CODEWITHSUNDEEP
pythonarraysnumpy
Jack_Bandit
Jack_Bandit

Reputation: 87

Finding first 0 in first row of numpy array

I am trying to write a script using python to solve a maze problem by the right hand method. I have written the below script to read in a file of the maze and put it into numpy 2D array. Now, I would like to search the first row of the array and find the 0. This 0 is the starting point of the maze. From here I would apply my maze algorithm to check the squares sourronding that point on whether or not they have a 1 or 0.

Maze_matrix is the matrix containing my maze and I want to find the index of the first 0 in the first row.

#!/usr/bin/python

import sys
import numpy as np
import itertools

if len(sys.argv) == 3:
        maze_file = sys.argv[1]
        soln_file = sys.argv[2]
        rows = []
        columns = []

        with open(maze_file) as maze_f:
                for line in maze_f:
                        row, column = line.split()
                        row = int(row)
                        column = int(column)
                        rows.append(row)
                        columns.append(column)
                maze_matrix = np.zeros((rows[0], columns[0]))
                for line1, line2 in zip(rows[1:], columns[1:]):
                        maze_matrix[line1][line2]  = 1

        print maze_matrix

else:
        print('Usage:')
        print('  python {} <maze file> <solution file>'.format(sys.argv[0]))
        sys.exit()

Upvotes: 1

Views: 1428

Answers (2)

daniel451
daniel451

Reputation: 10992

I would suggest using numpy.where(). It has a very good performance and searches through your whole array (or a subset) at once. If the condition is true for an element, it returns an array with the index of that element.

In [1]: import numpy as np

In [2]: a = np.random.randint(0, 9, (4,4))

In [8]: a
Out[8]: 
array([[6, 5, 0, 3],
       [4, 5, 8, 6],
       [0, 3, 4, 4],
       [6, 4, 6, 7]])

In [9]: np.where(a == 0)
Out[9]: (array([0, 2]), array([2, 0]))  # two 0's found
                                        # first at a[0, 2] (row 0, column 2)
                                        # second at a[2, 0] (row 2, column 0)

Upvotes: 1

Erik
Erik

Reputation: 898

I would suggest taking a look at the numpy.array method argmin:

>>> n = numpy.ones(100)
>>> n[50] = 0
>>> n.argmin()
50

Upvotes: 1

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