Scala partition python equivalent

Question

I am currently porting some Scala code to Python, and I am wondering what is the most pythonic way to do something similar to the Scala partition ? In particular, in the Scala code, I have a situation where I break up the list of elements based on whether they believe true or false from some filter predicate that I pass:

 val (inGroup,outGroup) = items.partition(filter) 

What is the best way to do something like this in Python?

Answer 1

Using a filter (requires two iterations):

 >>> items = [1,2,3,4,5] >>> inGroup = filter(is_even, items) # list(filter(is_even, items)) in Python 3.x >>> outGroup = filter(lambda n: not is_even(n), items) >>> inGroup [2, 4] >>> outGroup 

Simple loop:

 def partition(item, filter_): inGroup, outGroup = [], [] for n in items: if filter_(n): inGroup.append(n) else: outGroup.append(n) return inGroup, outGroup 

Example:

 >>> items = [1,2,3,4,5] >>> inGroup, outGroup = partition(items, is_even) >>> inGroup [2, 4] >>> outGroup [1, 3, 5] 

Answer 2

Scala

 val (inGroup,outGroup) = items.partition(filter) 

Python - Using List Comprehension

 inGroup = [e for e in items if _filter(e)] outGroup = [e for e in items if not _filter(e)] 

Answer 3

This version is lazy and does not apply the predicate twice to a single element:

 def partition(it, pred): buf = [[], []] it = iter(it) def get(t): while True: while buf[t]: yield buf[t].pop(0) x = next(it) if t == bool(pred(x)): yield x else: buf[not t].append(x) return get(True), get(False) 

Example:

 even = lambda x: x % 2 == 0 e, o = partition([1,1,1,2,2,2,1,2], even) print list(e) print list(o) 

Answer 4

Scala has rich api list handling, and Python as such.

You should read the itertools document. And you can find the registry key.

 from itertools import ifilterfalse, ifilter, islice, tee, count def partition(pred, iterable): ''' >>> is_even = lambda i: i % 2 == 0 >>> even, no_even = partition(is_even, xrange(11)) >>> list(even) [0, 2, 4, 6, 8, 10] >>> list(no_even) [1, 3, 5, 7, 9] # Lazy evaluation >>> infi_list = count(0) >>> ingroup, outgroup = partition(is_even, infi_list) >>> list(islice(ingroup, 5)) [0, 2, 4, 6, 8] >>> list(islice(outgroup, 5)) [1, 3, 5, 7, 9] ''' t1, t2 = tee(iterable) return ifilter(pred, t1), ifilterfalse(pred, t2)