How to find Run length encoding in python

Question

I have an array ar = [2,2,2,1,1,2,2,3,3,3,3]. For this array, I want to find the lengths of consecutive same numbers like:

 values: 2, 1, 2, 3
l         


        

Answer 1

Here is an answer using the high-performance pyrle library for run length arithmetic:

# pip install pyrle
# (pyrle >= 0.0.25)

from pyrle import Rle

v = [2,2,2,1,1,2,2,3,3,3,3]

r = Rle(v)
print(r)
# +--------+-----+-----+-----+-----+
# | Runs   | 3   | 2   | 2   | 4   |
# |--------+-----+-----+-----+-----|
# | Values | 2   | 1   | 2   | 3   |
# +--------+-----+-----+-----+-----+
# Rle of length 11 containing 4 elements

print(r[4])
# 1.0

print(r[4:7])
# +--------+-----+-----+
# | Runs   | 1   | 2   |
# |--------+-----+-----|
# | Values | 1.0 | 2.0 |
# +--------+-----+-----+
# Rle of length 3 containing 2 elements

r + r + 0.5
# +--------+-----+-----+-----+-----+
# | Runs   | 3   | 2   | 2   | 4   |
# |--------+-----+-----+-----+-----|
# | Values | 4.5 | 2.5 | 4.5 | 6.5 |
# +--------+-----+-----+-----+-----+
# Rle of length 11 containing 4 elements

Answer 2

Here is an answer for pure numpy:

import numpy as np


def find_runs(x):
    """Find runs of consecutive items in an array."""

    # ensure array
    x = np.asanyarray(x)
    if x.ndim != 1:
        raise ValueError('only 1D array supported')
    n = x.shape[0]

    # handle empty array
    if n == 0:
        return np.array([]), np.array([]), np.array([])

    else:
        # find run starts
        loc_run_start = np.empty(n, dtype=bool)
        loc_run_start[0] = True
        np.not_equal(x[:-1], x[1:], out=loc_run_start[1:])
        run_starts = np.nonzero(loc_run_start)[0]

        # find run values
        run_values = x[loc_run_start]

        # find run lengths
        run_lengths = np.diff(np.append(run_starts, n))

        return run_values, run_starts, run_lengths

Credit goes to https://github.com/alimanfoo

Answer 3

You can do this with groupby

In [60]: from itertools import groupby
In [61]: ar = [2,2,2,1,1,2,2,3,3,3,3]
In [62]: print [(k, sum(1 for i in g)) for k,g in groupby(ar)]
[(2, 3), (1, 2), (2, 2), (3, 4)]