python how to record the number of elements in listA that appear in the same order as in listB-CodePudding

I am trying to trace to what extent is listA, listB, listC... similar to the original list. How do I print the number of elements that occur in the same sequence in listA as they occur in the original list?

original_list = ['I', 'live', 'in', 'space', 'with', 'my', 'dog']

listA = ['my', 'name', 'my', 'dog', 'is', 'two', 'years', 'old']

listB = ['how', 'where', 'I', 'live', 'in', 'space', 'with']

listC = ['I', 'live', 'to', 'the' 'in', 'space', 'with', 'my', 'football', 'my','dog']

Output: 
listA: Count = 2     #'my', 'dog'

listB: Count = 5     #'I',  'live', 'in', 'space', 'with'

listC: Count = 2,4,2     #'I',  'live'
                     #'in', 'space', 'with', 'my'
                     #'my', 'dog'

CodePudding user response：

I wrote a function that does the job I think. It might be a bit too complex, but I can't see an easier way at the moment:

original = ['I', 'live', 'in', 'space', 'with', 'my', 'dog']

listA = ['my', 'name', 'my', 'dog', 'is', 'two', 'years', 'old']
listB = ['how', 'where', 'I', 'live', 'in', 'space', 'with']
listC = ['I', 'live', 'to', 'the', 'in', 'space', 'with', 'my', 'football', 'my', 'dog']


def get_sequence_lengths(original_list, comparative_list):

    original_options = []
    for i in range(len(original_list)):
        for j in range(i   1, len(original_list)):
            original_options.append(original_list[i:j   1])

    comparative_options = []
    for i in range(len(comparative_list)):
        for j in range(i 1, len(comparative_list)):
            comparative_options.append(comparative_list[i:j 1])
    comparative_options.sort(key=len, reverse=True)

    matches = []
    while comparative_options:
        for option in comparative_options:
            if option in original_options:
                matches.append(option)
                new_comparative_options = comparative_options.copy()
                for l in comparative_options:
                    counter = 0
                    for v in option:
                        counter = counter   1 if v in l else 0
                        if counter == len(l):
                            new_comparative_options.remove(l)
                            break
                comparative_options = new_comparative_options
                break
        if option == comparative_options[-1]:
            break

    matches = [option for option in original_options if option in matches]
    lengths = [len(option) for option in matches]
    print(lengths)
    print(matches)

    return lengths

If you call it with the original list and example lists, it prints the following.
get_sequence_lengths(original, listA) prints [2] [['my', 'dog']].
get_sequence_lengths(original, listB) prints [5] [['I', 'live', 'in', 'space', 'with']].
get_sequence_lengths(original, listC) prints [2, 4, 2] [['I', 'live'], ['in', 'space', 'with', 'my'], ['my', 'dog']].

CodePudding user response：

Off the top of my head, I thought of using the in keyword.

Essentially you could convert your original list into a string list_str = ''.join(original_list)

Then iterate over the other lists and by combining items from each list, such as items = listA[2] listA[3] # items = "mydog", you could then see if they match matched = items in list_str.

I am unaware of other solutions but I'm sure given enough time you would be able to find something better, or someone else would know of something.