Question

是否有可能获得导致IndexError异常的索引？

示例代码：

arr = [0, 2, 3, 4, 5, 6, 6]
try:  
   print arr[10] # This will cause IndexError
except IndexError as e:
    print e.args # Can I get which index (in this case 10) caused the exception?

Answer 1

没有直接的方法，因为与KeyError不同，IndexError尚未提供此信息。您可以使用您想要的参数将内置list子类化为IndexError：

class vist(list): # Verbose list
    def __getitem__(self, item):
        try:
            v = super().__getitem__(item) # Preserve default behavior
        except IndexError as e:
            raise IndexError(item, *e.args) # Construct IndexError with arguments

        return v

arr = [0, 2, 3, 4, 5, 6, 6] # list
arr = vist(arr) # vist

try:
    arr[10]
except IndexError as e:
    print(e.args) # (10, 'list index out of range')

实际上，您甚至不需要将其转换回正常list。

Answer 2

仅手动;例如：

arr = [1,2,3]
try:
    try_index = 42
    print(arr[try_index])
except IndexError:
    print 'Index', try_index, 'caused an IndexError'

Answer 3

除了手动跟踪您访问的索引外，我不相信，至少不是2.7。除非我误读了提案，there is a proposal for this in 3.5。

获取导致IndexError异常的索引

3 个答案: