Comparable code to array[++i] and array[i++] in Python?

Question

In C, as well as in C++, one can in a for-loop change the index (for example i). This can be useful to, for example, compare a current element and based on that comparison compare the next element:

for(int i = 0; i < end; i++)
  if(array[i] == ':')
    if(array[++i] == ')')
      smileyDetected = true;

Now I know this cannot be done in Python (for various reasons). However, I cannot help wondering if there are short alternatives for Python? I can come up with:

while i < end:
    if array[i] == ':':
        i += 1
        if array[i] == ')':
            smileyDetected = True;

However, this costs me an extra line, which doesn't sound so bad until you do the same multiple times ('less readable' did not mean having a long file). So to have it in one line, I would think of something like array[i += 1], but this is invalid syntax as it seems.

Is there Python equivalent which does the incrementation of the index in the same line as reading out that incremented index?

EDIT:
As most answers mention using in to find a substring, as an alternative for the particular example, let me add another example which wouldn't be solvable in such a way:

j = 0;
for(int i = 0; i < end; i++)
  if(array[i] == ':')
    if(array[++i] == ')')
      anotherArray[j++] = array[++i];

With that I mean it's about the incrementing of the index, not the finding of a particular string.

The C/C++ will show undefined behavior if the array is no C-string and ':' is the last character. — , May 14 '16 at 15:45
If `array` is a string you could just write `smiley_detected = ':)' in array`. — kennytm, May 14 '16 at 16:00
SImplify like this (non-pythonic way and langauge agnostic) while i < end: if array[i] == ':' and array[i+1]==')': smileyDetected = True i += 1 — manugupt1, May 14 '16 at 16:04
@DieterLücking Not if `end` is smaller than the last index of `array` ;) @jonrsharpe I want to find what's after the smiley, not if the smiley is there. — Simon Klaver, May 14 '16 at 18:21

cdlane · Answer 1 · 2016-05-14T17:39:17.617

2

Perhaps:

smileyDetected = ':)' in "".join(array)

or per @jonrsharpe:

from itertools import tee

# pairwise() from "Itertools Recipes"

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)

for a, b in pairwise(array):
    if a == ':' and b == ')':
         smileyDetected = True

edited May 14 '16 at 17:39

answered May 14 '16 at 16:02

cdlane

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The first one is wrong with `array = ['foo:', ')bar']` – Valentin Lorentz May 14 '16 at 16:04
1

Please don't `import *`. – L3viathan May 14 '16 at 16:05
@ValentinLorentz your example doesn't match OP's example code but my first solution does in fact handle it correctly. Better posted as a follow up comment to jonrsharpe's pairwise commet to OP. – cdlane May 14 '16 at 16:08
@cdlane oh sorry, I misunderstood the question (I thought it was an array of strings instead of an array of chars) – Valentin Lorentz May 14 '16 at 16:09

score 1 · Answer 2 · answered May 14 '16 at 16:54

1

If array is a string, the shortest pythonic version would obviously be:

smileyDetected = ':)' in array

A more generic / non-string-specific way of doing this would be using iterators:

smileyDetected = (':', ')') in zip(array, array[1:])

answered May 14 '16 at 16:54

aldanor

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score 0 · Answer 3 · answered May 14 '16 at 16:05

0

In this special case, you could do:

for i, char in enumerate(array[:-1]):
    if char == ":" and array[i+1] == ")":
        smiley_detected = True

However, in the more general case, if you need to skip elements, you could modify the raw iterator:

iterator = iter(array)
for char in iterator:
    if char == ":" and next(iterator) == ")":
        smiley_detected = True

Here you need to take more care about the array bounds; if the last element is :, you will get a StopIteration exception from calling next on the depleted iterator. You would need to catch that.

answered May 14 '16 at 16:05

Jonas Schäfer

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The second one is wrong for `array = ['a', ':', ')', 'b']`, because it calls next() twice on the **same** iterator every loop. – Valentin Lorentz May 14 '16 at 16:12
@ValentinLorentz No, it only calls next twice when the current element is ``:``, as the ``and`` operator is short-circuiting. For the given search pattern this works, because it cannot start with ``)``, thus it does not matter if ``)`` is skipped. – Jonas Schäfer May 15 '16 at 08:15

Comparable code to array[++i] and array[i++] in Python?

3 Answers3