Say I have the following loop:
i = 0
l = [0, 1, 2, 3]
while i < len(l):
if something_happens:
l.append(something)
i += 1
Will the len(i) condition being evaluated in the while loop be updated when something is appended to l?
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3If that's your code, it'll never exit for a different reason: at the beginning of the loop, i < len(l). As the loop continues, l can only get bigger, and i stays the same.ojrac– ojrac2009-05-14 17:23:08 +00:00Commented May 14, 2009 at 17:23
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@orjac i think he omitted the "i" increment on purpose for the sake of brevetyalbertein– albertein2009-05-14 17:27:18 +00:00Commented May 14, 2009 at 17:27
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yes... i know that you have to increment i.Allan Lavell– Allan Lavell2009-05-14 17:29:33 +00:00Commented May 14, 2009 at 17:29
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3I assumed that, if it's worth asking SO, he'd tried it and the loop never exited -- which is why I was looking for other bugs. ;)ojrac– ojrac2009-05-14 19:02:14 +00:00Commented May 14, 2009 at 19:02
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2 Answers
Yes it will.
Comments
Your code will work, but using a loop counter is often not considered very "pythonic". Using for works just as well and eliminates the counter:
>>> foo = [0, 1, 2]
>>> for bar in foo:
if bar % 2: # append to foo for every odd number
foo.append(len(foo))
print bar
0
1
2
3
4
If you need to know how "far" into the list you are, you can use enumerate:
>>> foo = ["wibble", "wobble", "wubble"]
>>> for i, bar in enumerate(foo):
if i % 2: # append to foo for every odd number
foo.append("appended")
print bar
wibble
wobble
wubble
appended
appended
