1

I have following matrix:

W = [['a', 'b', 'b', 'b', 'a'],
     ['a', 'a', 'a', 'a', 'b'],
     ['a', 'b', 'a', 'b', 'b'],
     ['a', 'a', 'a', 'b', 'b'],
     ['b', 'b', 'b', 'b', 'b'],
     ['b', 'b', 'b', 'b', 'b']]

If I choose W[x][y] from matrix, how can I count a-characters around that spesific point? For example let's say I choose W[4][1] which is b. I can see there is three a's around that point. But how can I determine it by coding? I ended up really messy peace of code, and realized it would not have worked if we change the dimension of matrix. Problem was also that if I choose the point near rounds, W[y-1][(x):(x+1)].count("a") -kind of thinking doesn't work.

Again, help would be appreciated! Thanks!

UPDATE:

W     = [['K', ' ', ' ', ' ', ' '],
         ['K', 'K', 'K', 'K', ' '],
         ['K', ' ', 'K', ' ', ' '],
         ['K', 'K', 'K', ' ', ' '],
         [' ', ' ', ' ', ' ', ' '],
         [' ', ' ', ' ', ' ', ' ']]

def count_matrix(W, y, x, ch="K"):
    ref = (-1, 0, 1)
    occ = []
    a = "K"
    b = " "
    for dy, dx in [(a, b) for a in ref for b in ref if (a,b)!=(0, 0)]:
        if (x+dx) >= 0 and (y+dy) >= 0:
            try:
                occ.append(W[x+dx][y+dy])
            except IndexError:
                pass
    return occ.count(ch)
print count_matrix(W,0,0)

This returns 0.

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2
  • 1
    typo: ref = (-1, 0, -1) Commented Nov 2, 2013 at 8:30
  • Your assignment of a='K' and b=" " is not doing what you think. The a and b inside the list comprehension are just temp variables with namespace scope limited to the comprehension. Setting a and b to other values outside the comprehension does not change anything. Commented Nov 2, 2013 at 16:30

3 Answers 3

Reset to default
2

This works:

W = [['a', 'b', 'b', 'b', 'a'],
     ['a', 'a', 'a', 'a', 'b'],
     ['a', 'b', 'a', 'b', 'b'],
     ['a', 'a', 'a', 'b', 'b'],
     ['b', 'b', 'b', 'b', 'b'],
     ['b', 'b', 'b', 'b', 'b']]

def count_matrix(W, r,c, ch):
    ref=(-1,0,1)
    matches=[]
    for dr, dc in [(a, b) for a in ref for b in ref if (a,b)!=(0,0)]:
        if r+dr>=0 and c+dc>=0:
            try:
                matches.append(W[r+dr][c+dc])
            except IndexError:
                pass
    return matches.count(ch)

print count_matrix(W,0,0,'a')   # correctly handles upper LH 
# 2
print count_matrix(W,4,1,'a') 
# 3

Remove the if (a,b)!=(0,0) if you want to count the character in the square itself. ie, With the square 0,0 do you count the 'a' there or not?

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1 Comment

No I do not count it. But I ended up in troubles again, if I let user to determine a and b themselves. Let's say user chooses a = "K" and b = " ". Still I would like to be able to count "K":s in the matrix without errors. Any suggestions?
1

See code comments for explanation:

from itertools import product

W = [['a', 'b', 'b', 'b', 'a'],
     ['a', 'a', 'a', 'a', 'b'],
     ['a', 'b', 'a', 'b', 'b'],
     ['a', 'a', 'a', 'b', 'b'],
     ['b', 'b', 'b', 'b', 'b'],
     ['b', 'b', 'b', 'b', 'b']]

def count_chars(W, x, y, ch):
    return sum(W[x+dx][y+dy] == ch
                   for dx, dy in product([-1,0,1], [-1,0,1])
                       if 0 <= x+dx < len(W) and 0 <= y+dy < len(W[0]))
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3 Comments

Ok, so it's just about ignoring IndexError's it is causing.. Funny though, I thought this could be done more easily. But will try this, thanks!
This is incorrect in several ways: 1) any top row or left col will switch to relative indexing since W[0-1] or W[][0-1] will now give the end of that list 2) Because of '1' the try is never triggered; there is never an index error 3) it counts the square with no offset W[x][y] since product([-1,0,1], [-1,0,1] will produce the tuple (0,0) as one of the results.
1. agree, removed wrong answer since it's modification to being correct make it also a bit messy. 2. index error is not triggerd only for 0's row, but it is not a point in view of (1) 3. see sample code where (0,0) is counted as well
0

Regarding to problem with the original matrix: 

W = [['a', 'b', 'b', 'b', 'a'],
     ['a', 'a', 'a', 'a', 'b'],
     ['a', 'b', 'a', 'b', 'b'],
     ['a', 'a', 'a', 'b', 'b'],
     ['b', 'b', 'b', 'b', 'b'],
     ['b', 'b', 'b', 'b', 'b']]

How about this?

def count_matrix(W, x, y):
    occ = 0
    for n in range(max(0, y - 1), min(len(W), y + 2)):
        for m in range(max(0, x - 1), min(len(W[0]), x + 2)):
            if W[n][m] == "a":
                occ += 1
    return occ

print count_matrix(W, 0, 0)

retunrs 3. Only problem left is how to exclude the value of square itself?

EDIT:

def count_matrix(W, x, y):
occ = 0
for n in range(max(0, y - 1), min(len(W), y + 2)):
    for m in range(max(0, x - 1), min(len(W[0]), x + 2)): 
        if n == y and m == x:
            pass
        else: 
            if W[n][m] == "a":
                occ += 1
return occ

W = [['a', 'b', 'b', 'b', 'a'],
     ['a', 'a', 'a', 'a', 'b'],
     ['a', 'b', 'a', 'b', 'b'],
     ['a', 'a', 'a', 'b', 'b'],
     ['b', 'b', 'b', 'b', 'b'],
     ['b', 'b', 'b', 'b', 'b']]

print count_matrix(W, 1, 4)

Returns 3 as expected.

print count_matrix(W, 0, 0)

Returns 2 as expected. Is this good practise?

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1 Comment

This doesn't cause any errors even if matrix is 1x1 matrix. So this just the code I was looking for!

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