python: comparing values in a list if exists in another list

Question

So now I have 2 lists,

list1 = [[0,1],[0,2],[0,10]]
list2 = [[1, ['miniwok', 'food1']], [2,['chicken', 'food2']], [3,['duck', 'food3']], ..... , [10, ['pizza', 'food10']]]

I want to compare the all 2nd element in list1 and if it exists in list2, print the corresponding list. so the result I want is something like this:

[[1, 'miniwok'],[2, 'chicken'],[10,'pizza']]

I tried using nested for loop but I think I'm doing something wrong

for x in range(len(list1)):
    for y in range(1, len(list2)+1):
        if(list1[x][1] == list2[y]):
            result = [y, list2[y][0]]
            fstore.append(result)

Is it possible to modify the code which generates those lists to use dictionaries, which will be both more performant (accessing the "random" element with O(1)), and simpler to use? This is assuming your keys (2nd integers) are unique. — Jan Christoph Terasa
– Jan Christoph Terasa, Commented Nov 10, 2019 at 8:23

Ajax1234 · Accepted Answer · 2019-11-10 03:54:17Z

2

You can convert list2 to a dictionary for faster lookup:

list1 = [[0,1],[0,2],[0,10]]
list2 = [[1, ['miniwok', 'food1']], [2,['chicken', 'food2']], [3,['duck', 'food3']], [10, ['pizza', 'food10']]]
new_l2 = dict(list2)
result = [[b, k[a]] for a, b in list1 if (k := new_l2.get(b)) is not None]

Output:

[[1, 'miniwok'], [2, 'chicken'], [10, 'pizza']]

answered Nov 10, 2019 at 3:54

Ajax1234

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1 Comment

Aryerez Over a year ago

It's worth mentioning that the := operator is new from python 3.8.

Chetan Goyal · Accepted Answer · 2019-11-10 08:17:41Z

1

Your code had some problems with accessing values via indexing and you haven't assigned fstore as empty list before using it.

The corrected version of your answer is here-

list1 = [[0,1],[0,2],[0,10]]
list2 = [[1, ['miniwok', 'food1']], [2,['chicken', 'food2']], [3,['duck', 'food3']], [10, ['pizza', 'food10']]]
fstore = []
for x in range(len(list1)):
    for y in range(len(list2)):
        if(list1[x][1] == list2[y][0]):
            result = [list2[y][0], list2[y][1][0] ]
            fstore.append(result)
            break

Contents of fstore:

[[1, 'miniwok'], [2, 'chicken'], [10, 'pizza']]

I hope it might help you. If you have any doubt, you can ask in comments. :)

answered Nov 10, 2019 at 8:17

Chetan Goyal

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Comments

Aryerez · Accepted Answer · 2019-11-10 07:56:06Z

0

You can do:

list1 = [[0,1],[0,2],[0,10]]
list2 = [[1, ['miniwok', 'food1']], [2,['chicken', 'food2']], [3,['duck', 'food3']],... , [10, ['pizza', 'food10']]]
numbers = [number[1] for number in list1]
[(item[0], item[1][0]) for item in list2 if item[0] in numbers]

Output:

[(1, 'miniwok'), (2, 'chicken'), (10, 'pizza')]

Of course the "()" in the list comprehension that creates a list of tuples can be switched with "[]" to create a list of lists, if you prefer.

edited Nov 10, 2019 at 7:56

answered Nov 10, 2019 at 7:49

Aryerez

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