Python - Finding multiple values in python array list in order

One solution:

arr = [('var1', 'val1'), ('var2', 'val2'),
       ('var3', 'val3'), ('var4', 'val4'),
       ('var1', 'val5'), ('var3', 'val6'),
       ('var5', 'val7'), ('var1', 'val8'),
       ('var3', 'val9'), ('var6', 'val10')]

finds = ['var1', 'var1 var3', 'var1 var3 var6']

keys, values = zip(*arr)
for find in finds:
    b = tuple(find.split())

    # find the first index of the matching sequence
    index = next(i for i in range(len(arr)) if keys[i:i + len(b)] == b)

    # join the result from the matching index to the required length
    res = " ".join(values[index:index + len(b)])
    print(res)

Output

val1
val5 val6
val8 val9 val10

As an alternative use difflib.SequenceMatcher:

from difflib import SequenceMatcher

arr = [('var1', 'val1'), ('var2', 'val2'),
       ('var3', 'val3'), ('var4', 'val4'),
       ('var1', 'val5'), ('var3', 'val6'),
       ('var5', 'val7'), ('var1', 'val8'),
       ('var3', 'val9'), ('var6', 'val10')]

finds = ['var1', 'var1 var3', 'var1 var3 var6']

keys, values = zip(*arr)
for find in finds:
    b = find.split()
    s = SequenceMatcher()
    s.set_seqs(keys, b)
    match = s.find_longest_match()
    res = " ".join(values[match.a:match.a+match.size])
    print(res)

Output

val1
val5 val6
val8 val9 val10

I would recommend transforming your list into the dictionary:

dct = {i[0]:i[1] for i in arr}

Then use this function to return multiplies values from your dictionary as your described in your question:

def findMultiple(dct, find):
    find = find.split(" ")
    results = [dct[key] for key in find]
    return " ".join(results)

Usage:

>>> findMultiple(dct, "var1")
'val8'
>>> findMultiple(dct, "var1 var3")
'val8 val9'
>>> findMultiple(dct, "var1 var3 var6")
'val8 val9 val10'