How to use data type in an if statement

Question

I am writing a program to check if a matrix is square (The number of rows and columns are equal ex: 2x2, 3x3, etc)

I thought it would be best to count the elements using the built in size function and take the square root. I want to write an if statement where if the square root does not result in a whole number it prints an error statement, but I'm not sure how to specify a whole number in my statement.

here is what I've tried

    import numpy as np
    A = np.array([[1,2,3],[4,5,6]])
    check = A.size
    if check**.5 ...

and I don't know what to put in the rest of the statement

To check if a value is a whole number, you can use x % 1 == 0, but this approach will fail for checking for square arrays. For example, a matrix of size (18, 2) will return True if you check that the square root of its size is an integer. — hoffee
– hoffee, Commented Jul 19, 2019 at 16:55

Jmonsky · Accepted Answer · 2019-07-19 16:49:54Z

3

The shape attribute is probably what you want to check.

if A.shape[0] == A.shape[1]:
    # Is square
else:
    # Is not square

answered Jul 19, 2019 at 16:49

Jmonsky

1,5291 gold badge9 silver badges16 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Marcus Grass · Accepted Answer · 2019-07-19 16:52:48Z

1

You could check if it's an integer.

if check**.5 == int(check**.5):
    ...

answered Jul 19, 2019 at 16:52

Marcus Grass

1,0832 gold badges18 silver badges41 bronze badges

5 Comments

chepner Over a year ago

This is a bad idea in general; for certain large near-squares, this check will return true due to floating-point precision issues.

Marcus Grass Over a year ago

Granted, the best solution to the problem is provided to Jmonsky. Although I often feel that SO answers questions that weren't asked. OP wanted to know how to check if it's a whole number.

chepner Over a year ago

And like I said, this is not a good way to do that, because floating-point values have limited precision. Consider x = 99999999; y = ((x+1)*(x-1))**.5. x**2 != (x+1)*(x-1), but y == int(y).

Marcus Grass Over a year ago

I fold, if OP constructs a 99999999x99999999 dimensional matrix this solution will not work.

chepner Over a year ago

I didn't downvote because in the context of the question, it's probably OK, but I wouldn't rule out the existence of smaller, less-square matrices that give the same false positive.

Collectives™ on Stack Overflow

How to use data type in an if statement

2 Answers 2

Comments

5 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

5 Comments

Linked

Related