Python loop through list levels

Question

I am trying to loop through and merge of two lists that look like this:

T1 = [[['a',  1], ['c', 3]], [['e',  5], ['g', 7]], [['i',  9],['j', 11]]]
T2 = [[['m',  1], ['n', 5]], [['q',  7], ['r', 7]], [['t',  9],['u', 11]]]

So for example, T1[0] = [['a', 1], ['c', 3]] T2[0] = [['m', 1], ['n', 5]]

I am writing what is essentially a merge algorithm, where I compare the second element in each list in T1[0][0] (the 1), with the second value in T2[0][0] (so 1 again). Essentially what I want is T1[0][0][0] and T2[0][0][0]. If they match, I merge the two values into a tuple {'a', 1, 'm'} (can also be a list if it makes things easier).

My confusion is stemming from the fact that I don't know how to best loop through T1 and T2 without running it 3 times: (T1[0], T2[0]), (T1[1], T2[1]), (T1[2], T2[2])

Is there a way to elegantly loop through T1 and T2 so that I can go through each sublist in T1 and T2 and perform the check and merge?

Note: the data is already sorted, and I am not looking to use libraries. I want to know if I can do this myself.

##Update## So I know it is a for loop, however the below code is as far as i can get, and it is throwing me a list index out of range as shown below

result = []
i = j = 0

while True:
    for s_T1 in T1:
        for s_T2 in T2:
            if s_T1[i][1] < s_T2[j][1]: ### error at this line
                i = i + 1

            elif  s_T1[i][1] > s_T2[j][1]:
                j = j + 1

            elif s_T1[i][1] == s_T2[j][1]:
                result.append(s_T1[i][0], str(s_T1[i][0), s_T2[j][0])
                i = i + 1
                j = j + 1
                
            if i == len(T1) or j == len(T2):
                break

return result

What the code should return is:

[[['a', 1, 'm']], [['g', 7, 'q'], ['g', 7, 'r']], [['i', 9, 't'], ['j', 11, 'u']]]

So, look at the integer in ['e', 5] in T1, compare it to ['q', 7] in T2, since 5 is smaller than 7, move to the next value in T1. if they match, append. And the other way around - if the value in T1 is greater than the value in T2, move to the next value in T1.

Also update your question with a required output given the sample input you have. — quamrana
– quamrana, Commented Sep 8, 2020 at 10:00
The result ['e', 5, 'n'] are not in the same "index" in both of the lists, is this a mistake, or should each index in T1 be compared against each index of T2? — Hampus Larsson
– Hampus Larsson, Commented Sep 8, 2020 at 10:38
@quamrana i have updated my question. Any help with my loop is also appreciated — soapycat
– soapycat, Commented Sep 8, 2020 at 10:38
@soapycat Then I assume that ['g', 7, 'q'] shouldn't be there either correct? — Hampus Larsson
– Hampus Larsson, Commented Sep 8, 2020 at 10:41

tzaman · Accepted Answer · 2020-09-08 10:53:31Z

2

zip ties together multiple iterables so you can iterate through them in lock-step, which is what it seems you're looking for:

>>> result = []
>>> for i, j in zip(T1, T2):
...   sub = []
...   for x, y in zip(i, j):
...     if x[1] == y[1]:
...       sub.append((x[0], x[1], y[0]))
...   if sub:
...     result.append(sub)
...
>>> result
[[('a', 1, 'm')], [('g', 7, 'r')], [('i', 9, 't'), ('j', 11, 'u')]]

answered Sep 8, 2020 at 10:53

tzaman

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1 Comment

soapycat Over a year ago

Not quite, i apologise i wasn't very clear in how i am looping. I have updated my loop and match requirements.

IoaTzimas · Accepted Answer · 2020-09-08 11:28:05Z

1

This should work:

def task(T1,T2):
    result = []
    for i in range(len(T1)):
        for k in range(2):
            for l in range(2):
                if T1[i][k][1]==T2[i][l][1]:
                    result.append([T1[i][k][0], T1[i][k][1],T2[i][l][0]]) 
    return result

task(T1,T2)

When run it results to:

[['a', 1, 'm'], ['g', 7, 'q'], ['g', 7, 'r'], ['i', 9, 't'], ['j', 11, 'u']]

as indicated in your question

edited Sep 8, 2020 at 11:28

answered Sep 8, 2020 at 11:09

IoaTzimas

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2 Comments

soapycat Over a year ago

you can do things like T2[i][l][1] ?? Every time I tried to do it, it would throw me an error! Must not have been doing it right. Is this a standard practice?

IoaTzimas Over a year ago

You can definitely do it as long the data stucture (the depth of it) allows it. Have you tested the answers? If yes, please accept some, in case it answers the question properly. Cheers

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