2

This program tests if a matrix is an identity matrix or not.

I have pasted my code beneath, and would like to know ways in which I can optimize the efficiency of this code. Also I am new to python programming, are there some built in functions that can solve the purpose too?

    def is_identity_matrix(test):
    if (test == []):
        return False
    i = 0
    while (i < len(test)):
        if (len(test[i]) == len(test)):
            j = 0
            while(j < len(test[i])):
                if (j != i):
                    if(test[i][j] != 0):
                        return False
                else:
                    if(test[i][j] != 1):
                        return False
                if(j == (len(test[i]) - 1)):
                    break
                j += 1
            if(i == (len(test) - 1)):
                break
            i += 1
        else:
            return False
    if(i == j and i == (len(test) - 1)):
        return True

# Test Cases:

matrix1 = [[1,0,0,0],
           [0,1,0,0],
           [0,0,1,0],
           [0,0,0,1]]
print is_identity_matrix(matrix1)
#>>>True

matrix2 = [[1,0,0],
           [0,1,0],
           [0,0,0]]

print is_identity_matrix(matrix2)
#>>>False

matrix3 = [[2,0,0],
           [0,2,0],
           [0,0,2]]

print is_identity_matrix(matrix3)
#>>>False

matrix4 = [[1,0,0,0],
           [0,1,1,0],
           [0,0,0,1]]

print is_identity_matrix(matrix4)
#>>>False

matrix5 = [[1,0,0,0,0,0,0,0,0]]

print is_identity_matrix(matrix5)
#>>>False

matrix6 = [[1,0,0,0],  
           [0,1,0,2],  
           [0,0,1,0],  
           [0,0,0,1]]

print is_identity_matrix(matrix6)
#>>>False
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5
  • 8
    One word: numpy. Commented May 17, 2013 at 16:49
  • 1
    or two, numpy and identity Commented May 17, 2013 at 16:50
  • 3
    Do you realy need to work with matrices? If you do, should use numpy. If you are just coding to learn: import this Commented May 17, 2013 at 16:53
  • 1
    Sounds like a homework-type question to me. Commented May 17, 2013 at 16:54
  • @Aya it is :P But I wasn't satisfied with my code... Commented May 17, 2013 at 16:57

6 Answers 6

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13 
def is_identity_matrix(listoflist):
    return all(val == (x == y) 
        for y, row in enumerate(listoflist)  
            for x, val in enumerate(row))

(though, this does not check if the matrix is square, and it returns True for an empty list)

Explanation: Inside all we have a generator expression with nested loops where val loops over each value in the matrix. x == y evaluates to True on the diagonal and False elsewhere. In Python, True == 1 and False == 0, so you can compare val == (x == y). The parentheses are important: val == x == y would be a chained comparison equivalent to val == x and x == y

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6 Comments

I think I needed to think through this one for about a full minute before deciding that it seems like it would work. Ultimately, I think this deserves kudos (and an upvote) for being able to pull off a not-too-ugly 1-liner in pure python with short-circuiting behavior.
I think val == 1 for ... for ... if x == y would be clearer. I sometimes use booleans as integers too, but generally it's not very obvious.
@Janne Karila: Hey can you help me understand what this code is doing? I would really really appreciate. This code is awesome- its clean :D
@delnan: but you also have to verify that val == 0 where x != y! So in the end I'm not sure Janne's answer can really be improved.
@icecrime D'oh! You're right. I think val == 1 if x == y else val == 0 is still competitive, but not obviously better.
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6

I'd use numpy:

(np.array(matrix1) == np.identity(len(matrix1))).all()

Of course, it'd be better if you were storing matrix1 as a numpy array in the first place to avoid the conversion.

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3 Comments

Hey Thank you! +1 I didn't know of numpy :)
@AbrarKhan -- It's not in the standard library, but it's a pretty common 3rd party extension because it is extremely useful.
Put dtype=np.bool in the call to np.identity for roughly twice the speed! (If the matrix is already a numpy array)
1

Multiply by the all ones vector and check that the result is the same

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1 Comment

There are many non-identity matrices that will pass this test.
1

Check the size of your matrix, make sure it is n x n, then create an actual identity matrix using np.identity(n), then compare your matrix with the new one you created.

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Comments

1 
def is_identity_matrix(test):
    if not test : return False
    le = len(test[0])
    for i,x in enumerate(test):
        if len(x) == le:
            if any(y!=1 if j==i else y!=0 for j,y in enumerate(x)):
                return False
        else:
            return False
    return True if len(test) == le else False
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1 Comment

I don't understand your first line -- if len(test)%2: return False. Why does the number of rows have to be odd?
0

If speed is important you should probably look at Numpy, but in a situation where you can't use it or perhaps it isn't worth it to have the matrix as a Numpy array you could also use the following:

def is_identity(mat):
    for i, row in enumerate(mat):
        if any(row[:i]) or row[i]!=1 or any(row[i+1:]):
            return False
    return True

About 12x faster than the code of the currently accepted answer! For a matrix of 2000x2000 at least...

The above does not check the matrix dimensions, but you can easily add something like:

n = len(matrix)
if not all(len(row) == n for row in matrix):
    return False
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