如何验证一个列表是否是另一个列表的子集？

I need to verify if a list is a subset of another – a boolean return is all I seek.

Is testing equality on the smaller list after an intersection the fastest way to do this? Performance is of utmost importance given the number of datasets that need to be compared.

Adding further facts based on discussions:

1. Will either of the lists be the same for many tests? It does as one of them is a static lookup table.

2. Does it need to be a list? It does not – the static lookup table can be anything that performs best. The dynamic one is a dict from which we extract the keys to perform a static lookup on.

What would be the optimal solution given the scenario?

The performant function Python provides for this is set.issubset. It does have a few restrictions that make it unclear if it’s the answer to your question, however.

A list may contain items multiple times and has a specific order. A set does not. Additionally, sets only work on hashable objects.

Are you asking about subset or subsequence (which means you’ll want a string search algorithm)? Will either of the lists be the same for many tests? What are the datatypes contained in the list? And for that matter, does it need to be a list?

Your other post intersect a dict and list made the types clearer and did get a recommendation to use dictionary key views for their set-like functionality. In that case it was known to work because dictionary keys behave like a set (so much so that before we had sets in Python we used dictionaries). One wonders how the issue got less specific in three hours.

>>> a = [1, 3, 5]
>>> b = [1, 3, 5, 8]
>>> c = [3, 5, 9]
>>> set(a) <= set(b)
True
>>> set(c) <= set(b)
False

>>> a = ['yes', 'no', 'hmm']
>>> b = ['yes', 'no', 'hmm', 'well']
>>> c = ['sorry', 'no', 'hmm']
>>> 
>>> set(a) <= set(b)
True
>>> set(c) <= set(b)
False

one = [1, 2, 3]
two = [9, 8, 5, 3, 2, 1]

all(x in two for x in one)

Explanation: Generator creating booleans by looping through list one checking if that item is in list two. all() returns True if every item is truthy, else False.

There is also an advantage that all return False on the first instance of a missing element rather than having to process every item.

Assuming the items are hashable

>>> from collections import Counter
>>> not Counter([1, 2]) - Counter([1])
False
>>> not Counter([1, 2]) - Counter([1, 2])
True
>>> not Counter([1, 2, 2]) - Counter([1, 2])
False

If you don’t care about duplicate items eg. [1, 2, 2] and [1, 2] then just use:

>>> set([1, 2, 2]).issubset([1, 2])
True

Is testing equality on the smaller list after an intersection the fastest way to do this?

.issubset will be the fastest way to do it. Checking the length before testing issubset will not improve speed because you still have O(N + M) items to iterate through and check.

One more solution would be to use a intersection.

one = [1, 2, 3]
two = [9, 8, 5, 3, 2, 1]

set(one).intersection(set(two)) == set(one)

The intersection of the sets would contain of set one

(OR)

one = [1, 2, 3]
two = [9, 8, 5, 3, 2, 1]

set(one) & (set(two)) == set(one)

one = [1, 2, 3]
two = [9, 8, 5, 3, 2, 1]

set(x in two for x in one) == set([True])

If list1 is in list 2:

• (x in two for x in one) generates a list of True.

• when we do a set(x in two for x in one) has only one element (True).

Set theory is inappropriate for lists since duplicates will result in wrong answers using set theory.

For example:

a = [1, 3, 3, 3, 5]
b = [1, 3, 3, 4, 5]
set(b) > set(a)

has no meaning. Yes, it gives a false answer but this is not correct since set theory is just comparing: 1,3,5 versus 1,3,4,5. You must include all duplicates.

Instead you must count each occurrence of each item and do a greater than equal to check. This is not very expensive, because it is not using O(N^2) operations and does not require quick sort.

#!/usr/bin/env python

from collections import Counter

def containedInFirst(a, b):
  a_count = Counter(a)
  b_count = Counter(b)
  for key in b_count:
    if a_count.has_key(key) == False:
      return False
    if b_count[key] > a_count[key]:
      return False
  return True


a = [1, 3, 3, 3, 5]
b = [1, 3, 3, 4, 5]
print "b in a: ", containedInFirst(a, b)

a = [1, 3, 3, 3, 4, 4, 5]
b = [1, 3, 3, 4, 5]
print "b in a: ", containedInFirst(a, b)

Then running this you get:

$ python contained.py 
b in a:  False
b in a:  True

Pardon me if I am late to the party. ;)

To check if one set A is subset of set B, Python has A.issubset(B) and A <= B. It works on set only and works great BUT the complexity of internal implementation is unknown. Reference: https://docs.python.org/2/library/sets.html#set-objects

I came up with an algorithm to check if list A is a subset of list B with following remarks.

