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Opposite of set.intersection in python?
Solving 'is one away' with set()Get complement of three listsHow do I check whether a file exists without exceptions?Calling an external command in PythonWhat are metaclasses in Python?Finding the index of an item given a list containing it in PythonHow can I safely create a nested directory in Python?Does Python have a ternary conditional operator?How to know if an object has an attribute in PythonDoes Python have a string 'contains' substring method?Delete an element from a dictionaryMost elegant way to check if the string is empty in Python?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
In Python you can use a.intersection(b) to find the items common to both sets.
Is there a way to do the disjoint opposite version of this? Items that are not common to both a and b; the unique items in a unioned with the unique items in b?
python set
edited Apr 29 '15 at 15:18
Martijn Pieters♦
730k14625622361
asked Apr 29 '15 at 15:13
user4847061user4847061
275137
add a comment |
In Python you can use a.intersection(b) to find the items common to both sets.
Is there a way to do the disjoint opposite version of this? Items that are not common to both a and b; the unique items in a unioned with the unique items in b?
python set
edited Apr 29 '15 at 15:18
Martijn Pieters♦
730k14625622361
asked Apr 29 '15 at 15:13
user4847061user4847061
275137
add a comment |
In Python you can use a.intersection(b) to find the items common to both sets.
Is there a way to do the disjoint opposite version of this? Items that are not common to both a and b; the unique items in a unioned with the unique items in b?
python set
edited Apr 29 '15 at 15:18
Martijn Pieters♦
730k14625622361
asked Apr 29 '15 at 15:13
user4847061user4847061
275137
In Python you can use a.intersection(b) to find the items common to both sets.
Is there a way to do the disjoint opposite version of this? Items that are not common to both a and b; the unique items in a unioned with the unique items in b?
python set
python set
edited Apr 29 '15 at 15:18
Martijn Pieters♦
730k14625622361
asked Apr 29 '15 at 15:13
user4847061user4847061
275137
edited Apr 29 '15 at 15:18
Martijn Pieters♦
730k14625622361
asked Apr 29 '15 at 15:13
user4847061user4847061
275137
edited Apr 29 '15 at 15:18
Martijn Pieters♦
730k14625622361
edited Apr 29 '15 at 15:18
Martijn Pieters♦
730k14625622361
edited Apr 29 '15 at 15:18
Martijn Pieters♦
730k14625622361
730k14625622361
asked Apr 29 '15 at 15:13
user4847061user4847061
275137
asked Apr 29 '15 at 15:13
user4847061user4847061
275137
asked Apr 29 '15 at 15:13
user4847061user4847061
275137
275137
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
You are looking for the symmetric difference; all elements that appear only in set a or in set b, but not both:
a.symmetric_difference(b)
From the set.symmetric_difference() method documentation:
Return a new set with elements in either the set or other but not both.
You can use the ^ operator too, if both a and b are sets:
a ^ b
while set.symmetric_difference() takes any iterable for the other argument.
The output is the equivalent of (a | b) - (a & b), the union of both sets minus the intersection of both sets.
answered Apr 29 '15 at 15:15
Martijn Pieters♦Martijn Pieters
730k14625622361
Isn't ^ normaly XOR operator?
– user4847061
Apr 29 '15 at 15:31
1
@user4847061: it is, but sets have overloaded several such operators.|and&are normally bitwise OR and bitwise AND, but on sets they give you the union and the intersection. The comparison operators<,<=,>and>=have been overloaded too.
– Martijn Pieters♦
Apr 29 '15 at 15:34
add a comment |
a=1,2,4,5,6
b=5,6,4,9
c=(a^b)&b
print(c) # you got 9
answered May 28 '18 at 4:39
Kaung Yar ZarKaung Yar Zar
111
add a comment |
The best way is a list comprehension.
a = [ 1,2,3,4]
b = [ 8,7,9,2,1]
c = [ element for element in a if element not in b]
d = [ element for element in b if element not in a]
print(c)
# output is [ 3,4]
print(d)
# output is [8,7,9]
You can join both lists
answered Mar 22 at 17:24
ApeasantApeasant
213
add a comment |
Try this code for (set(a) - intersection(a&b))
a = [1,2,3,4,5,6]
b = [2,3]
for i in b:
if i in a:
a.remove(i)
print(a)
the output is [1,4,5,6]
I hope, it will work
answered Mar 11 '18 at 11:00
Muhammad Ammar FauzanMuhammad Ammar Fauzan
614
It's usually bad to mutate lists you are iterating over (in this case, there is no real consequence, unless I only care about returning a new list and not modifyinga). Alsocheck = i in ais redundant since you can alwaysif i in a:
– cowbert
Mar 14 '18 at 20:26
@cowbert thanks for your advise. I have fixed it. I will learn more about that.
