How to find index from a tuple by using elements from another tuple

Question

I have following info tuple, how do i find the index of the of tuple[?] from name tuple?

info  = [('Water SPF 15', '12B', '#f2c9b5', 31, '-100.1', '-100.1', '-100.1', '12N', '', '', '27B'), ('Water Spectrum SPF 15', '12B', '#f2c9b5', 31, '7..7', '7..7', '7..7', '12N', '', '', '27B'), ('Water Foundation SPF 15', '12N', '#e5be9d', 44, ' ', ' ', ' ', '', '12B', '', '15N'), ('FACE15', '12N', '#e5be9d', 44, '**', '**', '**', '12S', '14H', '', '16N'), ('FACE', '12B', '#f2c9b5', 31, 'asd', 'asd', 'asd', '14H', '', '8B', '18B'), ('hydrator', '10N', '#e5be9d', 44, '', '', '', '', '', '', '13N'), ('APE', '12B', '#e7cbb3', 39, '-100', '-100', '-100', '', '', '', '')]
name = [('Water Spectrum SPF 15', '12B', '#f2c9b5'),('FACE15', '12N', '#e5be9d')]

matched_index = []
n = 0
for i in info :
    if name[n][0] and name[n][1] and name[n][2] == i[n][0] and i[n][1] and i[n][2]:
        print("match")
        matched_index.append(n)
        n = n + 1
    else:
        n = n + 1
print("DONE")

Is there a way to match only the first 3 elements like above? I will get error when n reached 3...

Answer 1

something like the below

infos = [('Water SPF 15', '12B', '#f2c9b5', 31, '-100.1', '-100.1', '-100.1', '12N', '', '', '27B'),
         ('Water Spectrum SPF 15', '12B', '#f2c9b5', 31, '7..7', '7..7', '7..7', '12N', '', '', '27B'),
         ('Water Foundation SPF 15', '12N', '#e5be9d', 44, ' ', ' ', ' ', '', '12B', '', '15N'),
         ('FACE15', '12N', '#e5be9d', 44, '**', '**', '**', '12S', '14H', '', '16N'),
         ('FACE', '12B', '#f2c9b5', 31, 'asd', 'asd', 'asd', '14H', '', '8B', '18B'),
         ('hydrator', '10N', '#e5be9d', 44, '', '', '', '', '', '', '13N'),
         ('APE', '12B', '#e7cbb3', 39, '-100', '-100', '-100', '', '', '', '')]
names = [('Water Spectrum SPF 15', '12B', '#f2c9b5'), ('FACE15', '12N', '#e5be9d')]

for name in names:
    for idx, info in enumerate(infos):
        if name[0] == info[0] and name[1] == info[1] and name[2] == info[2]:
            print(f'{name} {idx}')

output

('Water Spectrum SPF 15', '12B', '#f2c9b5') 1
('FACE15', '12N', '#e5be9d') 3

Answer 2

You could also consider using list comprehension which is preferred over traditional looping in Python as:

Get only indices:

lst = [idx for idx, info in enumerate(infos) for name in names if name == info[:3]]
print(lst)

Output: [1, 3]

As a list of dictionary:

lst = [{name:idx} for idx, info in enumerate(infos) for name in names if name == info[:3]]
print(lst)

Output:

[{('Water Spectrum SPF 15', '12B', '#f2c9b5'): 1}, {('FACE15', '12N', '#e5be9d'): 3}]

Answer 3

You can avoid nested loops by first converting the tuples in info into a dictionary keyed by the first 3 items of each tuple with its index as the value. Then a simple dictionary lookup provides the results. Overall that will give a run time complexity of O(n) vs O(n ² ) for a nested loop:

info  = [('Water SPF 15', '12B', '#f2c9b5', 31, '-100.1', '-100.1', '-100.1', '12N', '', '', '27B'), ('Water Spectrum SPF 15', '12B', '#f2c9b5', 31, '7..7', '7..7', '7..7', '12N', '', '', '27B'), ('Water Foundation SPF 15', '12N', '#e5be9d', 44, ' ', ' ', ' ', '', '12B', '', '15N'), ('FACE15', '12N', '#e5be9d', 44, '**', '**', '**', '12S', '14H', '', '16N'), ('FACE', '12B', '#f2c9b5', 31, 'asd', 'asd', 'asd', '14H', '', '8B', '18B'), ('hydrator', '10N', '#e5be9d', 44, '', '', '', '', '', '', '13N'), ('APE', '12B', '#e7cbb3', 39, '-100', '-100', '-100', '', '', '', '')]
name = [('Water Spectrum SPF 15', '12B', '#f2c9b5'),('FACE15', '12N', '#e5be9d')]

lookup = {t[:3]: i for i,t in enumerate(info)}
for n in name:
    if n in lookup:
        print(n, lookup[n])

Output:

('Water Spectrum SPF 15', '12B', '#f2c9b5') 1
('FACE15', '12N', '#e5be9d') 3

If you only want the indices stored in a list:

lookup = {t[:3]: i for i,t in enumerate(info)}
matched_index = [lookup[n] for n in name if n in lookup]
print(matched_index)

Output

[1, 3]