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[英]Elegant way to extract a tuple from list of tuples with minimum value of element
[英]Elegant way to find mutually exclusive tuples in the list based on the element present in each tuple
我想从彼此中减去以下两个元组以获得所需的结果(也包含在下面)。 请注意,减法仅基于(a,b)中的a。
# the two tuples
first = [(('the',), 431), (('and',), 367)]
second = [(('the',), 100), (('hello',), 100)]
# the desired result
first = [(('and',), 367)]
second = [(('hello',), 100)]
我试过map(operation.sub, first, second)
,没用。 试过b = map(sub, first, second)
,没用。 表示不支持的操作数类型 - :'tuple'和'tuple 。
感谢您的帮助和提前的时间。
编辑 :最能帮助我的解决方案包括创建两个元组的交集并从每个元组中减去它。
编辑:我想基于常见项目减去。 希望澄清一下。
这是你在找什么 -
# 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)
交叉点(编辑)
intersection = [ ((item,),interim1[item]) for item in interim1 if item in interim2]
print(intersection)
联盟(额外)
union = set().union(*[ op1, op2, intersection])
print(union)
产量
[(('the',), 431)]
[(('and',), 367)]
[(('hello',), 100)]
{(('hello',), 100), (('and',), 367), (('the',), 431)}
您可以使用带有以下列表理解的 set
来获得所需的结果:
>>> 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)]
说明:
在这里,我使用set通过在两个集合上执行XOR(也称为异或)来获取两个列表中的非公共单词。 例如:
>>> 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',)])
然后,我在列表理解中的列表上进行迭代,并检查它们是否存在于上面的非常用单词元组中。 如果它们存在,那么我们将它作为新列表的一部分保存为:
>>> 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)
也许以下就是你想要的:
# 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]
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