– Muhammad Ammar Fauzan
Mar 16 '18 at 23:11
add a comment |
e, f are two list you want to check disjoint
a = [1,2,3,4]
b = [8,7,9,2,1]
c = []
def loop_to_check(e,f):
for i in range(len(e)):
if e[i] not in f:
c.append(e[i])
loop_to_check(a,b)
loop_to_check(b,a)
print(c)
## output is [3,4,8,7,9]
This loops around to list and returns the disjoint list
answered Aug 31 '18 at 5:16
chandra singhchandra singh
133
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are looking for the symmetric difference; all elements that appear only in set a or in set b, but not both:
a.symmetric_difference(b)
From the set.symmetric_difference() method documentation:
Return a new set with elements in either the set or other but not both.
You can use the ^ operator too, if both a and b are sets:
a ^ b
while set.symmetric_difference() takes any iterable for the other argument.
The output is the equivalent of (a | b) - (a & b), the union of both sets minus the intersection of both sets.
answered Apr 29 '15 at 15:15
Martijn Pieters♦Martijn Pieters
730k14625622361
Isn't ^ normaly XOR operator?
– user4847061
Apr 29 '15 at 15:31
1
@user4847061: it is, but sets have overloaded several such operators.|and&are normally bitwise OR and bitwise AND, but on sets they give you the union and the intersection. The comparison operators<,<=,>and>=have been overloaded too.
– Martijn Pieters♦
Apr 29 '15 at 15:34
add a comment |
You are looking for the symmetric difference; all elements that appear only in set a or in set b, but not both:
a.symmetric_difference(b)
From the set.symmetric_difference() method documentation:
Return a new set with elements in either the set or other but not both.
You can use the ^ operator too, if both a and b are sets:
a ^ b
while set.symmetric_difference() takes any iterable for the other argument.
The output is the equivalent of (a | b) - (a & b), the union of both sets minus the intersection of both sets.
answered Apr 29 '15 at 15:15
Martijn Pieters♦Martijn Pieters
730k14625622361
Isn't ^ normaly XOR operator?
– user4847061
Apr 29 '15 at 15:31
1
@user4847061: it is, but sets have overloaded several such operators.|and&are normally bitwise OR and bitwise AND, but on sets they give you the union and the intersection. The comparison operators<,<=,>and>=have been overloaded too.
– Martijn Pieters♦
Apr 29 '15 at 15:34
add a comment |
You are looking for the symmetric difference; all elements that appear only in set a or in set b, but not both:
a.symmetric_difference(b)
From the set.symmetric_difference() method documentation:
Return a new set with elements in either the set or other but not both.
You can use the ^ operator too, if both a and b are sets:
a ^ b
while set.symmetric_difference() takes any iterable for the other argument.
The output is the equivalent of (a | b) - (a & b), the union of both sets minus the intersection of both sets.
answered Apr 29 '15 at 15:15
Martijn Pieters♦Martijn Pieters
730k14625622361
You are looking for the symmetric difference; all elements that appear only in set a or in set b, but not both:
a.symmetric_difference(b)
From the set.symmetric_difference() method documentation:
Return a new set with elements in either the set or other but not both.
You can use the ^ operator too, if both a and b are sets:
a ^ b
while set.symmetric_difference() takes any iterable for the other argument.
The output is the equivalent of (a | b) - (a & b), the union of both sets minus the intersection of both sets.
answered Apr 29 '15 at 15:15
Martijn Pieters♦Martijn Pieters
730k14625622361
edited May 9 '17 at 14:27
answered Apr 29 '15 at 15:15
Martijn Pieters♦Martijn Pieters
730k14625622361
answered Apr 29 '15 at 15:15
Martijn Pieters♦Martijn Pieters
730k14625622361
answered Apr 29 '15 at 15:15
Martijn Pieters♦Martijn Pieters
730k14625622361
730k14625622361
Isn't ^ normaly XOR operator?
– user4847061
Apr 29 '15 at 15:31
1
@user4847061: it is, but sets have overloaded several such operators.|and&are normally bitwise OR and bitwise AND, but on sets they give you the union and the intersection. The comparison operators<,<=,>and>=have been overloaded too.
– Martijn Pieters♦
Apr 29 '15 at 15:34
add a comment |
Isn't ^ normaly XOR operator?
– user4847061
Apr 29 '15 at 15:31
1
@user4847061: it is, but sets have overloaded several such operators.|and&are normally bitwise OR and bitwise AND, but on sets they give you the union and the intersection. The comparison operators<,<=,>and>=have been overloaded too.
– Martijn Pieters♦
Apr 29 '15 at 15:34
Isn't ^ normaly XOR operator?
– user4847061
Apr 29 '15 at 15:31
Isn't ^ normaly XOR operator?
