I am getting an indexerror: list index out of range when I try to run this program with any sort of list.
Any ideas?
def binaryListSort(aList):
'''takes a list of binary numbers and puts them in ascending order
inputs: a list of binary integers, aList Outputs: a new list with numbers
in ascending order.'''
if aList[0] == 0:
return aList[0] + binaryListSort(aList[1:])
else:
return binaryListSort(aList[1:]) + aList[0]
This doesn't seem to be doing any kind of sorting at all. All you're doing is checking that the first element of the list is 0, then chopping off the first and checking again. If it's never 0, eventually it will try to check the first element of an empty list, hence the error.
In every recursive, you will decrease the list length by 1, so you will get a list empty which length is just 1, so when you use the aList[1:], it will throw the indexerror. and you should consider the aList is empty list.
you can give test with an example, and you can use the following code, it will work well.
def binaryListSort(aList):
'''takes a list of binary numbers and puts them in ascending order
inputs: a list of binary integers, aList Outputs: a new list with numbers
in ascending order.'''
if len(aList) == 0:
return
if len(aList) == 1：
return aList
if aList[0] == 0:
return [aList[0]] + binaryListSort(aList[1:])
else:
return binaryListSort(aList[1:]) + [aList[0]]
With this loop, you'll finally call binaryListSort with a list containing just 1 item. Then is aList[1] not defined and thus aList[1:] will cause an indexerror You need to insert code into your white line that will return aList if the length of aList is one.
if len(aList) == 1： return aList