in python: difference between two lists

Question

I have two lists like so

found = ['CG', 'E6', 'E1', 'E2', 'E4', 'L2', 'E7', 'E5', 'L1', 'E2BS', 'E2BS', 'E2BS', 'E2', 'E1^E4', 'E5']
expected = ['E1', 'E2', 'E4', 'E1^E4', 'E6', 'E7', 'L1', 'L2', 'CG', 'E2BS', 'E3']

I want to find the differences between both lists.
I have done

list(set(expected)-set(found))

and

list(set(found)-set(expected))

Which returns ['E3'] and ['E5'] respectively.

However, the answers I need are:

'E3' is missing from found.
'E5' is missing from expected.
There are 2 copies of 'E5' in found.
There are 3 copies of 'E2BS' in found.
There are 2 copies of 'E2' in found.

Any help/suggestions are welcome!

Duy Nguyen · Accepted Answer · 2015-05-09 18:54:39Z

8

The collections.Counter class will excel at enumerating the differences between multisets:

>>> from collections import Counter
>>> found = Counter(['CG', 'E6', 'E1', 'E2', 'E4', 'L2', 'E7', 'E5', 'L1', 'E2BS', 'E2BS', 'E2BS', 'E2', 'E1^E4', 'E5'])
>>> expected = Counter(['E1', 'E2', 'E4', 'E1^E4', 'E6', 'E7', 'L1', 'L2', 'CG', 'E2BS', 'E3'])
>>> list((found - expected).elements())
['E2', 'E2BS', 'E2BS', 'E5', 'E5']
>>> list((expected - found).elements())

You might also be interested in difflib.Differ:

>>> from difflib import Differ
>>> found = ['CG', 'E6', 'E1', 'E2', 'E4', 'L2', 'E7', 'E5', 'L1', 'E2BS', 'E2BS', 'E2BS', 'E2', 'E1^E4', 'E5']
>>> expected = ['E1', 'E2', 'E4', 'E1^E4', 'E6', 'E7', 'L1', 'L2', 'CG', 'E2BS', 'E3']
>>> for d in Differ().compare(expected, found):
...     print(d)

+ CG
+ E6
  E1
  E2
  E4
+ L2
+ E7
+ E5
+ L1
+ E2BS
+ E2BS
+ E2BS
+ E2
  E1^E4
+ E5
- E6
- E7
- L1
- L2
- CG
- E2BS
- E3

edited May 9, 2015 at 18:54

Duy Nguyen

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answered Apr 21, 2013 at 3:19

Raymond Hettinger

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1 Comment

COCO Over a year ago

Great answer on the difflib.Differ. I began using that sometimes instead of sets.

Community · Accepted Answer · 2017-05-23 11:59:55Z

4

Leverage the Python set class and Counter class instead of rolling your own solution:

symmetric_difference: finds elements that are either in one set or the other, but not both.
intersection: finds elements in common with the two sets.
difference: which is essentially what you did by subtracting one set from another

Code examples

found.difference(expected) # set(['E5'])

expected.difference(found) # set(['E3'])

found.symmetric_difference(expected) # set(['E5', 'E3'])

Finding copies of objects: this question was already referenced. Using that technique gets you all duplicates, and using the resultant Counter object, you can find how many duplicates. For example:
```
collections.Counter(found)['E5'] # 2
```

edited May 23, 2017 at 11:59

CommunityBot

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answered Apr 21, 2013 at 3:01

Colonel Panic

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Comments

Community · Accepted Answer · 2017-05-23 11:59:55Z

2

You've already answered the first two:

print('{0} missing from found'.format(list(set(expected) - set(found)))
print('{0} missing from expected'.format(list(set(found) - set(expected)))

The second two require you to look at counting duplicates in lists, for which there are many solutions to be found online (including this one: Find and list duplicates in a list?).

edited May 23, 2017 at 11:59

CommunityBot

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answered Apr 21, 2013 at 2:59

Mel Boyce

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