How do I manipulate a multi dimension array in a python not using a library?

Question

I have this method:

def encode(message, rails):
    result = ""
    dataRails = []
    activeRail = 0
    # allocate multi-dimensional array...seems inefficient
    for r in range(rails):
        dataRails.append([])

    # copy characters from message into one of the arrays, or .
    for m in message:
        count = 0
        # for each rail either give it the character or a .
        while count <= rails:
            print("Count: ", str(count), " Active Rail:", str(activeRail))
            if activeRail == count:
                dataRails[activeRail][count] = m
            else:
                dataRails[activeRail][count] = "."
            count += 1

        activeRail += 1
        if activeRail >= rails:
            activeRail = 0

    for r in dataRails:
        print(r)
    # ill calculate result once I get the above right    
    return result

Message is a string of text. Rails is a number. The encode method would be called like this:

encode('XOXOXOXOXOXOXOXOXO', 2)

In this case, I want to allocate my dataRails as 2 elements, each the same length as input message. I would replace elements in the arrays with either a letter from message or a ".".

But, it seems rather inefficient to use a loop to allocate dataRails.

When I run this, I get the following error:

dataRails[activeRail][count] = m
IndexError: list assignment index out of range

The print statement looks like valid index values:

Count:  0  Active Rail: 0

It feels like there are two problems with my multi-dimensional array. 1) allocation and 2) access.

If there is a duplicate question, please post a link. I searched through stackoverflow. Answers that seemed relevant relied on a library, which I did not want to use. I have just started learning python and want to learn the basics of python, not using libraries.

Are arrays really that difficult in python or did I make it more difficult than it needs to me?

Thanks Matt

Edit: Ultimately, if the message input was this "WE ARE DIS", the return of this method would be "WEERDSAI". I am working through a tutorial on exercism.io. I haven't completed the method yet. I am stuck on the array I defined.

Can you show some desired inputs/outputs? For your multidimensional arrays, you can do this: dataRails = [[] for _ in range(rails)] — user3483203
– user3483203, Commented Mar 1, 2018 at 18:19
Well, for starters you're dealing with lists. Arrays are a separate data type that can't change size dynamically. However, even with lists you can't access elements beyond the list size, so you probably change the assignment to ... .apppend(m) — user2699
– user2699, Commented Mar 1, 2018 at 18:24
Why is that the expected output? You've used while and counters instead of for loops, multiple checks on one variable, and skipped string slices altogether. A simple Caesar cipher would be much shorter, so I'm not sure what you're trying to accomplish; the example has too many duplicate letters to be intuitive. — Prune
– Prune, Commented Mar 1, 2018 at 18:28
@Prune the purpose is to implement a rail fence cipher, not a Caesar cipher. A Caesar cipher may be simpler, but that's not the exercise. I did reduce the example. Hopefully that helps. However, the purpose of this post isn't to solve the problem for me. The purpose is to help me fix my array manipulation. — tatmanblue
– tatmanblue, Commented Mar 1, 2018 at 19:53
Got it; the cypher name was enough of a clue. All you need is an array of strings, one string per rail. The above append comments should get you to the next set of problems. :-) — Prune
– Prune, Commented Mar 1, 2018 at 20:06

Prune · Accepted Answer · 2018-03-01 20:11:35Z

1

I suggest that you keep a list of strings, rather than a multi-dimensional list. Just a couple of simple changes:

# Initialize empty rails
dataRails = [""] * rails
...
        if activeRail == count:
            dataRails[activeRail] += m
        else:
            dataRails[activeRail] += '.'

I tested with "We Are Dis":

W...A... ...s...
.e...r...D..
.. ...e...i.

Now you can deal with removing non-letters and properly reversing direction. Is that good for now?

answered Mar 1, 2018 at 20:11

Prune

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1 Comment

tatmanblue Over a year ago

yes, that is clear. I was over complicating the data structure. thank you

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