Question

我有一个程序，该程序从用户处获取输入，并使用Population()函数显示输入的多种变化。 store_fit函数将这些不同的变体添加到列表中，然后将其删除，因此列表一次只能填充一个变体。

我希望能够从列表中获取变体并使用它来更新我的文字。但是，我的程序仅在Population函数完成后才更新文本。如何运行Population函数并同时更新文本？

代码：

fit = []
...

def store_fit(fittest): # fittest is each variation from Population
    clear.fit()
    fit.append(fittest)
...

pg.init()
...
done = False

while not done:
...
    if event.key == pg.K_RETURN:
        print(text)
        target = text
        Population(1000) #1000 variations
        store_fit(value)
        # I want this to run at the same time as Population
        fittest = fit[0]
...
top_sentence = font.render(("test: " + fittest), 1, pg.Color('lightskyblue3'))
screen.blit(top_sentence, (400, 400))

Answer 1

我建议使Population成为生成器函数。参见The Python yield keyword explained：

def Populate(text, c):
    for i in range(c):

        # compute variation
        # [...]

        yield variation

创建一个迭代器并使用next()来检索循环中的下一个变量，因此您可以打印每个变量：

populate_iter = Populate(text, 1000)

final_variation = None
while not done:

    next_variation = next(populate_iter, None)
    if next_variation :
        final_variation = next_variation 

        # print current variation
        # [...]

    else:
        done = True

根据评论进行编辑：

为了简化我的问题，我没有提到Population是一类[...]

当然还有Populate can be a class。在这种情况下，您必须实现object.__iter__(self)方法。例如：

class Populate:
    def __init__(self, text, c):
        self.text = text
        self.c    = c

    def __iter__(self):
        for i in range(self.c):

            # compute variation
            # [...]

            yield variation

通过iter()创建一个迭代器。例如：

populate_iter = iter(Populate(text, 1000))

final_variation = None
while not done:

    next_variation = next(populate_iter, None)
    if next_variation :
        final_variation = next_variation 

        # print current variation
        # [...]

    else:
        done = True

通过调用两个函数Pygame实时更新文本

1 个答案: