Counting consecutive numbers in a list and pick corresponding numbers at the same index from another list

Question

I want to count if consecutive numbers are the same in the first list. If the count is 1, I want to pick the corresponding number in the second list.

lst1 = [[500],[500], [500], [300], [500], [300], [300], [200]]
lst2 = [[10], [10.5], [10.7], [9], [10.1], [97], [10.2], [10.9]]



def trains(lst):
    element = []
    freque = []
    if not lst:
        return element
    count = 1
    for i in range(len(lst)-1):
        if lst[i][0] == lst[i+1][0]:
            count += 1
        else:
            freque.append(count)
            element.append(lst[i][0])
            count = 1
    freque.append(count)
    element.append(lst[i+1][0])

    return element,freque

print(trains(lst1)) # = ([500, 300, 500, 300, 200], [3, 1, 1, 2, 1])

Eventually, I want the result to look like this:

[[300, 9], [500, 10.1], [200, 10.9]]

Asked By: Alghai29

||

Source

Answer 1

Here is a fixed version of your code:

lst1 = [[500],[500], [500], [300], [500], [300], [300], [200]]
lst2 = [[10], [10.5], [10.7], [9], [10.1], [97], [10.2], [10.9]]

def trains(lst1, lst2):
    result = []
    count = 1
    for i in range(1, len(lst1)):
        if lst1[i][0] == lst1[i-1][0]:
            count += 1
        else:
            if count == 1:
                result.append([lst1[i-1][0], lst2[i-1][0]])
            count = 1
    
    if count == 1:
        result.append([lst1[-1][0], lst2[-1][0]])
    return result

print(trains(lst1, lst2))

[[300, 9], [500, 10.1], [200, 10.9]]

Answered By: Jamiu S.

Answer 2

You can also use itertools.zip_longest:

from itertools import zip_longest

zippy = zip_longest(lst1, lst1[1:], lst1[2:], lst2, lst2[1:], lst[2:])
out = [[i1[0], j1[0]] for i0, i1, i2, j0, j1, j2 in zippy if (i1 != i0) & (i1 != i2)]
print(out)

# Output
[[300, 9], [500, 10.1], [200, 10.9]]

Answered By: Corralien

Answer 3

Two input lists can be pairwised (zip), the obtained pairs are then grouped by the adjacent 1st value and filtered by "non-repeated" groups.
With itertools.chain and itertools.groupby operations:

from itertools import chain, groupby

gen_groups = (list(gr) for _, gr in
              groupby(zip(chain.from_iterable(lst1), chain.from_iterable(lst2)), lambda x: x[0]))
res = [list(gr[0]) for gr in gen_groups if len(gr) == 1]
print(res)

[[300, 9], [500, 10.1], [200, 10.9]]

Answered By: RomanPerekhrest

Counting consecutive numbers in a list and pick corresponding numbers at the same index from another list

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