Set values around a specific position in a 2D array using Python-CodePudding

So say I have a 2D array such as:

[
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 3, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]

And I want to set all the values 2 levels out around the 3 to a specific number like:

[
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
    [0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
    [0, 0, 0, 0, 1, 1, 3, 1, 1, 0],
    [0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
    [0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]

Note that the 3 could be in any position in the list, I'm using a random generator to get it. So how could I achieve this? Maybe using a for loop?

CodePudding user response：

Carrying on from the comment - I find numpy super useful for slicing like this;

import numpy as np

arr = np.array([
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 3, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
])

xs, ys = np.where(arr == 3)
arr[xs[0] - 2: xs[0]   3, ys[0] - 2: ys[0]   3] = 1
arr[xs[0], ys[0]] = 3

Obviously possible in pure python/list form as well but you will be knee deep in double iteration probably

CodePudding user response：

Here's a pure Python approach (albeit rather clumsy):

A = [
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 3, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]

for j, a in enumerate(A):
    try:
        i = a.index(3)
        P = A[j-1] if j > 0 else None
        Q = A[j 1] if j < len(a) - 1 else None
        for k in range(max(0, i-2), min(i 3, len(a))):
            if P:
                P[k] = 1
            if Q:
                Q[k] = 1
            a[k] = 1
        a[i] = 3
        break
    except ValueError:
        pass
print(A)

CodePudding user response：

pure python

A = [
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 3, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
]


location = [(i, j) for i, row in enumerate(A) for j, item in enumerate(row) if item == 3][0]
for i, row in enumerate(A):
    for j, item in enumerate(row):
        if (abs(i - location[0]) <= 2) and (abs(j - location[1]) <= 2) and not ((i, j) == location):
            A[i][j] = 1