CODEWITHSUNDEEP
pythonlist
jamiek
jamiek

Reputation: 65

Python-compare and sort two lists of numbers and return them in ascending order

I have two lists:

list_1 = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
list_2 = [5, 5, 5, 6, 6, 7]

I want to return the list of elements that appear only in the first list but not in the second and the list should be sorted ascending so the result is like this:

[1, 3, 2, 4]

So far I have this:

def sorted_nums(list_1,list2_2):
    c = (set(list_1) - set(list_2))
    d = dict.fromkeys(c, 0)
    for index in list_1:
        if index in c:
            d[index]+=1
    return d
a = sorted_nums(list_1,list_2)
b = sorted(a.items(), key = lambda x: x[1])
print(b)

and it returns this:

[(1,2), (3,2), (2,3), (4,5)]

Could you help me to change the last part of the code so that I get the result I want?

Upvotes: 0

Views: 616

Answers (8)

Adon Bilivit
Adon Bilivit

Reputation: 26976

My interpretation of the question leads me to this:

list_1 = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
list_2 = [5, 5, 5, 6, 6, 7]

d = dict()
list_2_s = set(list_2)
for e in list_1:
    if e not in list_2_s:
        d[e] = d.setdefault(e, 0) + 1
lst = [t[1] for t in sorted((v, k) for k, v in d.items())]
print(lst)

Output:

[1, 3, 2, 4]

Upvotes: 0

Remzak Kacper
Remzak Kacper

Reputation: 11

list(dict.fromkeys(list_1)) this is working same as set but keep order of elements

def sorted_nums(list_1,list_2):
return tuple(filter(lambda x: x not in set(list_2), list(dict.fromkeys(list_1))))

Upvotes: -1

Fong Sow Leong
Fong Sow Leong

Reputation: 48

You already have it sorted. So you need to extract the first element?

print([x[0] for x in b])

Upvotes: 0

Abhyuday Vaish
Abhyuday Vaish

Reputation: 2379

Method 1:

By using sorted(array, key = list_1.count) to sort by occurrence but this will be slow.

list_1 =  [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
list_2 = [5, 5, 5, 6, 6, 7]

def sorted_nums(list_1,list2_2):
    return sorted(list(set(list_1) - set(list_2)), key = list_1.count)

sorted_nums(list_1, list_2)

Method 2 (Credits):

By using Counter. This is a faster approach. :

from collections import Counter

list_1 =  [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
list_2 = [5, 5, 5, 6, 6, 7]
counter_1 = Counter(list1)
def sorted_nums(list_1,list2_2):
    return sorted(counter_1.keys() - set(list2), key=counter_1.get)

sorted_nums(list_1, list_2)

Output:

[1, 3, 2, 4]

Upvotes: 1

chthonicdaemon
chthonicdaemon

Reputation: 19760

If you're OK with the default order for tied counts, this is quite succinct:


from collections import Counter

list_1 = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
list_2 = [5, 5, 5, 6, 6, 7]

blacklist = set(list_2)

c = Counter(i for i in list_1 if i not in blacklist)

[n for n,_ in reversed(c.most_common())]

The output is [3, 1, 2, 4] (note the counts for 1 and 3 are the same).

If you can rely on the ordering of your list_1 to be as it is in your example (all clusters are consecutive) you can do a bit better using itertools.groupby:

import itertools.groupby

c = sorted((len(list(g)), i) for i, g in itertools.groupby(list_1) if i not in blacklist)

[i for _, i in c]

Output is [1, 3, 2, 4]

Upvotes: 0

juanpa.arrivillaga
juanpa.arrivillaga

Reputation: 95948

You just want to sort by the occurance (counts) of lis1, taking the set-difference from the second list:

>>> list1 = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
>>> list2 = [5, 5, 5, 6, 6, 7]
>>> from collections import Counter
>>> counts1 = Counter(list1)
>>> sorted(counts1.keys() - set(list2), key=counts1.get)
[1, 3, 2, 4]

Upvotes: 2

Kuldeep Singh Sidhu
Kuldeep Singh Sidhu

Reputation: 3856

All you need is a map

def sorted_nums(list_1,list2_2):
    c = (set(list_1) - set(list_2))
    d = dict.fromkeys(c, 0)
    for index in list_1:
        if index in c:
            d[index]+=1
    return d

list_1 = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5] 
list_2 = [5, 5, 5, 6, 6, 7]
a = sorted_nums(list_1,list_2)
b = list(map(lambda x:x[0],sorted(a.items(), key = lambda x: x[1]))) ## changed here
print(b)

[1, 3, 2, 4]

Apply function lambda x:x[0] over each and every elements using map

Upvotes: 0

logi
logi

Reputation: 66

use set operations

a =  [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
b =  [5, 5, 5, 6, 6, 7]

c = sorted(list(set(a) - set(b)))
print(c)

Upvotes: 1

Related Questions