Counting Sequential Booleans

Question

This function takes a list/array of booleans and converts them to an array that counts the number of either True/False values found next to each other.

I'd like to see this optimized for performance. It's not too slow, but I do use multiple loops with embedded if-else statements, I'm wondering if they're absolutely necessary.

import numpy as np

x = np.random.uniform(1,100,100)
b = x > x.mean()


#function start, input is b
endarray = []
count = 0
instance = True
while True:
    subarray = 0
    while True:
        if count >= len(b):
            endarray.append(subarray)
            break
        if b[count] == instance:
            subarray += 1
            count += 1
        else:
            endarray.append(subarray)
            instance = not instance
            break
    if count >= len(b):
        break

if len(endarray) % 2 != 0:
    endarray = np.append(endarray, 0)
else:
    endarray = np.asarray(endarray)

endarray = endarray.reshape(-1,2)

The output is a Nx2 array, where the left-hand values are always a count of Trues, and the right-hand values are always a count of Falses.

After a sequence of False values are no longer continuous(a True value pops up), the next count of True values begin, and vice versa.

Example input

b
Out[31]: 
array([ True,  True,  True, False,  True,  True,  True,  True, False,
       False,  True, False, False,  True, False, False, False, False,
        True, False, False, False,  True,  True,  True,  True, False,
       False,  True, False, False, False, False, False, False,  True,
        True, False,  True,  True, False, False,  True, False, False,
        True, False, False,  True, False,  True, False,  True, False,
        True,  True,  True, False,  True, False,  True,  True,  True,
        True, False, False,  True, False,  True,  True,  True,  True,
        True,  True, False,  True,  True, False,  True,  True, False,
       False,  True, False,  True, False, False,  True,  True,  True,
        True, False, False, False, False, False,  True,  True,  True,
        True])

Example output

endarray
Out[32]: 
array([[3, 1],
       [4, 2],
       [1, 2],
       [1, 4],
       [1, 3],
       [4, 2],
       [1, 6],
       [2, 1],
       [2, 2],
       [1, 2],
       [1, 2],
       [1, 1],
       [1, 1],
       [1, 1],
       [3, 1],
       [1, 1],
       [4, 2],
       [1, 1],
       [6, 1],
       [2, 1],
       [2, 2],
       [1, 1],
       [1, 2],
       [4, 5],
       [4, 0]])

Edit: I wanted to add an updated version of this code, the one in the answer below is not technically correct in all regards. But this was entirely derived from it:

m = np.append(b[0], np.diff(b))
_, c = np.unique(m.cumsum(), return_index=True)
out = np.diff(np.append(c, len(b)))
if b[0] == False:
    out = np.append(0, out)
if len(out) % 2:
    out = np.append(out, 0)
out = out.reshape(-1, 2)

Answer 1

Using itertools.groupby

What you are looking for is itertools.groupby. When there is an odd number of groups then we use try-except block here.

from itertools import groupby

get_grp_len = lambda grp: len([*grp])


def transform(b):
    if len(b) == 0:  # if not b wouldn't work since your `b` is ndarray
        return []
    it = groupby(b)
    out = []
    for _, grp in it:
        try:
            t_size = get_grp_len(grp)
            f_size = get_grp_len(next(it)[1])
            out.append([t_size, f_size])
        except StopIteration:
            out.append([t_size, 0])
    return out


print(transform(b)) # `b` taken from the question itself.
# same output as expected output posted in question.

NumPy has a lot of vectorized operations

NumPy has many vectorized operations, you have to find the correct ones. Not an expert but the below approach should do good.

The idea here is to find the index of the first value from each group and then take the difference.

• We check if ith element is not equal to i+1th element.
• Now use np.ndarray.cumsum to give each group a unique sequential number.
• Then use np.unique to get first index of value from each group.
• Find the difference between i+1th value and ith to get size of each grp. We can use np.diff

---

m = b != (np.r_[np.nan, b[:-1]]) 
_, c = np.unique(m.cumsum(), return_index=True)
#print(c)
# array([ 0,  3,  4,  8, 10, 11, 13, 14, 18, 19, 22, 26, 28, 29, 35, 37, 38,
#        40, 42, 43, 45, 46, 48, 49, 50, 51, 52, 53, 54, 57, 58, 59, 60, 64,
#        66, 67, 68, 74, 75, 77, 78, 80, 82, 83, 84, 85, 87, 91, 96])

# np.unique gives the index of the first occurrence, our `b` len is 100
# and `c` stopped at 96. we need to add the last index + 1 to it.

out = np.diff(np.r_[c, len(b)])

# Now, reshape your array, if it's of odd length add a 0 at end.
if len(out) % 2:
    out = np.r_[out, 0].reshape(-1, 2)
out = out.reshape(-1, 2)
print(out)
# same output mentioned in the question.

Using Pandas's GroupBy

A lot of times pandas and NumPy are used together so might as well post pandas code too.

import pandas as pd
s = pd.Series(b)
g = s.ne(s.shift()).cumsum()
out = s.groupby(g).size().to_numpy()
# repeat same step as we did in above solution add a 0 if len is odd.

if len(out)%2:
    out = np.r_[out, 0].reshape(-1, 2)
out = out.reshape(-1, 2)
print(out)

---

Code Review

Everything's good as far as I can see, well-named variables, proper indentation but too many if :p