symmetric_difference Output as Two Separate Lists
Let's say we have two lists of values, whereby each list only contains unique values unto itself. There will never be duplicate values in a single list.
L1 | L2
-------
a | a
b | d
c | e
d | g
e | h
f | i
| j
We can get the differences of these lists using set(L1).symmetric_difference(L2), but unfortunately that lumps the results together in a single list. For example, the output of list(set(L1).symmetric_difference(L2)) is ['c', 'b', 'h', 'i', 'j', 'f', 'g'].
Is there a way to obtain two separate lists of output from list(set(L1).symmetric_difference(L2)) like ['c', 'b', 'f',] and ['h', 'i', 'j', 'g'] instead?
Or is there a way to obtain two separate lists as output while comparing the two sets/lists against each other only once?
arrays python-3.x list symmetric-difference
asked Nov 14 '18 at 23:31
AnthonyAnthony
1,375527
add a comment |
Let's say we have two lists of values, whereby each list only contains unique values unto itself. There will never be duplicate values in a single list.
L1 | L2
-------
a | a
b | d
c | e
d | g
e | h
f | i
| j
We can get the differences of these lists using set(L1).symmetric_difference(L2), but unfortunately that lumps the results together in a single list. For example, the output of list(set(L1).symmetric_difference(L2)) is ['c', 'b', 'h', 'i', 'j', 'f', 'g'].
Is there a way to obtain two separate lists of output from list(set(L1).symmetric_difference(L2)) like ['c', 'b', 'f',] and ['h', 'i', 'j', 'g'] instead?
Or is there a way to obtain two separate lists as output while comparing the two sets/lists against each other only once?
arrays python-3.x list symmetric-difference
asked Nov 14 '18 at 23:31
AnthonyAnthony
1,375527
add a comment |
Let's say we have two lists of values, whereby each list only contains unique values unto itself. There will never be duplicate values in a single list.
L1 | L2
-------
a | a
b | d
c | e
d | g
e | h
f | i
| j
We can get the differences of these lists using set(L1).symmetric_difference(L2), but unfortunately that lumps the results together in a single list. For example, the output of list(set(L1).symmetric_difference(L2)) is ['c', 'b', 'h', 'i', 'j', 'f', 'g'].
Is there a way to obtain two separate lists of output from list(set(L1).symmetric_difference(L2)) like ['c', 'b', 'f',] and ['h', 'i', 'j', 'g'] instead?
Or is there a way to obtain two separate lists as output while comparing the two sets/lists against each other only once?
arrays python-3.x list symmetric-difference
asked Nov 14 '18 at 23:31
AnthonyAnthony
1,375527
Let's say we have two lists of values, whereby each list only contains unique values unto itself. There will never be duplicate values in a single list.
L1 | L2
-------
a | a
b | d
c | e
d | g
e | h
f | i
| j
We can get the differences of these lists using set(L1).symmetric_difference(L2), but unfortunately that lumps the results together in a single list. For example, the output of list(set(L1).symmetric_difference(L2)) is ['c', 'b', 'h', 'i', 'j', 'f', 'g'].
Is there a way to obtain two separate lists of output from list(set(L1).symmetric_difference(L2)) like ['c', 'b', 'f',] and ['h', 'i', 'j', 'g'] instead?
Or is there a way to obtain two separate lists as output while comparing the two sets/lists against each other only once?
arrays python-3.x list symmetric-difference
arrays python-3.x list symmetric-difference
asked Nov 14 '18 at 23:31
AnthonyAnthony
1,375527
asked Nov 14 '18 at 23:31
AnthonyAnthony
1,375527
edited Nov 15 '18 at 0:05
Anthony
asked Nov 14 '18 at 23:31
AnthonyAnthony
1,375527
asked Nov 14 '18 at 23:31
AnthonyAnthony
1,375527
asked Nov 14 '18 at 23:31
AnthonyAnthony
1,375527
1,375527
add a comment |
add a comment |
1 Answer
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You can simply do the following:
dif_1_from_2 = list(set(L1) - set(L2))
dif_2_from_1 = list(set(L2) - set(L1))
And you can create a function to do that like this:
def get_symmetric_difference(L1, L2):
return list(set(L1)-set(L2)), list(set(L2)-set(L1))
and then you call it like this:
print(get_symmetric_difference(L1, L2))
Hope this helps.
answered Nov 14 '18 at 23:41
TeeKeaTeeKea
3,22851731
ok, thx, but is there no way to get the two lists of output while comparing the sets only once??
