Nested List and Count()

Nested List and count()

Here is yet another approach to flatten a nested sequence. Once the sequence is flattened it is an easy check to find count of items.

def flatten(seq, container=None):
    if container is None:
        container = []

    for s in seq:
        try:
            iter(s)  # check if it's iterable
        except TypeError:
            container.append(s)
        else:
            flatten(s, container)

    return container

c = flatten([(1,2),(3,4),(5,[6,7,['a','b']]),['c','d',('e',['f','g','h'])]])
print(c)
print(c.count('g'))

d = flatten([[[1,(1,),((1,(1,))), [1,[1,[1,[1]]]], 1, [1, [1, (1,)]]]]])
print(d)
print(d.count(1))

The above code prints:

[1, 2, 3, 4, 5, 6, 7, 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
1
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
12

Count occurrences of list element in nested list

This is probably most easily achieved just using a Counter on the first element of each list:

from collections import Counter

c =Counter([l[0] for l in list_all])
list(c.items())

Output:

[('car', 3), ('bicycle', 2), ('taxi', 3), ('motorcycle', 4)]

If you really want a list of lists (as opposed to a list of tuples) in the output, use a list comprehension over c.items():

[list(i) for i in c.items()]

Output:

[['car', 3], ['bicycle', 2], ['taxi', 3], ['motorcycle', 4]]

Recursively counting occurrences in a nested list of numbers

def count(lst, target):
    n = 0
    for i in lst:
        if i == target:
            n += 1
        elif type(i) is list:
            n += count(i, target)
    return n

Counting number of elements in nested list

If you're looking for the number of authors per year, you could use this:

# Authors per year
authors_per_year = { year: len(authors) for year, authors in the_list }

Gives you this:

{1940: 5, 1941: 9, 1942: 9, 1943: 6, 1944: 1}

Or, if you looking for a count of unique authors then you could use this:

# Unique authors
unique_authors = set([ a for year, authors in the_list 
                           for a in authors])

Gives you this set:

set(['Bousfield G',
     'Bruner DW',
     'Burr GO',
     'Crawford GN',
     'Dong L',
     'Edwards PR',
     'Faber HK',
     'Fishbein M',
     'Gaffron H',
     'Gardner FT',
     'Gates RR',
     'Gould BS',
     'Greene HS',
     'Haas O',
     'Hatch MH',
     'Hill DK',
     'Howorth MB',
     'Hughes H',
     'Lake NC',
     'Lewis WH',
     'McClung CE',
     'Myers J',
     'Ratner B',
     'Ritter WE',
     'Rubin J',
     'Silverberg RJ',
     'Sumner FB',
     'Tytell AA'])

So len(unique_authors) gives you a count of 28.

In any case, I think the way forward for you may well be to use some combination of list comprehensions or dict comprehension.

Count the number of terms that are not atoms in a nested list

One possible solution would be the following:

count(C) :-
    findall(X, vehicle(_, blue, X), Ls),
    countOpt(Ls, 0, C).

countOpt([], X, X) :- !.
countOpt([H|T], C, NewC) :-
    countOpt(T, C, NewC1),
    findall(Opt, member(optional(Opt), H), Opts),
    printOpts(Opts),
    length(Opts, Length),
    NewC is NewC1 + Length, !.

printOpts([]).
printOpts([H|T]) :-
    print(H),
    nl,
    printOpts(T).

As in your approach, first gather all lists of features (I guess?) for each vehicle and save it in a List of lists called Ls.

Then select in each sublist of Ls all Optional values (Opt) and add all the lengths of them.

I also added the predicate to print the results.

How to count a nested collection in a List with Java Streams

I think you are looking for something like

artifactList.stream()
            .flatMap(artifact -> artifact.getFindings().stream())
            .filter(finding -> finding.isChecked())
            .count();

Calculate count of all the elements in nested list

Here is one way.

from collections import Counter
from itertools import chain

test = [["P1", "P1", "P1", "P2", "P2", "P1", "P1", "P3"],
        ["P1", "P1", "P1"],
        ["P1", "P1", "P1", "P2"],
        ["P4"],
        ["P1", "P4", "P2"],
        ["P1", "P1", "P1"]]

c = Counter(chain.from_iterable(test))

for k, v in c.items():
    print(k, v)

# P1 15
# P2 4
# P3 1
# P4 2    

For output as dataframe:

df = pd.DataFrame.from_dict(c, orient='index').transpose()

#    P1 P2 P3 P4
# 0  15  4  1  2

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