Python: Progress the index for a for loop

Question

Is there some way to move the index forward in a for loop in python? For example, if I'm iterating through a string and find some char that I'm looking for, can I begin reading from that position on the next iteration of the for loop?

Like, after reading an x, read until a y and start from the position following the y on the next iteration:

for i in range(len(myString)):
  if myString[i] == 'x':
    j = i + 1
    while( myString[j] != '\n' ):
      if( myString[j] == 'y' ):
        i = j + 1  # start at position following y in string on the next iteration
      j = j + 1

Or do I have to use a while loop to achieve this?

I believe you'd have to use a while loop, or use the keyword continue in the case that you want to ignore. — jacob
– jacob, Commented Jan 18, 2016 at 23:00
Possible duplicate of python arbitrarily incrementing an iterator inside a loop — LinkBerest - SO sold our work
– LinkBerest - SO sold our work, Commented Jan 18, 2016 at 23:01
Nope, definitely not a duplicate, @JGreenwell - none of the examples there deal with reversing an iterator — Wayne Werner
– Wayne Werner, Commented Jan 18, 2016 at 23:06

Ali · Accepted Answer · 2016-01-18 23:22:54Z

2

You can use an string instead of list in this sample code. It finds the first x in the remaining part of the list, then finds the first y after that 'x' and so on

#mylist = [list to search]
i = 0
while true:
    list_to_search = mylist[i:]
    if 'x' not in list_to_search:
        break
    j = list_to_search.index('x')
    if 'y' not in list_to_search[j+1:]:
        break
    i = list_to_search.index('y')
    (x_pos,y_pos)  = (j,i)
    #your code

edited Jan 18, 2016 at 23:22

answered Jan 18, 2016 at 23:17

Ali

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1 Comment

Victor Brunell Over a year ago

@ alivar Nice. I didn't think to use splicing.

Christian Tapia · Accepted Answer · 2016-01-18 23:36:20Z

1

You could try using an infinite loop:

while true:
   if i >= len(myString): # Exit condition
      break

   # do something with myString[i]

   # set i as you want

edited Jan 18, 2016 at 23:36

answered Jan 18, 2016 at 23:07

Christian Tapia

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2 Comments

Victor Brunell Over a year ago

@ Christian That's a nice alternative. I'll give it a go. Thanks.

Victor Brunell Over a year ago

I think that should be if i >= len(myString). Otherwise it goes out of bounds when i == len(myString). Otherwise, this works well for my purposes.

Wayne Werner · Accepted Answer · 2016-01-18 23:26:55Z

1

Now, effectively you're still using a while loop here - but I think this comes closer in spirit to what you were thinking about:

def myrange(limit=0):
    counter = 0
    while counter < limit:
        next = yield counter
        try:
            counter = int(next) - 1
        except TypeError:
            counter += 1    


mylist = ['zero', 'one', 'two', 'five', 'four']       
my_iter = myrange(len(mylist))

for i in my_iter:
    print(mylist[i])
    if mylist[i] == 'five':
        print('-- three, sir! --')
        mylist[i] = 'three!'
        my_iter.send(i)
    elif mylist[i] == 'four':
        print("Don't count that number!")

This works because of the send function that generators have. They allow you to send a value to the generator that the generator can then use how it wishes. Note that when you call .send() on a generator that it will yield the next value (so when we call my_iter.send(i), it's actually yielding the next value. That's why we call counter = int(next) - 1. The alternative would be to put the -1 in your for loop.

answered Jan 18, 2016 at 23:26

Wayne Werner

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1 Comment

Victor Brunell Over a year ago

@ Wayne Werner I'm pretty new to python, so I think this example is a bit above my pay grade for now. I'll definitely play around with it though. Thanks for the feedback.

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