How to use numpy to get the cumulative count by unique values in linear time?

Question

Consider the following lists short_list and long_list

short_list = list('aaabaaacaaadaaac')
np.random.seed([3,1415])
long_list = pd.DataFrame(
    np.random.choice(list(ascii_letters),
                     (10000, 2))
).sum(1).tolist()

How do I calculate the cumulative count by unique value?

I want to use numpy and do it in linear time. I want this to compare timings with my other methods. It may be easiest to illustrate with my first proposed solution

def pir1(l):
    s = pd.Series(l)
    return s.groupby(s).cumcount().tolist()

print(np.array(short_list))
print(pir1(short_list))

['a' 'a' 'a' 'b' 'a' 'a' 'a' 'c' 'a' 'a' 'a' 'd' 'a' 'a' 'a' 'c']
[0, 1, 2, 0, 3, 4, 5, 0, 6, 7, 8, 0, 9, 10, 11, 1]

I've tortured myself trying to use np.unique because it returns a counts array, an inverse array, and an index array. I was sure I could these to get at a solution. The best I got is in pir4 below which scales in quadratic time. Also note that I don't care if counts start at 1 or zero as we can simply add or subtract 1.

Below are some of my attempts (none of which answer my question)

%%cython
from collections import defaultdict

def get_generator(l):
    counter = defaultdict(lambda: -1)
    for i in l:
        counter[i] += 1
        yield counter[i]

def pir2(l):
    return [i for i in get_generator(l)]

---

def pir3(l):
    return [i for i in get_generator(l)]

def pir4(l):
    unq, inv = np.unique(l, 0, 1, 0)
    a = np.arange(len(unq))
    matches = a[:, None] == inv
    return (matches * matches.cumsum(1)).sum(0).tolist()

Answer 1

setup

short_list = np.array(list('aaabaaacaaadaaac'))

functions

• dfill takes an array and returns the positions where the array changes and repeats that index position until the next change.

  # dfill
  # 
  # Example with short_list
  #
  #    0  0  0  3  4  4  4  7  8  8  8 11 12 12 12 15
  # [  a  a  a  b  a  a  a  c  a  a  a  d  a  a  a  c]
  #
  # Example with short_list after sorting
  #
  #    0  0  0  0  0  0  0  0  0  0  0  0 12 13 13 15
  # [  a  a  a  a  a  a  a  a  a  a  a  a  b  c  c  d]
  

• argunsort returns the permutation necessary to undo a sort given the argsort array. The existence of this method became know to me via this post.. With this, I can get the argsort array and sort my array with it. Then I can undo the sort without the overhead of sorting again.

• cumcount will take an array sort it, find the dfill array. An np.arange less dfill will give me cumulative count. Then I un-sort

  # cumcount
  # 
  # Example with short_list
  #
  # short_list:
  # [ a  a  a  b  a  a  a  c  a  a  a  d  a  a  a  c]
  # 
  # short_list.argsort():
  # [ 0  1  2  4  5  6  8  9 10 12 13 14  3  7 15 11]
  #
  # Example with short_list after sorting
  #
  # short_list[short_list.argsort()]:
  # [ a  a  a  a  a  a  a  a  a  a  a  a  b  c  c  d]
  # 
  # dfill(short_list[short_list.argsort()]):
  # [ 0  0  0  0  0  0  0  0  0  0  0  0 12 13 13 15]
  # 
  # np.range(short_list.size):
  # [ 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15]
  #
  # np.range(short_list.size) - 
  #     dfill(short_list[short_list.argsort()]):
  # [ 0  1  2  3  4  5  6  7  8  9 10 11  0  0  1  0]
  # 
  # unsorted:
  # [ 0  1  2  0  3  4  5  0  6  7  8  0  9 10 11  1]
  

• foo function recommended by @hpaulj using defaultdict
• div function recommended by @Divakar (old, I'm sure he'd update it)

code

def dfill(a):
    n = a.size
    b = np.concatenate([[0], np.where(a[:-1] != a[1:])[0] + 1, [n]])
    return np.arange(n)[b[:-1]].repeat(np.diff(b))

def argunsort(s):
    n = s.size
    u = np.empty(n, dtype=np.int64)
    u[s] = np.arange(n)
    return u

def cumcount(a):
    n = a.size
    s = a.argsort(kind='mergesort')
    i = argunsort(s)
    b = a[s]
    return (np.arange(n) - dfill(b))[i]

def foo(l):
    n = len(l)
    r = np.empty(n, dtype=np.int64)
    counter = defaultdict(int)
    for i in range(n):
        counter[l[i]] += 1
        r[i] = counter[l[i]]
    return r - 1

def div(l):
    a = np.unique(l, return_counts=1)[1]
    idx = a.cumsum()
    id_arr = np.ones(idx[-1],dtype=int)
    id_arr[0] = 0
    id_arr[idx[:-1]] = -a[:-1]+1
    rng = id_arr.cumsum()
    return rng[argunsort(np.argsort(l))]

demonstration

cumcount(short_list)

array([ 0,  1,  2,  0,  3,  4,  5,  0,  6,  7,  8,  0,  9, 10, 11,  1])

time testing

code

functions = pd.Index(['cumcount', 'foo', 'foo2', 'div'], name='function')
lengths = pd.RangeIndex(100, 1100, 100, 'array length')
results = pd.DataFrame(index=lengths, columns=functions)

from string import ascii_letters

for i in lengths:
    a = np.random.choice(list(ascii_letters), i)
    for j in functions:
        results.set_value(
            i, j,
            timeit(
                '{}(a)'.format(j),
                'from __main__ import a, {}'.format(j),
                number=1000
            )
        )

results.plot()

Answer 2

Here's a vectorized approach using custom grouped range creating function and np.unique for getting the counts -

def grp_range(a):
    idx = a.cumsum()
    id_arr = np.ones(idx[-1],dtype=int)
    id_arr[0] = 0
    id_arr[idx[:-1]] = -a[:-1]+1
    return id_arr.cumsum()

count = np.unique(A,return_counts=1)[1]
out = grp_range(count)[np.argsort(A).argsort()]

Sample run -

In [117]: A = list('aaabaaacaaadaaac')

In [118]: count = np.unique(A,return_counts=1)[1]
     ...: out = grp_range(count)[np.argsort(A).argsort()]
     ...: 

In [119]: out
Out[119]: array([ 0,  1,  2,  0,  3,  4,  5,  0,  6,  7,  8,  0,  9, 10, 11,  1])

For getting the count, few other alternatives could be proposed with focus on performance -

np.bincount(np.unique(A,return_inverse=1)[1])
np.bincount(np.fromstring('aaabaaacaaadaaac',dtype=np.uint8)-97)

Additionally, with A containing single-letter characters, we could get the count simply with -

np.bincount(np.array(A).view('uint8')-97)