如何在新列表中获取列表的唯一编号？

Question

I have this list:

list = [{1,2,3,4}, {3,4,5}, {2,6}] 

I want to have this as the output:

{1,6}

So I only want the unique numbers to have an output. This is what I tried, but it does not work:

list = [{1,2,3,4}, {3,4,5}, {2,6}] 
s1 = []
for number in list:
   if number not in s1:
      s1.append(number)
def unique(s1):
   return set.difference(*s1)
print (unique(s1))

My output is:

{1, 2, 3, 4}

I have no clue how to fix this? I am a python beginner so can anyone explain what the answer is and why that should be the solution? Many thanks in advance!

Answer 1

Use a Counter object.

>>> from collections import Counter
>>> from itertools import chain
>>> my_list = [{1,2,3,4}, {3,4,5}, {2,6}]
>>> set(k for k,v in Counter(chain.from_iterable(my_list)).items() if v == 1)
set([1, 5, 6])

chain.from_iterable "flattens" my_list . The Counter compiles how many times each element from the flattened list is seen, and the generator expression sends only the keys mapped to a value of 1 to set .

---

Some of the intermediate values involved:

>>> list(chain.from_iterable(my_list))
[1, 2, 3, 4, 3, 4, 5, 2, 6]
>>> Counter(chain.from_iterable(my_list))
Counter({2: 2, 3: 2, 4: 2, 1: 1, 5: 1, 6: 1})

Answer 2

I had an earlier answer posted that wasn't right. I don't like to give up, and I wanted to find an answer to this question that only used sets. I'm not saying that this is a better answer than the others, but it does achieve at least my own goal of not bringing any extra packages into the solution:

def unique(data):
    result = set()
    dups = set()
    for s1 in data:
        # accumulate everything we see more than once
        dups = dups | (result & s1)
        # accumulate everything
        result = (result | s1)
    # the result is everything we only saw once, or in other words,
    # everything we saw minus everything we saw more than once
    return result - dups

print(unique([{1,2,3,4}, {3,4,5}, {2,6}]))
print(unique([{1}, {1}, {1}]))

Output:

set([1, 5, 6])
set([])

There's probably a better solution even to using just sets. I always want to try to use the simplest set of tools possible because I think it is the most understandable to do so, and will often turn out to be as efficient as anything else as well.

Answer 3

Not a fancy python solution but what you probably want is to build a dict where the key is a number in the list and the value is the number of times it occurs.

This solution is by no way optimal since we are performing to many iterations:

list = [{1,2,3,4}, {3,4,5}, {2,6}] 
count_hash = {}

for dict in list:
    for number in dict:
        if number in count_hash:
            count_hash[number] += 1
        else:
            count_hash[number] = 1

for number in count_hash:
    if count_hash[number] == 1:
        print(number)

Answer 4

I agree with chepner on using a Counter. Using functools .reduce to flatten the list came from here .

import functools
import operator
from collections import Counter

my_list = [{1,2,3,4}, {3,4,5}, {2,6}]
flat_list = functools.reduce(operator.iconcat, my_list, [])

results = [k for (k, v) in Counter(flat_list).items() if v == 1]

print(results)

# OUTPUT
# [1, 5, 6]

Answer 5

I didn't see any of the answers given that were like my solution. I hope the symmetric difference of the sets given provides that correct manner (to this question).

>>> L = [{1,2,3,4}, {3,4,5}, {2,6}]
>>> M = L[0]
>>> for s in L[1:]:
    M ^= s

    
>>> M
{1, 5, 6}

Note: The question was about lists but the list he gave was a list of sets.