– user4847061
Apr 29 '15 at 15:31
1
1
@user4847061: it is, but sets have overloaded several such operators.
| and & are normally bitwise OR and bitwise AND, but on sets they give you the union and the intersection. The comparison operators <, <=, > and >= have been overloaded too.– Martijn Pieters♦
Apr 29 '15 at 15:34
@user4847061: it is, but sets have overloaded several such operators.
| and & are normally bitwise OR and bitwise AND, but on sets they give you the union and the intersection. The comparison operators <, <=, > and >= have been overloaded too.– Martijn Pieters♦
Apr 29 '15 at 15:34
add a comment |
a=1,2,4,5,6
b=5,6,4,9
c=(a^b)&b
print(c) # you got 9
answered May 28 '18 at 4:39
Kaung Yar ZarKaung Yar Zar
111
add a comment |
a=1,2,4,5,6
b=5,6,4,9
c=(a^b)&b
print(c) # you got 9
answered May 28 '18 at 4:39
Kaung Yar ZarKaung Yar Zar
111
add a comment |
a=1,2,4,5,6
b=5,6,4,9
c=(a^b)&b
print(c) # you got 9
answered May 28 '18 at 4:39
Kaung Yar ZarKaung Yar Zar
111
a=1,2,4,5,6
b=5,6,4,9
c=(a^b)&b
print(c) # you got 9
answered May 28 '18 at 4:39
Kaung Yar ZarKaung Yar Zar
111
answered May 28 '18 at 4:39
Kaung Yar ZarKaung Yar Zar
111
answered May 28 '18 at 4:39
Kaung Yar ZarKaung Yar Zar
111
answered May 28 '18 at 4:39
Kaung Yar ZarKaung Yar Zar
111
111
add a comment |
add a comment |
The best way is a list comprehension.
a = [ 1,2,3,4]
b = [ 8,7,9,2,1]
c = [ element for element in a if element not in b]
d = [ element for element in b if element not in a]
print(c)
# output is [ 3,4]
print(d)
# output is [8,7,9]
You can join both lists
answered Mar 22 at 17:24
ApeasantApeasant
213
add a comment |
The best way is a list comprehension.
a = [ 1,2,3,4]
b = [ 8,7,9,2,1]
c = [ element for element in a if element not in b]
d = [ element for element in b if element not in a]
print(c)
# output is [ 3,4]
print(d)
# output is [8,7,9]
You can join both lists
answered Mar 22 at 17:24
ApeasantApeasant
213
add a comment |
The best way is a list comprehension.
a = [ 1,2,3,4]
b = [ 8,7,9,2,1]
c = [ element for element in a if element not in b]
d = [ element for element in b if element not in a]
print(c)
# output is [ 3,4]
print(d)
# output is [8,7,9]
You can join both lists
answered Mar 22 at 17:24
ApeasantApeasant
213
The best way is a list comprehension.
a = [ 1,2,3,4]
b = [ 8,7,9,2,1]
c = [ element for element in a if element not in b]
d = [ element for element in b if element not in a]
print(c)
# output is [ 3,4]
print(d)
# output is [8,7,9]
You can join both lists
answered Mar 22 at 17:24
ApeasantApeasant
213
answered Mar 22 at 17:24
ApeasantApeasant
213
answered Mar 22 at 17:24
ApeasantApeasant
213
answered Mar 22 at 17:24
ApeasantApeasant
213
213
add a comment |
add a comment |
Try this code for (set(a) - intersection(a&b))
a = [1,2,3,4,5,6]
b = [2,3]
for i in b:
if i in a:
a.remove(i)
print(a)
the output is [1,4,5,6]
I hope, it will work
answered Mar 11 '18 at 11:00
Muhammad Ammar FauzanMuhammad Ammar Fauzan
614
It's usually bad to mutate lists you are iterating over (in this case, there is no real consequence, unless I only care about returning a new list and not modifyinga). Alsocheck = i in ais redundant since you can alwaysif i in a:
– cowbert
Mar 14 '18 at 20:26
@cowbert thanks for your advise. I have fixed it. I will learn more about that.
– Muhammad Ammar Fauzan
Mar 16 '18 at 23:11
add a comment |
Try this code for (set(a) - intersection(a&b))
a = [1,2,3,4,5,6]
b = [2,3]
for i in b:
if i in a:
a.remove(i)
print(a)
the output is [1,4,5,6]
I hope, it will work
answered Mar 11 '18 at 11:00
Muhammad Ammar FauzanMuhammad Ammar Fauzan
614
It's usually bad to mutate lists you are iterating over (in this case, there is no real consequence, unless I only care about returning a new list and not modifyinga). Alsocheck = i in ais redundant since you can alwaysif i in a:
– cowbert
Mar 14 '18 at 20:26
@cowbert thanks for your advise. I have fixed it. I will learn more about that.
