python - 从 try-except 创建函数

Question

根据此答案中提供的示例，我如何从以下位置创建函数：

from collections import Counter
s =  ['0', '0', '2', '1', '1', '0', '0', '0']
try:
    print(next(t[0] for t in Counter(s).most_common(2) if t[0] != '0'))
except StopIteration:
    print('0')

此代码不起作用：

def most_common_number(s):
    try:
        return next(t[0] for t in Counter(s).most_common(2) if t[0] != '0')
    except StopIteration:
        '0'

如果有可能在不尝试的情况下获得相同的结果 - 除了请告诉我

标签： python

Answer 1

您需要从except街区返回。

def most_common_number(s):
    try:
        return next(t[0] for t in Counter(s).most_common(2) if t[0] != '0')
    except StopIteration:
        return '0'