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This is a code to sort a list of lists (matrix of data) based on the names on second column (column 1) in a way equal names will be at the biggest possible distance of each other inside of the sorted matrix of data.

This subject has been discussed here and here seeking for a solution in Excel and Python, respectively. Despite of the good proposed solutions, none of them worked with my real data, and I worked-out this code that I am sharing with you. Feel free to comment, criticize and improve, if you like. It has a lot of print that are only for a better visualization of the steps. It worked well in the various tests I did so far.


import datetime

matrix = [[True, 'Tricha', 6, 'country', 'state', 'city', 'username', datetime.datetime(2020, 10, 25, 21, 0), 'comment'], [True, 'Joe', 6, 'country', 'state', 'city', 'username', datetime.datetime(2020, 10, 25, 21, 0), 'comment'], [True, 'July', 6, 'country', 'state', 'city', 'username', datetime.datetime(2020, 10, 25, 21, 0), 'comment'], [True, 'John', 8, 'country', 'state', 'city', 'username', datetime.datetime(2020, 10, 25, 21, 0), 'comment'], [True, 'John', 8, 'country', 'state', 'city', 'username', datetime.datetime(2020, 10, 25, 21, 0), 'comment'], [True, 'John', 8, 'country', 'state', 'city', 'username', datetime.datetime(2020, 10, 25, 21, 0), 'comment'], [True, 'Mary', 7, 'country', 'state', 'city', 'username', datetime.datetime(2020, 10, 25, 21, 0), 'comment'], [True, 'Mary', 7, 'country', 'state', 'city', 'username', datetime.datetime(2020, 10, 25, 21, 0), 'comment'], [True, 'Mike', 9, 'country', 'state', 'city', 'username', datetime.datetime(2020, 10, 25, 21, 0), 'comment'], [True, 'Mike', 9, 'country', 'state', 'city', 'username', datetime.datetime(2020, 10, 25, 21, 0), 'comment'], [True, 'Bob', 7, 'country', 'state', 'city', 'username', datetime.datetime(2020, 10, 25, 21, 0), 'comment']]


column_to_sort = 1  # column of the element that will be used to sort the matrix

lista = []

for item in matrix:
    lista.append(item[column_to_sort])

print(lista)


lista2 = []


for name in lista:
    if name not in lista2:
        lista2.append(name)

print (lista2)

dic = {}
for item in lista2:
    dic[item] = 0
    print(item)

print (f'dic with quantity equal to zero: {dic}')


for name in lista:
    n = dic[name]
    n = n + 1
    dic[name] = n

print (f'dic: {dic}')

# sort the names according to the number of times it appears in dic
dic2 = sorted(dic.items(), key=lambda x: x[column_to_sort])

print (f'dic2: {dic2}')

matrix2 = []



len_matrix = len(matrix)

for g in range (1, len_matrix+1):
    print(f'g: {g}')
    for item in dic2:
        if item[column_to_sort] == g:
            position = 0
            for i in range(0, len_matrix):
                print(f'i: {i}')
                print(f'g: {g}     matrix2 len: {len(matrix2)}')
                if item[0] == matrix[i][column_to_sort]:
                    print(f'Go to position: {position} for {matrix[i]}')
                    matrix2.insert(position, matrix[i])
                    position = int(position + len(matrix2) / g) + 1



print (f'matrix2: {matrix2}')

In my case it is used to rotate the vendor of digital information in a way that, if one vendor ("name") fail, then it is rotated to a different vendor, and will come back to the failed vendor at the latest possible time (that may give the vendor time enough to fix the issue).

Nanno
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  • *"Despite of the good proposed solutions, none of them worked with my real data"* <<< Why? What didn't work? – Stef Feb 22 '23 at 16:18
  • I'm sure there are other people who had the same problem before you, although it's not super easy to search. See for instance https://stackoverflow.com/search?q=shuffle+repeated+next ; [Python Shuffle two arrays of numbers so that adjacent ones differ](https://stackoverflow.com/questions/73058026/python-shuffle-two-arrays-of-numbers-so-that-adjacent-ones-differ); [How to shuffle a array of objects with no two duplicates next to each other?](https://stackoverflow.com/questions/59036679/how-to-shuffle-a-array-of-objects-with-no-two-duplicates-next-to-each-other); – Stef Feb 22 '23 at 16:24
  • Also [Evenly distributing all items in a list](https://stackoverflow.com/questions/3370554/evenly-distributing-all-items-in-a-list) – Stef Feb 22 '23 at 16:25

0 Answers0