Get difference between two lists

Question

I have two lists in Python, like these:

temp1 = [\'One\', \'Two\', \'Three\', \'Four\']
temp2 = [\'One\', \'Two\']

I need to create a third

Answer 1

Here's a Counter answer for the simplest case.

This is shorter than the one above that does two-way diffs because it only does exactly what the question asks: generate a list of what's in the first list but not the second.

from collections import Counter

lst1 = ['One', 'Two', 'Three', 'Four']
lst2 = ['One', 'Two']

c1 = Counter(lst1)
c2 = Counter(lst2)
diff = list((c1 - c2).elements())

Alternatively, depending on your readability preferences, it makes for a decent one-liner:

diff = list((Counter(lst1) - Counter(lst2)).elements())

Output:

['Three', 'Four']

Note that you can remove the list(...) call if you are just iterating over it.

Because this solution uses counters, it handles quantities properly vs the many set-based answers. For example on this input:

lst1 = ['One', 'Two', 'Two', 'Two', 'Three', 'Three', 'Four']
lst2 = ['One', 'Two']

The output is:

['Two', 'Two', 'Three', 'Three', 'Four']

Answer 2

We can calculate intersection minus union of lists:

temp1 = ['One', 'Two', 'Three', 'Four']
temp2 = ['One', 'Two', 'Five']

set(temp1+temp2)-(set(temp1)&set(temp2))

Out: set(['Four', 'Five', 'Three']) 

Answer 3

Here is an simple way to distinguish two lists (whatever the contents are), you can get the result as shown below :

>>> from sets import Set
>>>
>>> l1 = ['xvda', False, 'xvdbb', 12, 'xvdbc']
>>> l2 = ['xvda', 'xvdbb', 'xvdbc', 'xvdbd', None]
>>>
>>> Set(l1).symmetric_difference(Set(l2))
Set([False, 'xvdbd', None, 12])

Hope this will helpful.

Answer 4

You could use a naive method if the elements of the difflist are sorted and sets.

list1=[1,2,3,4,5]
list2=[1,2,3]

print list1[len(list2):]

or with native set methods:

subset=set(list1).difference(list2)

print subset

import timeit
init = 'temp1 = list(range(100)); temp2 = [i * 2 for i in range(50)]'
print "Naive solution: ", timeit.timeit('temp1[len(temp2):]', init, number = 100000)
print "Native set solution: ", timeit.timeit('set(temp1).difference(temp2)', init, number = 100000)

Naive solution: 0.0787101593292

Native set solution: 0.998837615564

Answer 5

The existing solutions all offer either one or the other of:

• Faster than O(n*m) performance.
• Preserve order of input list.

But so far no solution has both. If you want both, try this:

s = set(temp2)
temp3 = [x for x in temp1 if x not in s]

Performance test

import timeit
init = 'temp1 = list(range(100)); temp2 = [i * 2 for i in range(50)]'
print timeit.timeit('list(set(temp1) - set(temp2))', init, number = 100000)
print timeit.timeit('s = set(temp2);[x for x in temp1 if x not in s]', init, number = 100000)
print timeit.timeit('[item for item in temp1 if item not in temp2]', init, number = 100000)

Results:

4.34620224079 # ars' answer
4.2770634955  # This answer
30.7715615392 # matt b's answer

The method I presented as well as preserving order is also (slightly) faster than the set subtraction because it doesn't require construction of an unnecessary set. The performance difference would be more noticable if the first list is considerably longer than the second and if hashing is expensive. Here's a second test demonstrating this:

init = '''
temp1 = [str(i) for i in range(100000)]
temp2 = [str(i * 2) for i in range(50)]
'''

Results:

11.3836875916 # ars' answer
3.63890368748 # this answer (3 times faster!)
37.7445402279 # matt b's answer

Answer 6

temp3 = [item for item in temp1 if item not in temp2]