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Effectively determine if two of the three items in a list are the same

What is the most efficient way to determine if exactly two items in a list are the same? For instance:

>>> has1dup(["one", "one", "two"])
True
>>> has1dup(["one", "two", "three"])
False
>>> has1dup(["one", "one", "one"])
False

I have successfully done this using if / else statements. However, if the list was larger, the task of writing every opportunity for the couple would become very difficult and time-consuming. Is there a faster / easier way to do this?

Here is what I tried:

def has1dup(lst):
    if lst[0] == lst[1] and lst[1] != lst[2]:
        return True
    elif lst[1] == lst[2] and lst[2] != lst[0]:
        return True
    elif lst[0] == lst[2] and lst[2] != lst[1]:
        return True
    else:
        return False
+3
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12 answers

You can see how many unique values ​​there are with set. If the set has one more element than the list, it was a duplicate:

def has1dup(lst):
    return len(lst)-1 == len(set(lst))
+11
source 
>>> from collection import Counter
>>> 2 in Counter(["one", "one", "two"]).values()
True
>>> 2 in Counter(["one", "two", "three"]).values()
False

Update
If you want, there are only two identical elements.

Counter(seq).values().count(2) == 1

Counter Python 2.7+,

def counter(seq): 
    r = {}
    for x in seq:
        r[x] = r.setdefault(x, 0) + 1 # or defaultdict
    return r
+3

, any() :

def has_duplicates(seq):
    return any(seq.count(x) > 1 for x in seq)

:

>>> has_duplicates([1, 1, 2])
True
>>> has_duplicates([1, 2, 2])
True
>>> has_duplicates([1, 2, 3])
False

, , :

any(seq.count(x) == 2 for x in seq)

, , , , :

def any_n(iterable, n):
    seen = 0
    for value in iterable:
        if value:
            if seen >= n:
                return False
            else:
                seen += 1
    return seen == n

def has_one_value_repeated_n_times(seq, n):
    return any_n((seq.count(x) == n for x in seq), n)

:

tests = [
    [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5],
    [1,2,2,3,3,4,4,4,4,5,5,5,5,5],
    [1,2,2],
    [1,1,2],
    [1,2,3],
]

for test in tests:
    print(test, "-", has_one_value_repeated_n_times(test, 2))

:

[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5] - True
[1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5] - False
[1, 2, 2] - True
[1, 1, 2] - True
[1, 2, 3] - False
+2
2 in collections.Counter(yourList).values()

.

"", " ", 2 ", :

Counter(Counter(yourList).values()).get(2)==1
+1

Python, :

for i in 0 to length(list):
    j = indexOf(list, list[i], i + 1) > -1

    if j > -1 and indexOf(list, list[i], j + 1) == -1:
        // Found exactly two!

.

EDIT: , Python. , .

def exactlyTwo(l):
    for i in xrange(0, len(l)):
        try:
            j = l.index(l[i], i + 1)

            try:
                l.index(l[i], j + 1)
            except ValueError:
                return True
        except ValueError:
            # Do nothing. Not sure how to do that in Python.
            0

    return False

.

0

: len(set(my_list)) == 2

, (3 ), , : > python -c 'from timeit import Timer; t1 = Timer("a[0] == a[1] != a[2] or a[0] == a[2] != a[1] or a[1] == a[2] != a[0]", setup="""a=["one","one","two"]"""); t2 = Timer("len(set(a)) == 2", setup="""a=["one","one","two"]"""); print t1.timeit(), t2.timeit()'

0.893960952759 2.28438997269

0

, ( , ) - , , .

, , , , , .

, " ": -)

if list[0] == list[1] and list[1] != list[2]:
    return True

if list[1] == list[2] and list[2] != list[0]:
    return True

if list[0] == list[2] and list[2] != list[1]:
    return True

return False

, , / :

if list[0] == list[1] and list[1] != list[2]: return True
if list[1] == list[2] and list[2] != list[0]: return True
if list[0] == list[2] and list[2] != list[1]: return True
return False

, , :

>>> a = [1,2,3,4,4,5,6,7]
>>> a
[1, 2, 3, 4, 4, 5, 6, 7]
>>> len(a) == len (set(a)) + 1
True
0

/ , (, "" 1 dict)

def find_repeated_element(l):
    return reduce(lambda x,y: x^y, l + list(set(l)))

l = [1,1,2]
>>> find_repeated_element(l)
1
l = [1,2,3]
>>> find_repeated_element(l)
0
l = [1,1,1]
>>> find_repeated_element(l)
0
0

, itertools.product ?

from itertools import product

tests = """\
AAB
ABC
ABB
ABA
AAA""".splitlines()

def has_exactly_one_doubled_element(t):
    return sum(map(lambda a:a[0]==a[1], product(t,t))) == 5

for t in tests:
    print t, has_exactly_one_doubled_element(t)

:

AAB True
ABC False
ABB True
ABA True
AAA False

, 5? , , , . 3 , , , 3. 2 , 3 , , 2 3, 5. 3 , , 3 * 3 9.

