Question

I have a function. matchCondition(a), which takes an integer and either returns True or False.

I have a list of 10 integers. I want to return the first item in the list (in the same order as the original list) that has matchCondition returning True.

As pythonically as possible.

Answer 1

next(x for x in lst if matchCondition(x)) 

should work, but it will raise StopIteration if none of the elements in the list match. You can suppress that by supplying a second argument to next:

next((x for x in lst if matchCondition(x)), None)

which will return None if nothing matches.

Demo:

>>> next(x for x in range(10) if x == 7)  #This is a silly way to write 7 ...
7
>>> next(x for x in range(10) if x == 11)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration
>>> next((x for x in range(10) if x == 7), None)
7
>>> print next((x for x in range(10) if x == 11), None)
None

---

Finally, just for completeness, if you want all the items that match in the list, that is what the builtin filter function is for:

all_matching = filter(matchCondition,lst)

In python2.x, this returns a list, but in python3.x, it returns an iterable object.