CODEWITHSUNDEEP
pythonarraysgrid
Aaron165
Aaron165

Reputation: 59

Adding random integers to outer cells of a grid in python

I have a grid for a game set out as below:

grid = [(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)]

I need to iterate through the outer cells of the grid and randomly turn them into either "1" or "0".

Is there a way of doing this quickly whilst maintaining the ability to change the size of the grid and still perform the same thing?

thanks in advance!

Upvotes: 1

Views: 113

Answers (5)

mvelay
mvelay

Reputation: 1520

You should use arrays through numpy module to improve performances as follow:

import numpy as np
from random import randint

def random_grid():
    while True:  
        grid = np.array([randint(0, 1) for _ in range(35)]).reshape(5,7)
        yield grid

gen_grids = random_grid()
print gen_grids.next()

Upvotes: 0

JDurstberger
JDurstberger

Reputation: 4255

First of all you should use lists instead of tuples, tuples are immutable and cannot be changed.

Create grid as list with lists

grid = [[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]]

This should do the trick, though one of the python-gurus may have an even easier solution.

First Edition

#go through all the lines
for index, line in enumerate(grid):
    #is the line the first or the last one
    if index == 0 or index == len(grid)-1:
        #randomize all entries
        for i in range(0, len(line)):
            line[i] = randint(0, 1)
    #the line is one in the middle
    else:
        #randomize the first and the last
        line[0] = randint(0, 1)
        line[-1] = randint(0, 1)

After toying around some more I could replace the nested for with a list comprehension to make the code more readable

Second Edition

for index, line in enumerate(grid):
    if index == 0 or index == len(grid)-1:
        grid[index] = [randint(0, 1)for x in line]
    else:
        line[0] = randint(0, 1)
        line[-1] = randint(0, 1)

If someone points out an easier/more readable way to do the if I would be glad.

Upvotes: 2

Bharadwaj
Bharadwaj

Reputation: 745

Extending Jan's answer

from random import randint
grid = [[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 x, j in enumerate(grid):
    for y, i in enumerate(j):
        grid[x][y] = randint(0,1)
print(grid)

This is a simple solution independent of whatever is the size, but if performance is critical, then go for numpy.

Upvotes: 0

KIDJourney
KIDJourney

Reputation: 1220

You can use numpy , and you don't need to iterate at all . 

import numpy as np
grid = 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)])
grid[[0,-1]] = np.random.randint(2,size=(2,grid.shape[1])) 
grid[:,[0,-1]] = np.random.randint(2,size=(grid.shape[0],2))

Upvotes: 0

Jan Pomikálek
Jan Pomikálek

Reputation: 1439

If you represent your grid as a list of lists (rather than a list of tupples), then it's a matter of iterating over the outer cells and setting:

grid[x][y] = random.randint(0, 1)

... considering that by "randomly turn them" you mean "change them with 50% probability".

Upvotes: 1

Related Questions