For loops and multiple conditions

Question

In C++, it's possible to say:

for (int i = 0; i < 100 && !found; i++) {
  if (items[i] == "the one I'm looking for")
    found = true;
}

so you don't need to use the "break" statement.

In Python, I guess you need to write:

found = False

for item in items:
    if item == "the one I'm looking for"
        found = True
        break

I know that I can write a generator which has the same code in it so I can hide this break thing. But I wonder if there is any other way to implement the same thing (with the same performance) without using extra variables or while loop.

I know we can say:

found = "the one I'm looking for" in items

I'm just trying to learn if it's possible to use multiple conditions in for loops.

Thanks.

Answer 1

>>> from itertools import dropwhile
>>> try:
...     item = next(dropwhile(lambda x: x!="the one I'm looking for", items))
...     found = True
... except:
...     found = False

Of course you can also write this without the lambda function as

>>> from itertools import dropwhile
>>> try:
...     item = next(dropwhile("the one I'm looking for".__ne__, items))
...     found = True
... except:
...     found = False

Now it looks to me that it is the C version using extra variables

If you really just need the found variable set (and don't need to know the item), then simply use

found = any(item=="the one I'm looking for" for item in items)

Answer 2

Since for loops in Python iterate over a sequence rather than a condition and a mutation statement, the break is necessary to bail out early. In other words, for in python isn't a conditional loop. The Python equivalent to C++'s for would be a while loop.

i=0
found=False
while i < 100 and !found:
    if items[i] == "the one I'm looking for":
        found=True
    i += 1

Even in C++, for loops can be rewritten as while loops if they don't contain a continue statement.

{
    int i = 0;
    while (i < 100 && !found) {
        if (items[i] == "the one I'm looking for")
            found = true;
        i++;
    }
}