33
votes

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:

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', '8p8p8p', 9999], 
              ['Test4.doc', '4d4d4d', 4444]]  

Expected Output:

Differences = [['Test3.doc', '3c3c3c', 3333],
               ['Test3.doc', '8p8p8p', 9999], 
               ['Test4.doc', '4d4d4d', 4444]]
8
See the set-related documentation here: docs.python.org/3.8/library/…codeforester

8 Answers

39
votes

So you want the difference between two lists of items.

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', '8p8p8p', 9999], 
              ['Test4.doc', '4d4d4d', 4444]]

First I'd turn each list of lists into a list of tuples, so as tuples are hashable (lists are not) so you can convert your list of tuples into a set of tuples:

first_tuple_list = [tuple(lst) for lst in first_list]
secnd_tuple_list = [tuple(lst) for lst in secnd_list]

Then you can make sets:

first_set = set(first_tuple_list)
secnd_set = set(secnd_tuple_list)

EDIT (suggested by sdolan): You could have done the last two steps for each list in a one-liner:

first_set = set(map(tuple, first_list))
secnd_set = set(map(tuple, secnd_list))

Note: map is a functional programming command that applies the function in the first argument (in this case the tuple function) to each item in the second argument (which in our case is a list of lists).

and find the symmetric difference between the sets:

>>> first_set.symmetric_difference(secnd_set) 
set([('Test3.doc', '3c3c3c', 3333),
     ('Test3.doc', '8p8p8p', 9999),
     ('Test4.doc', '4d4d4d', 4444)])

Note first_set ^ secnd_set is equivalent to symmetric_difference.

Also if you don't want to use sets (e.g., using python 2.2), its quite straightforward to do. E.g., with list comprehensions:

>>> [x for x in first_list if x not in secnd_list] + [x for x in secnd_list if x not in first_list]
[['Test3.doc', '3c3c3c', 3333],
 ['Test3.doc', '8p8p8p', 9999],
 ['Test4.doc', '4d4d4d', 4444]]

or with the functional filter command and lambda functions. (You have to test both ways and combine).

>>> filter(lambda x: x not in secnd_list, first_list) + filter(lambda x: x not in first_list, secnd_list)

[['Test3.doc', '3c3c3c', 3333],
 ['Test3.doc', '8p8p8p', 9999],
 ['Test4.doc', '4d4d4d', 4444]]
3
votes

Not sure if there is a nice function for this, but the "manual" way to do it isn't difficult:

differences = []

for list in firstList:
    if list not in secondList:
        differences.append(list)
3
votes
>>> 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')])
1
votes

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')}
1
votes

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}]
0
votes

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.

0
votes

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.

0
votes

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)))