基于每个元组中存在的元素在列表中查找互斥元组的优雅方式

Question

I want to subtract the two following tuples from each other to get the desired result (also included below). Note that the subtraction is based on only the a of the (a, b).

# the two tuples
first = [(('the',), 431), (('and',), 367)]
second = [(('the',), 100), (('hello',), 100)]

# the desired result
first = [(('and',), 367)]
second = [(('hello',), 100)]

I tried map(operation.sub, first, second) , didn't work. Tried b = map(sub, first, second) , didn't work. Says unsupported operand type(s) for -: 'tuple' and 'tuple .

Thanks for your help and time in advance.

Edit : A solution that would help me most would include creating an intersection of the two tuples and subtracting that from each of the tuples.

Edit: I want to subtract based on common items. Hope that clarifies it.

Answer 1

Here is what you are looking for -

# the two tuples
first = [(('the',), 431), (('and',), 367)]
second = [(('the',), 100), (('hello',), 100)]

interim1 = {i[0][0]:i[1] for i in first}

interim2 = {i[0][0]:i[1] for i in second}

op1 = [ ((item,),interim1[item]) for item in interim1 if item not in interim2]
op2 = [ ((item,),interim2[item]) for item in interim2 if item not in interim1]
print(op1)
print(op2)

Intersection (Edit)

intersection = [ ((item,),interim1[item]) for item in interim1 if item in interim2]
print(intersection)

Union (Extra)

union = set().union(*[ op1, op2, intersection])
print(union)

Output

[(('the',), 431)]
[(('and',), 367)]
[(('hello',), 100)]
{(('hello',), 100), (('and',), 367), (('the',), 431)}

Answer 2

You can use set with below list comprehension to get the desired result as:

>>> first = [(('the',), 431), (('and',), 367)]
>>> second = [(('the',), 100), (('hello',), 100)]

>>> [t for t in first+second if t[0] in set(f[0] for f in first) ^ (set(s[0] for s in second))]
[(('and',), 367), (('hello',), 100)]

---

Explanation:

Here, I am using set to get the non-common words in the two list by performing XOR (also know as Exclusive OR) on two sets. For example:

>>> first_words = set(f[0] for f in first)
>>> second_words = set(s[0] for s in second)

>>> first_words ^ second_words
set([('hello',), ('and',)])

Then I am iterating on both the list within list comprehension and checking whether they are present in above set of non-common words tuples. If they are present, then we are keeping it as the part of new list as:

>>> result = [t for t in first+second if t[0] in first_words ^ second_words]
# where `result` will hold value `[(('and',), 367), (('hello',), 100)]`

## If your resultant lists contains only two variable, 
## then you may assign them directly to individual variable as:
#     f, s = [t for t in first+second if t[0] in first_words ^ second_words]

## where `f` first required tuple will hold
# >>> f
# (('and',), 367)

## and `s` second required tuple will hold
# >>> s
# (('hello',), 100)

Answer 3

Maybe follow is what you want：

# the two tuples
first = [(('the',), 431), (('and',), 367)]
second = [(('the',), 100), (('hello',), 100)]

first_keys = set(_[0][0] for _ in first)
second_keys = set(_[0][0] for _ in second)

first = [_ for _ in first if _[0][0] not in second_keys]
second = [_ for _ in second if _[0][0] not in first_keys]

---

multi = [
    [(('the',), 431), (('and',), 367)],
    [(('the',), 100), (('hello',), 100)]
]

def get_key(x):
    return x[0][0]

def occur_counts(set_x):
    cnt = {}
    for x in set_x:
        cnt[get_key(x)] = cnt.get(get_key(x), 0) + 1
    return cnt


def do_one(single, total_cnt):
    single_cnt = occur_counts(single)
    return [_ for _ in single if single_cnt[get_key(_)] == total_cnt[get_key(_)]]


total_cnt = occur_counts(sum(multi, []))

answer = [do_one(_, total_cnt) for _ in multi]