– Muhammad Ammar Fauzan
Mar 16 '18 at 23:11
add a comment |
Try this code for (set(a) - intersection(a&b))
a = [1,2,3,4,5,6]
b = [2,3]
for i in b:
if i in a:
a.remove(i)
print(a)
the output is [1,4,5,6]
I hope, it will work
answered Mar 11 '18 at 11:00
Muhammad Ammar FauzanMuhammad Ammar Fauzan
614
Try this code for (set(a) - intersection(a&b))
a = [1,2,3,4,5,6]
b = [2,3]
for i in b:
if i in a:
a.remove(i)
print(a)
the output is [1,4,5,6]
I hope, it will work
answered Mar 11 '18 at 11:00
Muhammad Ammar FauzanMuhammad Ammar Fauzan
614
edited Mar 16 '18 at 23:09
answered Mar 11 '18 at 11:00
Muhammad Ammar FauzanMuhammad Ammar Fauzan
614
answered Mar 11 '18 at 11:00
Muhammad Ammar FauzanMuhammad Ammar Fauzan
614
answered Mar 11 '18 at 11:00
Muhammad Ammar FauzanMuhammad Ammar Fauzan
614
614
It's usually bad to mutate lists you are iterating over (in this case, there is no real consequence, unless I only care about returning a new list and not modifyinga). Alsocheck = i in ais redundant since you can alwaysif i in a:
– cowbert
Mar 14 '18 at 20:26
@cowbert thanks for your advise. I have fixed it. I will learn more about that.
– Muhammad Ammar Fauzan
Mar 16 '18 at 23:11
add a comment |
It's usually bad to mutate lists you are iterating over (in this case, there is no real consequence, unless I only care about returning a new list and not modifyinga). Alsocheck = i in ais redundant since you can alwaysif i in a:
– cowbert
Mar 14 '18 at 20:26
@cowbert thanks for your advise. I have fixed it. I will learn more about that.
– Muhammad Ammar Fauzan
Mar 16 '18 at 23:11
It's usually bad to mutate lists you are iterating over (in this case, there is no real consequence, unless I only care about returning a new list and not modifying
a). Also check = i in a is redundant since you can always if i in a:– cowbert
Mar 14 '18 at 20:26
It's usually bad to mutate lists you are iterating over (in this case, there is no real consequence, unless I only care about returning a new list and not modifying
a). Also check = i in a is redundant since you can always if i in a:– cowbert
Mar 14 '18 at 20:26
@cowbert thanks for your advise. I have fixed it. I will learn more about that.
– Muhammad Ammar Fauzan
Mar 16 '18 at 23:11
@cowbert thanks for your advise. I have fixed it. I will learn more about that.
– Muhammad Ammar Fauzan
Mar 16 '18 at 23:11
add a comment |
e, f are two list you want to check disjoint
a = [1,2,3,4]
b = [8,7,9,2,1]
c = []
def loop_to_check(e,f):
for i in range(len(e)):
if e[i] not in f:
c.append(e[i])
loop_to_check(a,b)
loop_to_check(b,a)
print(c)
## output is [3,4,8,7,9]
This loops around to list and returns the disjoint list
answered Aug 31 '18 at 5:16
chandra singhchandra singh
133
add a comment |
e, f are two list you want to check disjoint
a = [1,2,3,4]
b = [8,7,9,2,1]
c = []
def loop_to_check(e,f):
for i in range(len(e)):
if e[i] not in f:
c.append(e[i])
loop_to_check(a,b)
loop_to_check(b,a)
print(c)
## output is [3,4,8,7,9]
This loops around to list and returns the disjoint list
answered Aug 31 '18 at 5:16
chandra singhchandra singh
133
add a comment |
e, f are two list you want to check disjoint
a = [1,2,3,4]
b = [8,7,9,2,1]
c = []
def loop_to_check(e,f):
for i in range(len(e)):
if e[i] not in f:
c.append(e[i])
loop_to_check(a,b)
loop_to_check(b,a)
print(c)
## output is [3,4,8,7,9]
This loops around to list and returns the disjoint list
answered Aug 31 '18 at 5:16
chandra singhchandra singh
133
e, f are two list you want to check disjoint
a = [1,2,3,4]
b = [8,7,9,2,1]
c = []
def loop_to_check(e,f):
for i in range(len(e)):
if e[i] not in f:
c.append(e[i])
loop_to_check(a,b)
loop_to_check(b,a)
print(c)
## output is [3,4,8,7,9]
This loops around to list and returns the disjoint list
answered Aug 31 '18 at 5:16
chandra singhchandra singh
133
answered Aug 31 '18 at 5:16
chandra singhchandra singh
133
answered Aug 31 '18 at 5:16
chandra singhchandra singh
133
answered Aug 31 '18 at 5:16
chandra singhchandra singh
133
133
add a comment |
add a comment |
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