Finding Sequence and splitting it to array with Python

Question

I have an array, I need to find the sequence and then split it into array of arrays.

what I thought if i find outlier and order it accordingly. I have tried the following

data = [ 347,  348,  349,  350,  351,  352,  353,  354,  355,  356, 2987,
   2988, 2989, 2990, 2991, 2992, 2993, 2994, 2995, 2996, 4992, 4993,
   4994, 4995, 5007, 5008, 5009, 5010, 5011, 5012, 5013, 5014, 5015,
   5016, 5987, 5988, 5989, 5990, 5991, 5992, 5993, 5994, 5995, 5996,
   6036, 6037, 6038, 6039, 6040, 6041, 6042, 6043, 6044, 6045, 6046,
   6047]

def reject_outliers(data, m = 2.):
    d = np.abs(data - np.median(data))
    mdev = np.median(d)
    s = d/mdev if mdev else 0.
    return data[s<m], data[s>m]

def sort_sequences(data):
    d = tuple()
    sorted_data = tuple()

    seq = reject_outliers(data)
    sorted_data =  + (seq[1])
    print(sorted_data)
    main_part = seq[0]
    another_part = seq[1]
    if len(another_part) != 0:
        sort_sequences(main_part)
    return sorted_data

Now if i apply this

`data_sorted = sort_sequences(actual)`

I get:

[347 348 349 350 351 352 353 354 355 356]

which is not what I am looking for

keepAlive · Accepted Answer · 2019-01-06 19:03:36Z

What about

>>> ddata = data[1:] - data[:-1] # faster than np.diff(data)
>>> ddata
array([   1,    1,    1,    1,    1,    1,    1,    1,    1, 2631,    1,
          1,    1,    1,    1,    1,    1,    1,    1, 1996,    1,    1,
          1,   12,    1,    1,    1,    1,    1,    1,    1,    1,    1,
        971,    1,    1,    1,    1,    1,    1,    1,    1,    1,   40,
          1,    1,    1,    1,    1,    1,    1,    1,    1,    1,    1])

and finally

>>> sequences = np.split(data, np.argwhere(ddata>1).flatten() + 1)
>>> sequences[0]
array([347, 348, 349, 350, 351, 352, 353, 354, 355, 356])
>>> sequences[2]
array([4992, 4993, 4994, 4995])
>>> sequences[-1]
array([6036, 6037, 6038, 6039, 6040, 6041, 6042, 6043, 6044, 6045, 6046,
       6047])

Execution times: data[1:] - data[:-1] VS np.diff(data) -- tested via repl.it

setup_="""
import numpy as np

MULT = 100
data = np.array(MULT*[ 347,  348,  349,  350,  351,  352,  353,  354,  355,  356, 2987,
  2988, 2989, 2990, 2991, 2992, 2993, 2994, 2995, 2996, 4992, 4993,
  4994, 4995, 5007, 5008, 5009, 5010, 5011, 5012, 5013, 5014, 5015,
  5016, 5987, 5988, 5989, 5990, 5991, 5992, 5993, 5994, 5995, 5996,
  6036, 6037, 6038, 6039, 6040, 6041, 6042, 6043, 6044, 6045, 6046,
  6047])
"""
import timeit

then

Python 3.6.1 (default, Dec 2015, 13:05:11)
[GCC 4.8.2] on linux
>>> timeit.Timer('data[1:] - data[:-1]', setup=setup_).timeit()
35.17186516011134    
>>> timeit.Timer('np.diff(data)', setup=setup_).timeit()
63.88404295803048

Employee · Accepted Answer · 2019-01-06 11:51:03Z

1

Write less code:

diffs = np.diff(data)
sequences = np.split(data, np.argwhere(diffs>1).flatten() + 1)

answered Jan 6, 2019 at 11:51

Employee

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