To test if each row in your input file contains magic square data you need to re-organize the code slightly. I've used a different technique to Francis to fill the matrix. It might be a bit harder to understand how zip(*[iter(seq)] * size) works, but it's a very useful pattern. Please let me know if you need an explanation for it.
My code uses a list of tuples for the matrix, rather than a list of lists, but tuples are more suitable here anyway, since the data in the matrix doesn't need to be modified. Also, I convert the input data from str into int, since you need to do arithmetic on the numbers to test if matrix is a magic square.
#! /usr/bin/env python
def make_square(seq, size):
return zip(*[iter(seq)] * size)
def main():
fname = 'mydata'
size = 3
with open(fname, 'r') as f:
for line in f:
nums = [int(s) for s in line.split()]
matrix = make_square(nums, size)
print matrix
#Now call the function to test if the data in matrix
#really is a magic square.
#test_square(matrix)
if __name__ == '__main__':
main()
Here's a modified version of make_square() that returns a list of lists instead of a list of tuples, but please bear in mind that a list of tuples is actually better than a list of lists if you don't need the mutability that lists give you.
def make_square(seq, size):
square = zip(*[iter(seq)] * size)
return [list(t) for t in square]
I suppose I should mention that there's actually only one possible 3 x 3 magic square that uses all the numbers from 1 to 9, not counting rotations and reflections. But I guess there's no harm in doing a brute-force demonstration of that fact. :)
Also, I have Python code that I wrote years ago (when I was first learning Python) which generates magic squares of size n x n for odd n >= 5. Let me know if you'd like to see it.
zip and iterator objects
Here's some code that briefly illustrates what the zip() and iter() functions do.
''' Fun with zip '''
numbers = [1, 2, 3, 4, 5, 6]
letters = ['a', 'b', 'c', 'd', 'e', 'f']
#Using zip to create a list of tuples containing pairs of elements of numbers & letters
print zip(numbers, letters)
#zip works on other iterable objects, including strings
print zip(range(1, 7), 'abcdef')
#zip can handle more than 2 iterables
print zip('abc', 'def', 'ghi', 'jkl')
#zip can be used in a for loop to process two (or more) iterables simultaneously
for n, l in zip(numbers, letters):
print n, l
#Using zip in a list comprehension to make a list of lists
print [[l, n] for n, l in zip(numbers, letters)]
#zip stops if one of the iterables runs out of elements
print [[n, l] for n, l in zip((1, 2), letters)]
print [(n, l) for n, l in zip((3, 4), letters)]
#Turning an iterable into an iterator object using the iter function
iletters = iter(letters)
#When we take some elements from an iterator object it remembers where it's up to
#so when we take more elements from it, it continues from where it left off.
print [[n, l] for n, l in zip((1, 2, 3), iletters)]
print [(n, l) for n, l in zip((4, 5), iletters)]
#This list will just contain a single tuple because there's only 1 element left in iletters
print [(n, l) for n, l in zip((6, 7), iletters)]
#Rebuild the iletters iterator object
iletters = iter('abcdefghijkl')
#See what happens when we zip multiple copies of the same iterator object.
print zip(iletters, iletters, iletters)
#It can be convenient to put multiple copies of an iterator object into a list
iletters = iter('abcdefghijkl')
gang = [iletters] * 3
#The gang consists of 3 references to the same iterator object
print gang
#We can pass each iterator in the gang to zip as a separate argument
#by using the "splat" syntax
print zip(*gang)
#A more compact way of doing the same thing:
print zip(* [iter('abcdefghijkl')]*3)
Here's the same code running in the interactive interpreter so you can easily see the output of each statement.
>>> numbers = [1, 2, 3, 4, 5, 6]
>>> letters = ['a', 'b', 'c', 'd', 'e', 'f']
>>>
>>> #Using zip to create a list of tuples containing pairs of elements of numbers & letters
... print zip(numbers, letters)
[(1, 'a'), (2, 'b'), (3, 'c'), (4, 'd'), (5, 'e'), (6, 'f')]
>>>
>>> #zip works on other iterable objects, including strings
... print zip(range(1, 7), 'abcdef')
[(1, 'a'), (2, 'b'), (3, 'c'), (4, 'd'), (5, 'e'), (6, 'f')]
>>>
>>> #zip can handle more than 2 iterables
... print zip('abc', 'def', 'ghi', 'jkl')
[('a', 'd', 'g', 'j'), ('b', 'e', 'h', 'k'), ('c', 'f', 'i', 'l')]
>>>
>>> #zip can be used in a for loop to process two (or more) iterables simultaneously
... for n, l in zip(numbers, letters):
... print n, l
...
1 a
2 b
3 c
4 d
5 e
6 f
>>> #Using zip in a list comprehension to make a list of lists
... print [[l, n] for n, l in zip(numbers, letters)]
[['a', 1], ['b', 2], ['c', 3], ['d', 4], ['e', 5], ['f', 6]]
>>>
>>> #zip stops if one of the iterables runs out of elements
... print [[n, l] for n, l in zip((1, 2), letters)]
[[1, 'a'], [2, 'b']]
>>> print [(n, l) for n, l in zip((3, 4), letters)]
[(3, 'a'), (4, 'b')]
>>>
>>> #Turning an iterable into an iterator object using using the iter function
... iletters = iter(letters)
>>>
>>> #When we take some elements from an iterator object it remembers where it's up to
... #so when we take more elements from it, it continues from where it left off.
... print [[n, l] for n, l in zip((1, 2, 3), iletters)]
[[1, 'a'], [2, 'b'], [3, 'c']]
>>> print [(n, l) for n, l in zip((4, 5), iletters)]
[(4, 'd'), (5, 'e')]
>>>
>>> #This list will just contain a single tuple because there's only 1 element left in iletters
... print [(n, l) for n, l in zip((6, 7), iletters)]
[(6, 'f')]
>>>
>>> #Rebuild the iletters iterator object
... iletters = iter('abcdefghijkl')
>>>
>>> #See what happens when we zip multiple copies of the same iterator object.
... print zip(iletters, iletters, iletters)
[('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('j', 'k', 'l')]
>>>
>>> #It can be convenient to put multiple copies of an iterator object into a list
... iletters = iter('abcdefghijkl')
>>> gang = [iletters] * 3
>>>
>>> #The gang consists of 3 references to the same iterator object
... print gang
[<iterator object at 0xb737eb8c>, <iterator object at 0xb737eb8c>, <iterator object at 0xb737eb8c>]
>>>
>>> #We can pass each iterator in the gang to zip as a separate argument
... #by using the "splat" syntax
... print zip(*gang)
[('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('j', 'k', 'l')]
>>>
>>> #A more compact way of doing the same thing:
... print zip(* [iter('abcdefghijkl')]*3)
[('a', 'b', 'c'), ('d', 'e', 'f'), ('g', 'h', 'i'), ('j', 'k', 'l')]
>>>
