Question

There must be a simpler, more pythonic way of doing this.

Given this list of pairs:

pp = [('a',1),('b',1),('c',1),('d',2),('e',2)]

How do I most easily find the first item in adjacent pairs where the second item changes (here, from 1 to 2). Thus I’m looking for [‘c’,’d’]. Assume there will only be one change in pair[1] for the entire list, but that it may be a string.

This code works but seems excruciatingly long and cumbersome.

for i, pair in enumerate(pp):
    if i == 0: 
        pInitial = pair[0] 
        sgInitial = pair[1]
    pNext = pair[0]
    sgNext = pair[1]
    if sgInitial == sgNext:
        sgInitial = sgNext
        pInitial = pNext
    else:
        pOne = pInitial
        pTwo = pNext
        x = [pOne, pTwo]
        print x
        break

Thanks
Tim

Answer 1

import itertools as it

pp = [('a',1),('b',1),('c',1),('d',2),('e',2)]

# with normal zip and slicing
for a,b in zip(pp,pp[1:]):
    if a[1] != b[1]:
        x=(a[0],b[0])
        print x
        break
# with generators and izip
iterfirst = (b for a,b in pp)
itersecond = (b for a,b in pp[1:])
iterfirstsymbol = (a for a,b in pp)
itersecondsymbol = (a for a,b in pp[1:])
iteranswer = it.izip(iterfirstsymbol, itersecondsymbol, iterfirst, itersecond)

print next((symbol1, symbol2)
           for symbol1,symbol2, first, second in iteranswer
           if first != second)

Added my readable generator version.