Question

This is one of those embarassing pieces of code that simply needs a second pair of eyes to spot the “obvious” mistake. I am processing a list of lists, and for some reason, I am hitting a ‘List Index out of range’ erorr – and I can’t spot it despite peering at the screen for a while.

Here is the snippet:

def group_ldata(data, freq, normal_grouping = True):
    if freq > 1 and len(data) > (SOME_SANITY_FACTOR*freq):
        i, output, subset, lastpos = (0, [],[], len(data)-freq)
        if not normal_grouping:
            for i in range(lastpos):
                pass
        else:
            while True:
                subset = data[i:freq]
                #print subset
                firstrow = subset[0]
                lastrow = subset[-1]
                output.append((firstrow[0], firstrow[1], lastrow[2]))
                i += freq
                if i >= lastpos:
                    break
        return output
    else:
        return data

Here’s some sample data to boot:

>>> a =[]
>>> a.append(range(0,5))
>>> a.append(range(5,10))
>>> a.append(range(10,15))
>>> a.append(range(15,20))
>>> a.append(range(20,25))
>>> a.append(range(25,30))
>>> a.append(range(30,35))

Here is the result when I run it:

>>> b = group_ldata(a,2)
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9]]
[]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 11, in group_ldata
IndexError: list index out of range

Can anyone spot what is causing the error?

Answer 1

I am not sure what you are trying to do. But, the reason for the error is obvious.

In the beginning, you do i=0.
Then, later in the while True loop, you do i += freq. So, now i equals freq.

Now, when the while True loop runs again (2nd iteration), in the line subset = data[i:freq], subset will be equal to a list with lenght 0. And you are trying to access the first element of that. Makes sense?