How to compare a list of lists/sets in python?

Question

What is the easiest way to compare the 2 lists/sets and output the differences? Are there any built in functions that will help me compare nested lists/sets?

Inputs:

Answer 1

By using set comprehensions, you can make it a one-liner. If you want:

to get a set of tuples, then:

Differences = {tuple(i) for i in First_list} ^ {tuple(i) for i in Secnd_list}

Or to get a list of tuples, then:

Differences = list({tuple(i) for i in First_list} ^ {tuple(i) for i in Secnd_list})

Or to get a list of lists (if you really want), then:

Differences = [list(j) for j in {tuple(i) for i in First_list} ^ {tuple(i) for i in Secnd_list}]

PS: I read here: https://stackoverflow.com/a/10973817/4900095 that map() function is not a pythonic way to do things.

Answer 2

i guess you'll have to convert your lists to sets:

>>> a = {('a', 'b'), ('c', 'd'), ('e', 'f')}
>>> b = {('a', 'b'), ('h', 'g')}
>>> a.symmetric_difference(b)
{('e', 'f'), ('h', 'g'), ('c', 'd')}

Answer 3

http://docs.python.org/library/difflib.html is a good starting place for what you are looking for.

If you apply it recursively to the deltas, you should be able to handle nested data structures. But it will take some work.

Answer 4

Note that with this method you will loose the order

first_set=set(map(tuple,S))
second_set=set(map(tuple,T))
print map(list,list(first_set.union(second_set)-(first_set&second_set)))

Answer 5

>>> First_list = [['Test.doc', '1a1a1a', '1111'], ['Test2.doc', '2b2b2b', '2222'], ['Test3.doc', '3c3c3c', '3333']] 
>>> Secnd_list = [['Test.doc', '1a1a1a', '1111'], ['Test2.doc', '2b2b2b', '2222'], ['Test3.doc', '3c3c3c', '3333'], ['Test4.doc', '4d4d4d', '4444']] 


>>> z = [tuple(y) for y in First_list]
>>> z
[('Test.doc', '1a1a1a', '1111'), ('Test2.doc', '2b2b2b', '2222'), ('Test3.doc', '3c3c3c', '3333')]
>>> x = [tuple(y) for y in Secnd_list]
>>> x
[('Test.doc', '1a1a1a', '1111'), ('Test2.doc', '2b2b2b', '2222'), ('Test3.doc', '3c3c3c', '3333'), ('Test4.doc', '4d4d4d', '4444')]


>>> set(x) - set(z)
set([('Test4.doc', '4d4d4d', '4444')])

Answer 6

Old question but here's a solution I use for returning unique elements not found in both lists.

I use this for comparing the values returned from a database and the values generated by a directory crawler package. I didn't like the other solutions I found because many of them could not dynamically handle both flat lists and nested lists.

def differentiate(x, y):
    """
    Retrieve a unique of list of elements that do not exist in both x and y.
    Capable of parsing one-dimensional (flat) and two-dimensional (lists of lists) lists.

    :param x: list #1
    :param y: list #2
    :return: list of unique values
    """
    # Validate both lists, confirm either are empty
    if len(x) == 0 and len(y) > 0:
        return y  # All y values are unique if x is empty
    elif len(y) == 0 and len(x) > 0:
        return x  # All x values are unique if y is empty

    # Get the input type to convert back to before return
    try:
        input_type = type(x[0])
    except IndexError:
        input_type = type(y[0])

    # Dealing with a 2D dataset (list of lists)
    try:
        # Immutable and Unique - Convert list of tuples into set of tuples
        first_set = set(map(tuple, x))
        secnd_set = set(map(tuple, y))

    # Dealing with a 1D dataset (list of items)
    except TypeError:
        # Unique values only
        first_set = set(x)
        secnd_set = set(y)

    # Determine which list is longest
    longest = first_set if len(first_set) > len(secnd_set) else secnd_set
    shortest = secnd_set if len(first_set) > len(secnd_set) else first_set

    # Generate set of non-shared values and return list of values in original type
    return [input_type(i) for i in {i for i in longest if i not in shortest}]