Compare lists in the same dictionary of lists

Question

In resume, I have two keys in the same dictionary where each one has their corresponding lists.

I try to compare both list to check common and differential elements. It means that the output I will count how many elements are identical or present in only one key's list.

from the beginning I am inserting the elements using the files as arguments and they are read in the function

def shared(list):
        dict_shared = {}
        for i in list:
                infile = open(i, 'r')
                if i not in dict_shared:
                        dict_shared[i] = []

                for line in infile:
                        dict_shared[spacer].append(record.id)

        return dict_shared

Now I am stuck trying to find a way to compare the lists created and present in the dictionary.

dict = {a:[1,2,3,4,5], b:[2,3,4,6]}

My intention is to compare the lists in order to have the lines shared between two texts.

a: [1,5]
b: [6]
a-b: [2,3,4]

From now I can't find a way to solve this. Any suggestion?

I'm not sure what you're doing with files, but you should pretty much always use ```open``` in a ```with open(...) as file:``` block, so that it automatically closes/releases the file when it is no longer necessary. — johannesack, Apr 16 '20 at 13:34
@JohannesAck it will not be a big problem but I will accept your suggestion. — F.Lira, Apr 16 '20 at 13:38
This also could be done with numpy (for performance issues) Check https://stackoverflow.com/questions/1613249/numpy-comparing-elements-in-two-arrays — Ivan Calderon, Apr 16 '20 at 14:06

score 5 · Answer 1 · answered Apr 16 '20 at 13:35

5

You could use set:

d = {'a':[1,2,3,4,5], 'b':[2,3,4,6]}
print(list(set(d['a'])-set(d['b'])))
print(list(set(d['b'])-set(d['a'])))
print(list(set(d['b'])&set(d['a'])))

result:

[1, 5]
[6]
[2, 3, 4]

answered Apr 16 '20 at 13:35

jizhihaoSAMA

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I was wondering if there is any way to do it writing less lines in the code – F.Lira Apr 16 '20 at 13:39
Hmmmm,I think three lines is not very much.Do you have any idea? – jizhihaoSAMA Apr 16 '20 at 13:44

score 5 · Answer 2 · edited Apr 16 '20 at 19:43

5

you can do that by utilising python inbuilt functions like union, difference, intersection. Note: These are for sets, you can convert a list to set by

1stset = set(a)

example:

print(1stset.difference(2ndset))

print(1stset.intersection(2ndset))

print(1stset.union(2ndset))

you can refer the following links for more information

https://www.geeksforgeeks.org/python-intersection-two-lists/

https://www.geeksforgeeks.org/python-union-two-lists/

https://www.geeksforgeeks.org/python-difference-two-lists/

edited Apr 16 '20 at 19:43

mohammed wazeem

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answered Apr 16 '20 at 13:37

Lavanya V

337
2
7

it is consider better to provide solution or stackoverflow link than other website link for other benefits – sahasrara62 Apr 16 '20 at 13:39
I saw it before to come here but in my case I have the lists in one dictionary, so I was trying to do it in some recursive way. – F.Lira Apr 16 '20 at 13:41

score 4 · Accepted Answer · answered Apr 16 '20 at 13:37

4

A solution with list comprehension would be:

dictionary = {'a':[1,2,3,4,5], 'b':[2,3,4,6]}

only_in_a = [x for x in dictionary['a'] if not x in dictionary['b']]
only_in_b = [x for x in dictionary['b'] if not x in dictionary['a']]
in_both = [x for x in dictionary['a'] if x in dictionary['b']]

Note that this is not especially wise in terms of complexity, for larger lists.

answered Apr 16 '20 at 13:37

johannesack

660
3
19

That's my case. Some files contains more than 3000 lines. – F.Lira Apr 16 '20 at 13:47

score 3 · Answer 4 · answered Apr 16 '20 at 13:36

Not sure if I understand correctly what you are trying to achieve but it seems like you'd need set operations:

dictionary = {"a":[1,2,3,4,5], "b":[2,3,4,6]}

#in a but not in b
set(dictionary["a"]) - set(dictionary["b"])

#in b but not in a
set(dictionary["b"]) - set(dictionary["a"])

#union of both
set(dictionary["b"]).union(set(dictionary["a"]))

#intersection of both
set(dictionary["b"]).intersection(set(dictionary["a"]))

mohammed wazeem · Answer 5 · 2020-04-16T14:52:18.750

3

You can try something like this

mydict = {'a': [1,2,3,4,5], 'b': [2,3,4,6]}

>>> list(set(mydict['a']).intersection(mydict['b']))  # common to both
 [2, 3, 4]
>>> list(set(mydict['a']).difference(mydict['b']))  # a - b
 [1, 5]
>>> list(set(mydict['b']).difference(mydict['a']))  # b - a
 [6]
>>> list(set(mydict['a']).union(mydict['b']))  # union of both
 [1, 2, 3, 4, 5, 6]

edited Apr 16 '20 at 14:52

answered Apr 16 '20 at 13:53

mohammed wazeem

1,310
1
10
26

score 2 · Answer 6 · answered Apr 16 '20 at 14:06

Try this

print("a - b : {} ".format(list(set(_dict['a']) - set(_dict['b']))))
print('b - a : {} '.format(list(set(_dict['b']) - set(_dict['a']))))
print('a \u2229 b : {} '.format(list(set(_dict['a']).intersection(set(_dict['b'])))))

Output

a - b : [1, 5] 
b - a : [6] 
a ∩ b : [2, 3, 4]

Compare lists in the same dictionary of lists

6 Answers6