:
- Generators
- What does the "yield" keyword do in Python?
- yield and Generators explain's
使用生成器的示例:
def simple_generator(n):
i = 0
while i < n:
yield i
i += 1
my_simple_gen = simple_generator(3) // Create a generator
first_return = my_simple_gen.next() // 0
second_return = my_simple_gen.next() // 1
此外，您可以在循环开始之前创建list并将append项目添加到该列表，然后返回该列表，因此可以将该列表视为循环内返回"的结果列表.
Also you could create a list before the loop starts and append items to that list, then return that list, so this list could be treated as list of results "returned" inside the loop.
使用列表返回值的示例:
Example of using list to return values:
def get_results_list(n):
results = []
i = 0
while i < n:
results.append(i)
i += 1
return results
first_return, second_return, third_return = get_results_list(3)
注意::在使用列表的方法中，您必须知道函数将在results列表中返回多少个值，以避免too many values to unpack错误
NOTE: In the approach with list you have to know how many values your function would return in results list to avoid too many values to unpack error
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