FAQ
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/5313900/generating-cyclic-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!

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  • Antoon Pardon at Sep 16, 2013 at 1:53 pm

    Op 16-09-13 15:43, Arturo B schreef:
    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/5313900/generating-cyclic-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!

    Just start your python interpreter and type the following

    [[(i,j) for i in range(3)] for j in range(3)]

    That should give you a clue.


    --
    Antoon Pardon
  • Michael Torrie at Sep 16, 2013 at 2:20 pm

    On 09/16/2013 07:43 AM, Arturo B wrote:
    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:

    It helps to convert it to a conventional for loop to see how it works:


    a = [1, 2, 3, 4]
    n = len(a)


    resultj = []


    for j in range(n):
         resulti = []


         for i in range(n):
             resulti.append(a[i-j])


         resultj.append(resulti)
  • Roy Smith at Sep 17, 2013 at 12:04 am
    In article <eae87c72-f62d-4815-bb69-ca862ff78f1e@googlegroups.com>,
      Arturo B wrote:

    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/5313900/generating-cyclic-permutations-redu
    ced-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)]

    You can re-write any list comprehension as a for loop. In this case you
    have to un-wrap this one layer at a time. First step:


         a = [1, 2, 3, 4]
         n = len(a)
         temp1 = []
         for j in range(n):
             temp2 = [a[i - j] for i in range(n)]
             temp1.append(item)


    then, unwrap the next layer:


         a = [1, 2, 3, 4]
         n = len(a)
         temp1 = []
         for j in range(n):
             temp2 = []
             for i in range(n):
                  temp3 = a[i - j]
                  temp2.append(temp3)
             temp1.append(item)


    Does that make it any easier to understand?

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