A
Arturo B
Hello, I'm making Python mini-projects and now I'm making a Latin Square
(Latin Square: http://en.wikipedia.org/wiki/Latin_square)
So, I started watching example code and I found this question on Stackoverflow:
http://stackoverflow.com/questions/...-permutations-reduced-latin-squares-in-python
It uses a list comprenhension to generate the Latin Square, I'm am a newbie to Python, and I've tried to figure out how this is evaluated:
a = [1, 2, 3, 4]
n = len(a)
[[a[i - j] for i in range(n)] for j in range(n)]
I don't understand how the "i" and the "j" changes.
On my way of thought it is evaluated like this:
[[a[0 - 0] for 0 in range(4)] for 0 in range(4)]
[[a[1 - 1] for 1 in range(4)] for 1 in range(4)]
[[a[2 - 2] for 2 in range(4)] for 2 in range(4)]
[[a[3 - 3] for 3 in range(4)] for 3 in range(4)]
But I think I'm wrong... So, could you explain me as above? It would help me a lot.
Thanks for reading!
(Latin Square: http://en.wikipedia.org/wiki/Latin_square)
So, I started watching example code and I found this question on Stackoverflow:
http://stackoverflow.com/questions/...-permutations-reduced-latin-squares-in-python
It uses a list comprenhension to generate the Latin Square, I'm am a newbie to Python, and I've tried to figure out how this is evaluated:
a = [1, 2, 3, 4]
n = len(a)
[[a[i - j] for i in range(n)] for j in range(n)]
I don't understand how the "i" and the "j" changes.
On my way of thought it is evaluated like this:
[[a[0 - 0] for 0 in range(4)] for 0 in range(4)]
[[a[1 - 1] for 1 in range(4)] for 1 in range(4)]
[[a[2 - 2] for 2 in range(4)] for 2 in range(4)]
[[a[3 - 3] for 3 in range(4)] for 3 in range(4)]
But I think I'm wrong... So, could you explain me as above? It would help me a lot.
Thanks for reading!