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I have an input list that look like this

input_list = ["a", "b2","d"]

and another list that look like this

ref_list = [['a'], ['b1', 'b2', 'b3'], ['c1', 'c2', 'c3', 'c4'], ['d']]

What I want to do is to get the list of value from input_list that not existed in ref_list So, the result based on these values should be

[['c1', 'c2', 'c3', 'c4']]

At first, my case only have something like

input_list = ["a","b","d"]
ref_list = ["a","b","c","d"]

which I can use

missing_value = list(set(ref_list) - set(input_list))

which will extract a result like this

["c"]

but for the case that a ref_list that each index is a list with single or multiple value in it. Is there a simple way to achieve the missing value?

emp
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2 Answers2

3

Looks like a job for set.isdisjoint:

input_list = ["a", "b2","d"]
ref_list = [['a'], ['b1', 'b2', 'b3'], ['c1', 'c2', 'c3', 'c4'], ['d']]

res = list(filter(set(input_list).isdisjoint, ref_list))

print(res)

Output (Try it online!):

[['c1', 'c2', 'c3', 'c4']]

Benchmark with your small data:

 2.6 μs   2.6 μs   2.7 μs  python_user
 2.2 μs   2.2 μs   2.2 μs  Uttam_Velani
 0.9 μs   0.9 μs   0.9 μs  dont_talk_just_code

Benchmark with a 1000 times longer ref_list:

2150 μs  2184 μs  2187 μs  python_user
1964 μs  1981 μs  1990 μs  Uttam_Velani
 292 μs   304 μs   306 μs  dont_talk_just_code

Benchmark code (Try it online!):

from timeit import repeat

def python_user(input_list, ref_list):
    return [i for i in ref_list if all(j not in i for j in input_list)]

def Uttam_Velani(input_list, ref_list):
    required_list = []
    for data in ref_list:
        if len(set(data) & set(input_list)) == 0:
            required_list.append(data)
    return required_list

def dont_talk_just_code(input_list, ref_list):
    return list(filter(set(input_list).isdisjoint, ref_list))

funcs = python_user, Uttam_Velani, dont_talk_just_code


def test(input_list, ref_list, number, format_time):
    expect = funcs[0](input_list, ref_list)
    for func in funcs:
        result = func(input_list, ref_list)
        print(result == expect, func.__name__)
    print()

    for _ in range(3):
        for func in funcs:
            times = sorted(repeat(lambda: func(input_list, ref_list), number=number))[:3]
            print(*(format_time(t / number) for t in times), func.__name__)
        print()

test(["a", "b2","d"],
     [['a'], ['b1', 'b2', 'b3'], ['c1', 'c2', 'c3', 'c4'], ['d']],
     10000,
     lambda time: '%4.1f μs ' % (time * 1e6))
test(["a", "b2","d"],
     [['a'], ['b1', 'b2', 'b3'], ['c1', 'c2', 'c3', 'c4'], ['d']] * 1000,
     20,
     lambda time: '%4d μs ' % (time * 1e6))
no comment
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  • @python_user Right, but at least the data they've shown as well as their own non-list solution also using sets suggest that their elements indeed are hashable. – no comment Sep 25 '21 at 13:25
  • Nice comparison! All solutions worked well. The ref_list that I worked has about 15+ indexes so I think it will not make a huge different in term of run time. For the input and ref list, they will come as nested list like the example shown. – emp Sep 25 '21 at 14:28
  • Anyway, I've never used `isdisjoint` before. I searched and found this function but isn't it come with the parameter like `set.isdisjoint(another_set)` ? Not sure if its has any difference? – emp Sep 25 '21 at 14:31
  • @Jamiewp Well, yes, it is called later. Not by me directly but by `filter`. That takes the `isdisjoint` method bound to the `set(input_list)` object and calls it with each element of `ref_list` as argument. – no comment Sep 25 '21 at 15:01
2

It quit simple and I can not think of any efficient way to solve your problem.

input_list = ['a', 'b2','d']

ref_list = [['a'], ['b1', 'b2', 'b3'], ['c1', 'c2', 'c3', 'c4'], ['d']]

required_list = []

for data in ref_list:
    if len(set(data) & set(input_list)) == 0:
        required_list.append(data)

print(required_list)

Output:

[['c1', 'c2', 'c3', 'c4']]

Also you can refer here for more clarity. Thanks!

Uttam Velani
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