问题
I\'m telling my program to print out line 53 of an output. Is this error telling me that there aren\'t that many lines and therefore can not print it out?
回答1:
If you have a list with 53 items, the last one is thelist[52] because indexing start at 0.
回答2:
Yes,
You are trying to access an element of the list that does not exist.
MyList = ["item1", "item2"]
print MyList[0] # Will work
print MyList[1] # Will Work
print MyList[2] # Will crash.
Have you got an off-by-one error?
回答3:
Yes. The sequence doesn't have the 54th item.
回答4:
The way Python indexing works is that it starts at 0, so the first number of your list would be [0]. You would have to print[52], as the starting index is 0 and
therefore line 53 is [52].
Subtract 1 from the value and you should be fine. :)
回答5:
That's right. 'list index out of range' most likely means you are referring to n-th element of the list, while the length of the list is smaller than n.
回答6:
Always keep in mind when you want to overcome this error, the default value of indexing and range starts from 0, so if total items is 100 then l[99] and range(99) will give you access up to the last element.
whenever you get this type of error please cross check with items that comes between/middle in range, and insure that their index is not last if you get output then you have made perfect error that mentioned above.
回答7:
If you read a list from text file, you may get the last empty line as a list element. You can get rid of it like this:
list.pop()
for i in list:
i[12]=....
来源：https://stackoverflow.com/questions/1098643/indexerror-list-index-out-of-range-and-python