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I'm new to python and programming in general, so I feel like I'm trying to wrap my head around the simplest of things...

Let's say I have a list of 12 items:

a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

and a variable that matches one of the items in the list:

b = 7

Now I want to find that match in the list and move every item before the match to the end of the list in the same order like so:

a = [7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6]

How would I go about doing this in python 3.4.2?

The match could be for any item at any index, but the number of items will always be the same (12).

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2 Answers 2

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Use slicing of list:

>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] >>> idx = a.index(7) >>> a = a[idx:] + a[:idx] >>> a [7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6]

Note that this will throw a ValueError if the value being searched is not found in the array, in which case you will need to capture it using try-except block

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Thank-you. I haven't really learned about .index() yet. I guess that's why I couldn't find a solution to my problem. 3 hours later and I've finished creating a small program that creates different musical chords depending on the key the music is in (defined by the user). :P Just for practice.
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Another option is to simultaneously use append and pop:

>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] >>> idx = a.index(7) >>> for _ in range(idx): ... a.append(a.pop(0)) >>> a [7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6]

I do not advocate using this over the slicing method because this will be very slow (lists are not optimized for poping from the front). I just mention it here to show another method that I believe is a little bit easier to understand at first glance.

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