splitting list into evenly size groups

Question

i tried to split list into groups with three elements. i found a code to do it in the web;

def chunks(l, chunk_size):
    result = []
    chunk = []
    for item in l:
        chunk.append(item)
        if len(chunk) == chunk_size:
            result.append(chunk)
            chunk = []
    if chunk:
        result.append(chunk)
    return result

print(chunks([1, 2, 3, 4, 5], 2))

However i could not understand the function of second "chunk=[]" in the 8th row.

Can anyone help me to understand?

Ty

Answer 1

result.append(chunk) #adds a chunk to the result
chunk = [] # chunk is now an empty list

If you were able to figure out the rest of the code then it would be a good exercise to think what would happen if you ommit that line.

Answer 2

Not exactly.
Let me explain the "for item in l" loop step by step:

1st iteration:
# item = 1
chunk.append(item) # adds 1 to the chunk list
len(chunk) is 1 and chunk_size is 2 so we ignore the if condition body.

2nd iteration
# item = 2
# after appending: chunk = [1, 2]
len(chunk) equals chunk_size so the if body runs:
result.append(chunk) # now result = [ [1,2] ]
chunk = [] # now chunk is an empty list, we already added 1 and 2 to the result and we don't need those values anymore. If we remove that line the chunk will keep growing and we don't want that.

The algorithm can be summarized this way:

1- add items to the chunk until the chunk has length "chunk_size"
2- If the chunk has the correct size:
2.1- Add the chunk to the result
2.2- Create a new chunk
3- Start again

chunk = [] represents the step 2.2 of the algorithm.

Sorry for being late, i missed your reply😅

Answer 3

ty kevin. so i suppose by putting an empty list (chunk =[]) at the end of first part of the code, we ensure loop to continue. otherwise it ll stop after the first chunk. am i right?

Answer 4

it is really a great explanation. i understand the logic behind it now. ty a lot. 😀