python list sequential numbers

Question

I am trying to figure out a way to find whether all the integers in my list are increasing sequentially. If any numbers are non-sequential I want to get the index.

mylist = [1,2,3,2]

So, is 2 more than 1 (yes), is 3 more than 2 (yes), is 2 more than 3(no, get index).

I can compare an unsorted list against a sorted list to work out if it is sequential, but I also want to establish the index. Is there a way to do this?

Thanks Toby

Answer 1

I guess you are looking for something like this:

>>> mylist = [1,2,3,2,3,5,6,3,7,8,6]
>>> [i+1 for i in range(len(mylist)-1) if mylist[i]>mylist[i+1]]
[3, 7, 10]

Answer 2

You can use enumerate to do this for you:

In [1]: mylist = [1,2,3,2]

In [2]: for idx, element in enumerate(mylist):
   ...:     if idx+1 != element:
   ...:         print element, idx
   ...:         
2 3

In case your sequence can start from any number,

In [1]: mylist = [2,3,4,2,7]

In [2]: for idx, element in enumerate(mylist):
   ....:    if idx + mylist[0] != element:
   ....:           print element, idx
   ....:         
2 3
7 4

Answer 3

This should do the trick:

def nonincr_index(l):
    if not l:
        return 0
    prev=l[0]
    for n,i in enumerate(l):
        if i<prev:
            return n
        prev=i
    return len(l)

>>> nonincr_index([1,2,3,2])
3

It returns the index of the non increasing item.