• To reduce complexity of finding subset, I find it appropriate to sort both lists first before comparing elements to qualify for subset.
• It helped me to break the loop when value of element of second list B[j] is greater than value of element of first list A[i].
• last_index_j is used to start loop over list B where it last left off. It helps avoid starting comparisons from the start of list B (which is, as you might guess unnecessary, to start list B from index 0 in subsequent iterations.)

• Complexity will be O(n ln n) each for sorting both lists and O(n) for checking for subset.
  O(n ln n) + O(n ln n) + O(n) = O(n ln n).

• Code has lots of print statements to see what’s going on at each iteration of the loop. These are meant for understanding only.

Check if one list is subset of another list

is_subset = True;

A = [9, 3, 11, 1, 7, 2];
B = [11, 4, 6, 2, 15, 1, 9, 8, 5, 3];

print(A, B);

# skip checking if list A has elements more than list B
if len(A) > len(B):
    is_subset = False;
else:
    # complexity of sorting using quicksort or merge sort: O(n ln n)
    # use best sorting algorithm available to minimize complexity
    A.sort();
    B.sort();

    print(A, B);

    # complexity: O(n^2)
    # for a in A:
    #   if a not in B:
    #       is_subset = False;
    #       break;

    # complexity: O(n)
    is_found = False;
    last_index_j = 0;

    for i in range(len(A)):
        for j in range(last_index_j, len(B)):
            is_found = False;

            print("i=" + str(i) + ", j=" + str(j) + ", " + str(A[i]) + "==" + str(B[j]) + "?");

            if B[j] <= A[i]:
                if A[i] == B[j]:
                    is_found = True;
                last_index_j = j;
            else:
                is_found = False;
                break;

            if is_found:
                print("Found: " + str(A[i]));
                last_index_j = last_index_j + 1;
                break;
            else:
                print("Not found: " + str(A[i]));

        if is_found == False:
            is_subset = False;
            break;

print("subset") if is_subset else print("not subset");

Output

[9, 3, 11, 1, 7, 2] [11, 4, 6, 2, 15, 1, 9, 8, 5, 3]
[1, 2, 3, 7, 9, 11] [1, 2, 3, 4, 5, 6, 8, 9, 11, 15]
i=0, j=0, 1==1?
Found: 1
i=1, j=1, 2==1?
Not found: 2
i=1, j=2, 2==2?
Found: 2
i=2, j=3, 3==3?
Found: 3
i=3, j=4, 7==4?
Not found: 7
i=3, j=5, 7==5?
Not found: 7
i=3, j=6, 7==6?
Not found: 7
i=3, j=7, 7==8?
not subset

Below code checks whether a given set is a “proper subset” of another set

 def is_proper_subset(set, superset):
     return all(x in superset for x in set) and len(set)<len(superset)

In python 3.5 you can do a [*set()][index] to get the element. It is much slower solution than other methods.

one = [1, 2, 3]
two = [9, 8, 5, 3, 2, 1]

result = set(x in two for x in one)

[*result][0] == True

or just with len and set

len(set(a+b)) == len(set(a))

Here is how I know if one list is a subset of another one, the sequence matters to me in my case.

def is_subset(list_long,list_short):
    short_length = len(list_short)
    subset_list = []
    for i in range(len(list_long)-short_length+1):
        subset_list.append(list_long[i:i+short_length])
    if list_short in subset_list:
        return True
    else: return False

Most of the solutions consider that the lists do not have duplicates. In case your lists do have duplicates you can try this:

def isSubList(subList,mlist):
    uniqueElements=set(subList)
    for e in uniqueElements:
        if subList.count(e) > mlist.count(e):
            return False     
    # It is sublist
    return True

It ensures the sublist never has different elements than list or a greater amount of a common element.

lst=[1,2,2,3,4]
sl1=[2,2,3]
sl2=[2,2,2]
sl3=[2,5]

print(isSubList(sl1,lst)) # True
print(isSubList(sl2,lst)) # False
print(isSubList(sl3,lst)) # False

Since no one has considered comparing two strings, here’s my proposal.

You may of course want to check if the pipe (“|”) is not part of either lists and maybe chose automatically another char, but you got the idea.

Using an empty string as separator is not a solution since the numbers can have several digits ([12,3] != [1,23])

def issublist(l1,l2):
    return '|'.join([str(i) for i in l1]) in '|'.join([str(i) for i in l2])

If you are asking if one list is “contained” in another list then:

>>>if listA in listB: return True

If you are asking if each element in listA has an equal number of matching elements in listB try:

all(True if listA.count(item) <= listB.count(item) else False for item in listA)

If a2 is subset of a1, then Length of set(a1 + a2) == Length of set(a1)

a1 = [1, 2, 3, 4, 5]
a2 = [1, 2, 3]

len(set(a1)) == len(set(a1 + a2))