Question

在迭代器/生成器抛出的异常之后，Python中有没有办法继续迭代？就像下面的代码一样，有没有办法跳过ZeroDivisionError并继续循环gener()而没有modyfying run()函数？

def gener():
    a = [1,2,3,4,0, 5, 6,7, 8, 0, 9]
    for i in a:
        yield 2/i

def run():
    for i in gener():
        print i

#---- run script ----#

try:
    run()
except ZeroDivisionError:
    print 'what magick should i put here?'

Answer 1

try/except的合理位置将是违规计算发生的地方：

def gener():
    a = [1,2,3,4,0, 5, 6,7, 8, 0, 9]
    for i in a:
        try:
            yield 2/i
        except ZeroDivisionError:
            pass

Answer 2

一种可能的解决方案是将问题代码包装到try ... except block：

def gener():
    a = [1,2,3,4,0, 5, 6,7, 8, 0, 9]
    for i in a:
        try:
            div_result = 2/i
        except ZeroDivisionError:
            div_result = None

        yield div_result

Answer 3

我不确定，但如果你想了解发生错误的地方，也许这更适合你：

In [1]: def gener():
   ...:     a = [1, 2, 0, 3, 4, 5, 6, 7, 8, 9]
   ...:     errors = []
   ...:     for idx, i in enumerate(a):
   ...:         try:
   ...:             yield 2 / i
   ...:         except ZeroDivisionError:
   ...:             errors.append('ZeroDivisionError occured at idx = {}'.for
   ...: mat(idx))
   ...:     if errors:
   ...:         raise RuntimeWarning('\n'.join(errors))
   ...:

在python中的迭代器/生成器中异常提升后继续

3 个答案: