Return the first item in a list matching a condition [duplicate]

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.

Answer 2

Use the break statement:

for x in lis:   if matchCondition(x):      print x      break            #condition met now break out of the loop 

now x contains the item you wanted.

proof:

>>> for x in xrange(10):    ....:     if x==5:    ....:         break    ....:           >>> x >>> 5