compare each row with each column in matrix using only pure python-CodePudding

I have a certain function that I made and I want to run it on each column and each row of a matrix, to check if there are rows and columns that produce the same output.

for example:

matrix = [[1,2,3],
          [7,8,9]]

I want to run the function, lets call it myfun, on each column [1,7], [2,8] and [3,9] separatly, and also run it on each row [1,2,3] and [7,8,9]. If there is a row and a column that produce the same result, the counter ct would go up 1. All of this is found in another function, called count_good, which basically counts rows and columns that produce the same result.

here is the code so far:

def count_good(mat):
    ct = 0
    for i in mat:
        for j in mat:
            if myfun(i) == myfun(j):
                ct  = 1
    return ct

However, when I use print to check my code I get this:

mat = [[1,2,3],[7,8,9]]

for i in mat:
    for j in mat:
        print(i,j)

[1, 2, 3] [1, 2, 3]
[1, 2, 3] [7, 8, 9]
[7, 8, 9] [1, 2, 3]
[7, 8, 9] [7, 8, 9]

I see that the code does not return what I need' which means that the count_good function won't work. How can I run a function on each row and each column? I need to do it without any help of outside libraries, no map,zip or stuff like that, only very pure python.

CodePudding user response：

Let's start by using itertools and collections for this, then translate it back to "pure" python.

from itertools import product, starmap, chain # combinations?
from collections import Counter

To iterate in a nested loop efficiently, you can use itertools.product. You can use starmap to expand the arguments of a function as well. Here is a generator of the values of myfun over the rows:

starmap(myfun, product(matrix, repeat=2))

To transpose the matrix and iterate over the columns, use the zip(* idiom:

starmap(myfun, product(zip(*matrix), repeat=2))

You can use collections.Counter to map all the repeats for each possible return value:

Counter(starmap(myfun, chain(product(matrix, repeat=2), product(zip(*matrix), repeat=2))))

If you want to avoid running myfun on the same elements, replace product(..., repeat=2) with combinations(..., 2).

Now that you have the layout of how to do this, replace all the external library stuff with equivalent builtins:

counter = {}
for i in range(len(matrix)):
    for j in range(len(matrix)):
        result = myfun(matrix[i], matrix[j])
        counter[result] = counter.get(result, 0)   1
for i in range(len(matrix[0])):
    for j in range(len(matrix[0])):
        c1 = [matrix[row][i] for row in range(len(matrix))]
        c2 = [matrix[row][j] for row in range(len(matrix))]
        result = myfun(c1, c2)
        counter[result] = counter.get(result, 0)   1

If you want combinations instead, replace the loop pairs with

for i in range(len(...) - 1):
    for j in range(i   1, len(...)):

CodePudding user response：

Using native python:

def count_good(mat):
    ct = 0
    for row in mat:
        for col_idx in range(len(mat[0])):
            column = [x[col_idx] for x in mat]
            if myfun(row) == myfun(column):
                ct  = 1
    return ct

However, this is very inefficient as it is a triple nested for-loop. I would suggest using numpy instead.

e.g.

def count_good(mat):
    ct = 0
    mat = np.array(mat)
    for row in mat:
        for column in mat.T:
            if myfun(row) == myfun(column):
                ct  = 1
    return ct