Unable to create a proper set / list

Question

I have two lists

a = [1,1,2,3]
b = [1,2]

I need a new list c to have the answer which is derived by removing common elements of list b from list a. Which means if b has 1,2 then after removing from a the new list should have 1,3 as elements.

[1,3]

I tried

d = set(a) & set(b)
c = set(a) - d
print c

but I got

[3]

instead of

[1,3]

I know this is because the moment i do set(a) the repeated elements are trucncated and I am left with only one copy of all elements in there.

But my problem is that I want the repeated elements should not be removed.

What to do?

related: http://stackoverflow.com/questions/2727650/common-elements-between-two-lists-not-using-sets-in-python?rq=1 — NightShadeQueen, Jul 24 '15 at 05:14
`repeated elements should not be removed` - then do not use set. The whole point of this datastructure is to remove duplicates — Konstantin, Jul 24 '15 at 05:15
Why would you expect that result rather than the one you got? `set(a)` contains 1, 2 and 3, `set(b)` contains 1 and 2 so `set(d) will contain 1 and 2. Removing 1 and 2 from `set(a)` would only leave 3 which makes `set(a)-d` contains precisely 3. — skyking, Jul 24 '15 at 05:50

Anand S Kumar · Answer 1 · 2015-07-24T05:24:42.303

You can use collections.Counter -

>>> a = [1,1,2,3]
>>> b = [1,2]
>>> from collections import Counter
>>> c = list((Counter(a) - Counter(b)).elements())
>>> c
[1, 3]

What this does is that it counts the elements in list a and then in list b , and then it decreases the count in list a by the count of similar elements in list b , and converts them back to a list and stores in c.

This would even work if b has elements that are not in a, example -

>>> a = [1,1,2,3]
>>> b = [1,2,4]
>>> list((Counter(a) - Counter(b)).elements())
[1, 3]

soumen · Answer 2 · 2015-07-24T05:30:43.683

1

You do not need set and other packages. Importing additional packages "may" be expensive in terms of resources.

Try this.

c = list(a)
for data in b:
    if data in c: 
        c.remove(data)

To ignore, you may use try-except block.

c = list(a)
for data in b:
    try:
        c.remove(data)
    except ValueError:
        pass

edited Jul 24 '15 at 05:30

answered Jul 24 '15 at 05:19

soumen

66
5

This would not work if b has an element not in a. Though I am not sure if that is a case for the OP. – Anand S Kumar Jul 24 '15 at 05:22

mongjong · Answer 3 · 2015-07-24T05:22:54.840

0

You can use remove() to remove just one occurrence and a loop. For example:

a = [1,1,2,3]
b = [1,2]
# Create a copy of a so you don't modify a itself
c = list(a)
for num in b:
    c.remove(num)

print c

edited Jul 24 '15 at 05:22

answered Jul 24 '15 at 05:18

mongjong

479
3
6

John La Rooy · Answer 4 · 2015-07-24T05:44:24.633

Counter is handy for this, but it has a reputation of being quite slow. Here is a version using defaultdict instead.

>>> from collections import defaultdict
>>> D = defaultdict(int)
>>> a = [1,1,2,3]
>>> b = [1,2]
>>> for k in a:
...     D[k] += 1
... 
>>> for k in b:
...     D[k] -= 1
... 
>>> [k for k in D for i in range(D[k])]
[1, 3]

Here is another way that retains the order of a in the result

>>> D = defaultdict(int)
>>> a = [1,1,2,3]
>>> b = [1,2]
>>> for k in b:
...     D[k] += 1
... 
>>> res = []
>>> for i in a:
...     if D[i]:
...         D[i] -= 1
...     else:
...         res.append(i)
... 
>>> res
[1, 3]

Unable to create a proper set / list

4 Answers4