4
\$\begingroup\$

As the title says, I want a function that takes a vector, as well as the number of bins, and splits the vector in that number of bins, with a minimum length of 1 for each bin.

def split_into_bins(nbin, vector):
    """
    Randomly split vector into nbin number of bins, each of random size 
    """
    permutation = list(np.random.permutation(vector))

    # Location of the splits
    splits = sorted(np.random.choice(range(1,len(vector)), nbin-1, replace=False))

    # Initializing empty bins
    bins = [[]]*nbin

    start = 0
    end = splits[0]

    for i in range(nbin):
        bins[i] = permutation[start:end]
        start = end
        try:
            end = splits[i+1]
        except IndexError:
            end = len(vector)

    return bins

I'm wondering if there's a cleaner way to split the vector into bins besides randomly selecting split locations. My method of splitting up the list given the location of the splits also seems pretty messy. Performance does matter, so I'm wondering if I should initialize the empty bins outside of the function.

I also don't want there to be any bias with regards to the size of the bins; they should all have the same size on average. I'm pretty sure this method isn't biased, however.

\$\endgroup\$
5
\$\begingroup\$

np.random.choice can just take an int, so that can simplify your code - no need to build up a range():

splits = sorted(1 + x 
    for x in np.random.choice(len(vector)-1, nbins-1, replace=True))

When you build up the bins, you don't need a try/except (you know up front what the size is). You just need to iterate over the actual splits:

bins = []

last = 0
for split in splits:
    bins.append(permutation[last:split])
    last = split
bins.append(permutation[split:]) 
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.