Is there a python way/instruction to make a function execute only in first iteration of while loop?

Question

i = SomeIndex()
while mylist[i] is not None:
   if mylist[i] == name:
       return
   foo()
   i+=1

I want foo() to always run on 1st iteration of loop, if mylist[i] isn't 'name', but never run if its any iteration but the first. I know I could the following, but I don't know if it's the most efficient and prettiest python code:

i = SomeIndex()
FirstIter = True
while mylist[i] is not None:
   if mylist[i] == name:
       return
   if FirstIter:
       foo()
       FirstIter = False
   i+=1

Answer 1

Personally I like mark_ends from the third party library more-itertools

from more_itertools import mark_ends

i = SomeIndex()
for first, last, elem in mark_ends(mylist[i:]):
    if elem == name:
        return
    if first:
        foo()

mark_ends gives you a 3-tuple for every element in your iterable, in this case the sliced mylist. The tuples are (True, False, elem_0), (False, False, elem_1), ..., (False, False, elem_n-2), (False, True, elem_n-1). In your use case you never use the middle element of the tuples.

If for some reason you can't or don't want to use the external library you can swipe the code from https://more-itertools.readthedocs.io/en/stable/_modules/more_itertools/more.html#mark_ends

Addendum:

In light of the OP's requirement to let foo change the list, here's a quick modification:

from more_itertools import mark_ends

i = SomeIndex()
for first, last, (j, elem) in mark_ends(enumerate(mylist[i:], start=i)):
    if elem == name:
        return
    if first:
        foo(mylist, j)

j now gives you the index that you need to tell foo what to change.

Answer 2

You are trying to emulate a do-while, take a look at this question if you want.

Since there is no do-while equivalent in Python, the simple idea is to move the first iteration out of the loop

i = SomeIndex()
foo()
while mylist[i] is not None:
   if mylist[i] == name:
       return
   foo()
   i+=1

Also, the way you iterate over the list is unnecessarily complex, you can simplify it like:

i = SomeIndex()
if myList[i] != name:
  foo()
for element in myList[i+1:]:
  if element == name:
    return    
  foo()

Answer 3

Simplify your logic to independent steps. Yes, you will make a Boolean value check a second time. This takes far less time than you spent with your design problem.

start = SomeIndex()
if mylist[start] != name:
    foo()

for idx in range(start, len(mylist)): 
   if mylist[idx] == name:
       return
   # remainder of loop

Answer 4

Let's "pythonize" your example, step by step.

1. Remove the first_index flag:

start_idx = SomeIndex()
i = start_idx

while mylist[i] is not None:
    if mylist[i] == name:
        return
    if i == start_idx:
        foo()
    i += 1

2. Convert to while True:

start_idx = SomeIndex()
i = start_idx

while True:
    if mylist[i] is not None:
        break
    if mylist[i] == name:
        return
    if i == start_idx:
        foo()
    i += 1

3. Convert to for i in range loop:

start_idx = SomeIndex()

for i in range(start_idx, len(mylist)):
    if mylist[i] is not None:
        break
    if mylist[i] == name:
        return
    if i == start_idx:
        foo()

Note that this also fixes a bug by bounding i < len(mylist).

---

Variations

These are not equivalent to your code sample, but they might be relevant patterns for you to use.

Variation 1:

Find the index where mylist[i] == name:

index = next(
    (
        i
        for i in range(start_idx, len(mylist))
        if mylist[i] == name
    ),
    None,  # If no index is found, set to None.
)