, , ( operator.__eq__ itertools.starmap, lambda map):

from operator import __eq__ as EQ
from itertools import product, starmap

def has_exactly_one_doubled_element(t):
    return sum(starmap(EQ, product(t,t))) == len(t)+2

args = ["ABCD"]*4
tests = map(''.join, product(*args))
for t in tests:
    print t, has_exactly_one_doubled_element(t)


AAAA False
AAAB False
AAAC False
AAAD False
AABA False
AABB False
AABC True
AABD True
AACA False
AACB True
AACC False
AACD True
AADA False
AADB True
AADC True
AADD False
ABAA False
ABAB False
ABAC True
ABAD True
ABBA False
ABBB False
ABBC True
ABBD True
ABCA True
ABCB True
ABCC True
ABCD False
ABDA True
ABDB True
ABDC False
ABDD True
ACAA False
ACAB True
ACAC False
ACAD True
ACBA True
ACBB True
ACBC True
ACBD False
ACCA False
ACCB True
ACCC False
ACCD True
ACDA True
ACDB False
ACDC True
ACDD True
ADAA False
ADAB True
ADAC True
ADAD False
ADBA True
ADBB True
ADBC False
ADBD True
ADCA True
ADCB False
ADCC True
ADCD True
ADDA False
ADDB True
ADDC True
ADDD False
BAAA False
BAAB False
BAAC True
BAAD True
BABA False
BABB False
BABC True
BABD True
BACA True
BACB True
BACC True
BACD False
BADA True
BADB True
BADC False
BADD True
BBAA False
BBAB False
BBAC True
BBAD True
BBBA False
BBBB False
BBBC False
BBBD False
BBCA True
BBCB False
BBCC False
BBCD True
BBDA True
BBDB False
BBDC True
BBDD False
BCAA True
BCAB True
BCAC True
BCAD False
BCBA True
BCBB False
BCBC False
BCBD True
BCCA True
BCCB False
BCCC False
BCCD True
BCDA False
BCDB True
BCDC True
BCDD True
BDAA True
BDAB True
BDAC False
BDAD True
BDBA True
BDBB False
BDBC True
BDBD False
BDCA False
BDCB True
BDCC True
BDCD True
BDDA True
BDDB False
BDDC True
BDDD False
CAAA False
CAAB True
CAAC False
CAAD True
CABA True
CABB True
CABC True
CABD False
CACA False
CACB True
CACC False
CACD True
CADA True
CADB False
CADC True
CADD True
CBAA True
CBAB True
CBAC True
CBAD False
CBBA True
CBBB False
CBBC False
CBBD True
CBCA True
CBCB False
CBCC False
CBCD True
CBDA False
CBDB True
CBDC True
CBDD True
CCAA False
CCAB True
CCAC False
CCAD True
CCBA True
CCBB False
CCBC False
CCBD True
CCCA False
CCCB False
CCCC False
CCCD False
CCDA True
CCDB True
CCDC False
CCDD False
CDAA True
CDAB False
CDAC True
CDAD True
CDBA False
CDBB True
CDBC True
CDBD True
CDCA True
CDCB True
CDCC False
CDCD False
CDDA True
CDDB True
CDDC False
CDDD False
DAAA False
DAAB True
DAAC True
DAAD False
DABA True
DABB True
DABC False
DABD True
DACA True
DACB False
DACC True
DACD True
DADA False
DADB True
DADC True
DADD False
DBAA True
DBAB True
DBAC False
DBAD True
DBBA True
DBBB False
DBBC True
DBBD False
DBCA False
DBCB True
DBCC True
DBCD True
DBDA True
DBDB False
DBDC True
DBDD False
DCAA True
DCAB False
DCAC True
DCAD True
DCBA False
DCBB True
DCBC True
DCBD True
DCCA True
DCCB True
DCCC False
DCCD False
DCDA True
DCDB True
DCDC False
DCDD False
DDAA False
DDAB True
DDAC True
DDAD False
DDBA True
DDBB False
DDBC True
DDBD False
DDCA True
DDCB True
DDCC False
DDCD False
DDDA False
DDDB False
DDDC False
DDDD False
0
def exactlyTwo(elements):
    return sum(elements.count(i)-1 for i in elements) == 2

works for non-displayable elements and where “exactly two elements” mean triplets and two doubles, should return False

0
source 
def checkk(lis):
  sett=set(lis)
  for ele in sett:
     if lis.count(ele)==2:
       return 'yes'
  else:
     return 'no'


x=list(["one", "one", "two"])
y=list(["one", "two", "three"])
z=list(["one", "one", "one"])
w=list(["one", "two", "two"])
print(checkk(x))
print(checkk(y))
print(checkk(z))
print(checkk(w))

output:

yes
no
no
yes
-1
source

You can try

list(["one", "one", "two"]).count("one") == 2
-2
source

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