Convert a list of two numbers to indices for an array-CodePudding

Short version: Given a list with two elements [i, j], how do I get the i, j -element of a 2d array instead of the i, j rows: arr[[i,j]] to arr[i,j].

I've seen similar cases where *list has been used, but I don't entirely understand how that operator works.

Deeper context:

I have a function that returns a nested list, where each sub-list is a pair of indices to be used in an array:


grid = np.full((3,3), 1)

def path():
   ...
   return [[i_1, j_1], [i_2, j_2], ...]

for i in path():
   grid[path()[i]] = 0

But since path()[i] is a list, grid[path()[i]] == 0 sets two rows equal to zero, and not a single element. How do I prevent that?
While not stricly necessary, a faster solution would be preferable as this operation is to be done many times.

CodePudding user response：

The thing that is confusing you is the difference in mathematic notation and indexing 2D (or n-dimensional) lists in Python/programming languages.

If you have a 2D matrix in mathematics, called X, and let's say you'd want to access the element in the first row and first column, you'd write X(1, 1).

If you have a 2D array, it's elements are lists. So, if you want to access the 1st row of an array called X you'd do:

X[0] # 0 because indexation starts at 0

Keep in mind that the previous statement returns a new list. You can index this list as well. Since we need the first column, we do:

X[0][0] # element in 1st row and 1st column of matrix X

Thus the answer is you need to successively index the matrix, i.e. apply chain indexing.

As for your original question, here is my answer. Let's say a is the 2-element list which contains the indices i and j which you want to access in the 2D array. I'll denote the 2D array as b. Then you apply chain indexing, the first index is the first element of a and the second index is the second element of a:

a = [0, 0]  # list with 2 elements containing the indices
b = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]  # 2D array in which you want to access b[0][0]
i, j = a[0], a[1]
print(b[i][j])

Obviously you don't need to use variables i, j, I just did so for clarity:

print(b[a[0]][a[1]])

Both statements print out the element in the 1st row and 1st column of b: