How Can I Create Word Count Output In Python Just By Using Reduce Function?

I have the following list of tuples: [('a', 1), ('a', 1), ('b', 1), ('c',1), ('a', 1), ('c', 1)] I would like to know if I can utilize python's reduce function to aggregate them an

Solution 1:

It seems difficult to achieve using reduce, because if both tuples that you "reduce" don't bear the same letter, you cannot compute the result. How to reduce ('a',1) and ('b',1) to some viable result?

Best I could do was l = functools.reduce(lambda x,y : (x[0],x[1]+y[1]) if x[0]==y[0] else x+y,sorted(l))

it got me ('a', 3, 'b', 1, 'c', 1, 'c', 1). So it kind of worked for the first element, but would need more than one pass to do the other ones (recreating tuples and make another similar reduce, well, not very efficient to say the least!).

Anyway, here are 2 working ways of doing it

First, using collections.Counter counting elements of the same kind:

l = [('a', 1), ('a', 1), ('b', 1), ('c',1), ('a', 1), ('c', 1)]

importcollectionsc= collections.Counter()
for a,i in l:
    c[a] += i

We cannot use listcomp because each element has a weight (even if here it is 1)

Result: a dictionary: Counter({'a': 3, 'c': 2, 'b': 1})

Second option: use itertools.groupby on the sorted list, grouping by name/letter, and performing the sum on the integers bearing the same letter:

print ([(k,sum(e for _,e in v)) for k,v in itertools.groupby(sorted(l),key=lambda x : x[0])])

result:

[('a', 3), ('b', 1), ('c', 2)]

Solution 2:

The alternative approach using defaultdict subclass and sum function:

import collections

l = [('a', 1), ('a', 1), ('b', 1), ('c',1), ('a', 1), ('c', 1)]
d = collections.defaultdict(list)
for t in l:
    d[t[0]].append(t[1])

result = [(k,sum(v)) for k,v in d.items()]
print(result)

The output:

[('b', 1), ('a', 3), ('c', 2)]

Solution 3:

Another way is that to create your custom reduce function. for example: l = [('a', 1), ('a', 1), ('b', 1), ('c',1), ('a', 1), ('c', 1)]

defmyreduce(func , seq):
    output_dict = {}
    for k,v in seq:
        output_dict[k] = func(output_dict.get(k,0),v)
    return output_dict  

myreduce((lambda sum,value:total+sum),l)

output: {'a': 3, 'b': 1, 'c': 2}

later on you can modify the generated output as a list of tuples.