knzhou knzhou - 1 year ago 102
Python Question

Quickly convert numpy arrays with index to dict of numpy arrays keyed on that index

I have a set of numpy arrays. One of these is a list of "keys", and I'd like to rearrange the arrays into a dict of arrays keyed on that key. My current code is:

for key, val1, val2 in itertools.izip(keys, vals1, vals2):

This is pretty slow, since the arrays involved are millions of entries long, and this happens many times. Is it possible to rewrite this in vectorized form? The set of possible keys is known ahead of time, and there are ~10 distinct keys.

Edit: If there are k distinct keys and the list is n long, the current answers are O(nk) (iterate once for each key) and O(n log n) (sort first). I'm still looking for an O(n) vectorized solution, though. This is hopefully possible; after all, the easiest possible nonvectorized thing (i.e. what I already have) is O(n).

Answer Source

Some timings:

import numpy as np
import itertools

def john1024(keys, v1, v2):
  d1 = {}; d2 = {};
  for k in set(keys):
    d1[k] = v1[k==keys]
    d2[k] = v2[k==keys]
  return d1,d2

def birico(keys, v1, v2):
  order = keys.argsort()
  keys_sorted = keys[order]
  diff = np.ones(keys_sorted.shape, dtype=bool)
  diff[1:] = keys_sorted[1:] != keys_sorted[:-1]
  key_change = diff.nonzero()[0]
  uniq_keys = keys_sorted[key_change]
  v1_split = np.split(v1[order], key_change[1:])
  d1 = dict(zip(uniq_keys, v1_split))
  v2_split = np.split(v2[order], key_change[1:])
  d2 = dict(zip(uniq_keys, v2_split))
  return d1,d2

def knzhou(keys, v1, v2):
  d1 = {k:[] for k in np.unique(keys)}
  d2 = {k:[] for k in np.unique(keys)}
  for key, val1, val2 in itertools.izip(keys, v1, v2):
  return d1,d2

I used 10 keys, 20 million entries:

import timeit

keys = np.random.randint(0, 10, size=20000000) #10 keys, 20M entries
vals1 = np.random.random(keys.shape)
vals2 = np.random.random(keys.shape)

timeit.timeit("john1024(keys, vals1, vals2)", "from __main__ import john1024, keys, vals1, vals2", number=3)
timeit.timeit("birico(keys, vals1, vals2)", "from __main__ import birico, keys, vals1, vals2", number=3)
timeit.timeit("knzhou(keys, vals1, vals2)", "from __main__ import knzhou, keys, vals1, vals2", number=3)

So, we see than the sorting technique is a bit faster than letting Numpy find the indices corresponding to each key, but of course both are much much faster than looping in Python. Vectorization is great!

This is on Python 2.7.12, Numpy 1.9.2

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