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## Small programming challenge No. 6 – nblocks

I came up with this challenge when I had to write a function to divide a sequence to percentiles. I needed this to calculate some statistics over trips for plnnr.com. “This sounds trivial” I thought, and reached for my simple blocks function:

```def blocks(seq, block_len): """blocks(range(5),2) -&gt; [[0, 1], [2, 3], [4]]""" seq_len = len(seq) if seq_len%block_len == 0: num_blocks = seq_len/block_len else: num_blocks = 1 + (seq_len/block_len)   result = [[] for i in xrange(num_blocks)] for idx, obj in enumerate(seq): result[idx/block_len].append(obj) return result```

So I set block_len to len(seq)/10 and called blocks(seq, block_len). Unfortunately, according to the docs of blocks (which I wrote…), when there is extra data, a “partial” block is added – which is exactly what we don’t want when calculating percentiles.
Instead, the behavior we want is nblocks(seq, number_of_blocks), for example: nblocks(range(10), 3) -> [0, 1, 2], [3, 4, 5], [6, 7, 8, 9].

This is a surprisingly hard to write function, or rather, harder than you’d expect. I’ll be especially glad if someone writes it elegantly. My solution works well enough, but isn’t the prettiest.

So, you have the definition – let’s see if you can do better than me. Once enough solutions are presented, I will present my own.

IMPORTANT EDIT: the required signature is nblocks(seq, num_blocks). So for seq(range(10), 3), the return value should be 3 blocks, with the last one having an extra item. As a general rule, the extra items should be spread as evenly as possible.