IndexError: list index out of range and python

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. sequence doesn't have the 54th item.

回答4:

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5:

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.

keep coding...

回答6:

The Way Python Indexing Works Is That It Starts At 0. So Your First Number Of Your List Would Be [0]. You Would Have to Print[52] as the first one is 0 and line 53 is [52].

Minus 1 And You Should be Fine :)