How to group consecutive data in 2d array in python

Question

I have a 2d NumPy array that looks like this:

array([[1, 1],
       [1, 2],
       [2, 1],
       [2, 2],
       [3, 1],
       [5, 1],
       [5, 2]])

and I want to group it and have an output that looks something like this:

         Col1 Col2
group 1: 1-2, 1-2
group 2: 3-3, 1-1
group 3: 5-5, 1-2

I want to group the columns based on if they are consecutive.

So, for a unique value In column 1, group data in the second column if they are consecutive between rows. Now for a unique grouping of column 2, group column 1 if it is consecutive between rows.

The result can be thought of as corner points of a grid. In the above example, group 1 is a square grid, group 2 is a a point, and group 3 is a flat line.

My system won't allow me to use pandas so I cannot use group_by in that library but I can use other standard libraries.

Any help is appreciated. Thank you

I don't understand how the output corresponds to the input - will you please elaborate? — user17242583
– user17242583, Commented Mar 18, 2022 at 22:08
@richardec I cannot seem to edit the question at the moment. For each unique value in column 1, group column 2 if column 2 is consecutive. Once this is grouped, for each grouped column 2, group column 1 if it is consecutive. You see that the first 4 rows form a square (if plotted). The 5th rows is a dot if plotted. And rows 6 and 7 form a flat line if plotted. Does this help? — Aidan O'Farrell
– Aidan O'Farrell, Commented Mar 18, 2022 at 22:15
I can't see why row [3,1] doesn't belong into group 1 according to your logic. — Michael Szczesny
– Michael Szczesny, Commented Mar 18, 2022 at 23:41
@MichaelSzczesny because column 2 is not consecutive with the row below with the same column 1 value. The row below isn't the same column value. — Aidan O'Farrell
– Aidan O'Farrell, Commented Mar 18, 2022 at 23:55
If [3,1] isn't a unique grouping according to step 1, shouldn't it belong to no group according to step 2 for a unique grouping of column 2, group column 1...? — Michael Szczesny
– Michael Szczesny, Commented Mar 19, 2022 at 0:17

constantstranger · Accepted Answer · 2022-03-18 23:24:39Z

3

Here you go ...

Steps are:

Get a list xUnique of unique column 1 values with sort order preserved.
Build a list xRanges of items of the form [col1_value, [col2_min, col2_max]] holding the column 2 ranges for each column 1 value.
Build a list xGroups of items of the form [[col1_min, col1_max], [col2_min, col2_max]] where the [col1_min, col1_max] part is created by merging the col1_value part of consecutive items in xRanges if they differ by 1 and have identical [col2_min, col2_max] value ranges for column 2.
Turn the ranges in each item of xGroups into strings and print with the required row and column headings.
Also package and print as a numpy.array to match the form of the input.

import numpy as np
data = np.array([
    [1, 1],
    [1, 2],    
    [2, 1],    
    [2, 2],
    [3, 1],
    [5, 1],
    [5, 2]])
xUnique = list({pair[0] for pair in data})
xRanges = list(zip(xUnique, [[0, 0] for _ in range(len(xUnique))]))
rows, cols = data.shape
iRange = -1
for i in range(rows):
    if i == 0 or data[i, 0] > data[i - 1, 0]:
        iRange += 1
        xRanges[iRange][1][0] = data[i, 1]
    xRanges[iRange][1][1] = data[i, 1]
xGroups = []
for i in range(len(xRanges)):
    if i and xRanges[i][0] - xRanges[i - 1][0] == 1 and xRanges[i][1] == xRanges[i - 1][1]:
        xGroups[-1][0][1] = xRanges[i][0]
    else:
        xGroups += [[[xRanges[i][0], xRanges[i][0]], xRanges[i][1]]]

xGroupStrs = [ [f'{a}-{b}' for a, b in row] for row in xGroups]

groupArray = np.array(xGroupStrs)
print(groupArray)

print()
print(f'{"":<10}{"Col1":<8}{"Col2":<8}')
[print(f'{"group " + str(i) + ":":<10}{col1:<8}{col2:<8}') for i, (col1, col2) in enumerate(xGroupStrs)]

Output:

[['1-2' '1-2']
 ['3-3' '1-1']
 ['5-5' '1-2']]

          Col1    Col2
group 0:  1-2     1-2
group 1:  3-3     1-1
group 2:  5-5     1-2

edited Mar 18, 2022 at 23:24

answered Mar 18, 2022 at 23:07

constantstranger

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5 Comments

Michael Szczesny Over a year ago

Breaks for [[1, 1],[2, 1]] and all consecutive 'point shapes'.

constantstranger Over a year ago

@Michael Szczesny [[1, 1], [2, 1]] gives [['1-2' '1-1']] as output. Is that not what you would expect?

Aidan O'Farrell Over a year ago

@constantstranger this is giving the answer I am looking for - thank you very much!!

constantstranger Over a year ago

@Aidan O'Farrell Thanks for confirming. Do you need any more help with your question?

Aidan O'Farrell Over a year ago

@constantstranger no I do not. That is exactly what I need, many thanks.

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