With itertools.chain() on the values This could be faster: from itertools import chain categories = list(chain.from_iterable(categories.values)) Performance from functools import reduce from itertools import chain categories = pd.Series([[‘a’, ‘b’], [‘c’, ‘d’, ‘e’]] * 1000) %timeit list(chain.from_iterable(categories.values)) 1000 loops, best of 3: 231 µs per loop %timeit list(chain(*categories.values.flat)) 1000 loops, best of 3: 237 µs per … Read more
It is no secret that reduce is not among the favored functions of the Pythonistas. Generically, reduce is a left fold on a list It is conceptually easy to write a fold in Python that will fold left or right on a iterable: def fold(func, iterable, initial=None, reverse=False): x=initial if reverse: iterable=reversed(iterable) for e in … Read more
There is no equivalent of foldLeft in Java 8’s Stream API. As others noted, reduce(identity, accumulator, combiner) comes close, but it’s not equivalent with foldLeft because it requires the resulting type B to combine with itself and be associative (in other terms, be monoid-like), a property that not every type has. There is also an … Read more
The return value of Array#push is the new length of the array after the push. This means that in the second iteration acc is a number, which doesn’t have the push method. The fix is simple – separate the push and return statements: const secondArray = [1, 2, 3].reduce((acc, item) => { acc.push(1); return acc; … Read more
Collectors.partitioningBy: Map<Boolean, List<Employee>> partitioned = listOfEmployees.stream().collect( Collectors.partitioningBy(Employee::isActive)); The resulting map contains two lists, corresponding to whether or not the predicate was matched: List<Employee> activeEmployees = partitioned.get(true); List<Employee> formerEmployees = partitioned.get(false); There are a couple of reasons to use partitioningBy over groupingBy (as suggested by Juan Carlos Mendoza): Firstly, the parameter of groupingBy is a Function<Employee, … Read more
You can make elem contain the value by splitting it up in 2 variables: H.reduce(0) {|memo, (key, val)| memo + val}
Seems like your problem is solved already, but there are plenty of easier methods to do it. Excellent one: .filter(Boolean); // will keep every truthy value in an array const arr = [true, false, true, false, true]; const count = arr.filter(Boolean).length; console.log(count); Good one: const arr = [true, false, true, false, true]; const count = … Read more
The other answers are great. I’ll simply add an illustrated example that I find pretty good to understand reduce(): >>> reduce(lambda x,y: x+y, [47,11,42,13]) 113 will be computed as follows: (Source) (mirror)