– Anthony
Nov 14 '18 at 23:46
Is my updated answer gives you what you asked in your comment? Please let me know.
– TeeKea
Nov 14 '18 at 23:48
no, unfortunately that's not what i meant. what i'm trying to get at it is thatdif_1_from_2compares the sets once, thendif_2_from_1compares the sets for a 2nd time. is there no way to compareL1andL2once, but get two different lists as the output? does that make sense?
– Anthony
Nov 14 '18 at 23:50
Not sure, actually. I believe you would really need two operations to do that at the end.
– TeeKea
Nov 15 '18 at 0:21
ok, interesting. thanks anyways!!
– Anthony
Nov 15 '18 at 0:50
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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oldest
votes
You can simply do the following:
dif_1_from_2 = list(set(L1) - set(L2))
dif_2_from_1 = list(set(L2) - set(L1))
And you can create a function to do that like this:
def get_symmetric_difference(L1, L2):
return list(set(L1)-set(L2)), list(set(L2)-set(L1))
and then you call it like this:
print(get_symmetric_difference(L1, L2))
Hope this helps.
answered Nov 14 '18 at 23:41
TeeKeaTeeKea
3,22851731
ok, thx, but is there no way to get the two lists of output while comparing the sets only once??
– Anthony
Nov 14 '18 at 23:46
Is my updated answer gives you what you asked in your comment? Please let me know.
– TeeKea
Nov 14 '18 at 23:48
no, unfortunately that's not what i meant. what i'm trying to get at it is thatdif_1_from_2compares the sets once, thendif_2_from_1compares the sets for a 2nd time. is there no way to compareL1andL2once, but get two different lists as the output? does that make sense?
– Anthony
Nov 14 '18 at 23:50
Not sure, actually. I believe you would really need two operations to do that at the end.
– TeeKea
Nov 15 '18 at 0:21
ok, interesting. thanks anyways!!
– Anthony
Nov 15 '18 at 0:50
add a comment |
You can simply do the following:
dif_1_from_2 = list(set(L1) - set(L2))
dif_2_from_1 = list(set(L2) - set(L1))
And you can create a function to do that like this:
def get_symmetric_difference(L1, L2):
return list(set(L1)-set(L2)), list(set(L2)-set(L1))
and then you call it like this:
print(get_symmetric_difference(L1, L2))
Hope this helps.
answered Nov 14 '18 at 23:41
TeeKeaTeeKea
3,22851731
ok, thx, but is there no way to get the two lists of output while comparing the sets only once??
– Anthony
Nov 14 '18 at 23:46
Is my updated answer gives you what you asked in your comment? Please let me know.
– TeeKea
Nov 14 '18 at 23:48
no, unfortunately that's not what i meant. what i'm trying to get at it is thatdif_1_from_2compares the sets once, thendif_2_from_1compares the sets for a 2nd time. is there no way to compareL1andL2once, but get two different lists as the output? does that make sense?
– Anthony
Nov 14 '18 at 23:50
Not sure, actually. I believe you would really need two operations to do that at the end.
– TeeKea
Nov 15 '18 at 0:21
ok, interesting. thanks anyways!!
– Anthony
Nov 15 '18 at 0:50
add a comment |
You can simply do the following:
dif_1_from_2 = list(set(L1) - set(L2))
dif_2_from_1 = list(set(L2) - set(L1))
And you can create a function to do that like this:
def get_symmetric_difference(L1, L2):
return list(set(L1)-set(L2)), list(set(L2)-set(L1))
and then you call it like this:
print(get_symmetric_difference(L1, L2))
Hope this helps.
answered Nov 14 '18 at 23:41
TeeKeaTeeKea
3,22851731
You can simply do the following:
dif_1_from_2 = list(set(L1) - set(L2))
dif_2_from_1 = list(set(L2) - set(L1))
And you can create a function to do that like this:
def get_symmetric_difference(L1, L2):
return list(set(L1)-set(L2)), list(set(L2)-set(L1))
and then you call it like this:
print(get_symmetric_difference(L1, L2))
Hope this helps.
answered Nov 14 '18 at 23:41
TeeKeaTeeKea
3,22851731
edited Nov 14 '18 at 23:46
answered Nov 14 '18 at 23:41
TeeKeaTeeKea
3,22851731
answered Nov 14 '18 at 23:41
TeeKeaTeeKea
3,22851731
answered Nov 14 '18 at 23:41
TeeKeaTeeKea
3,22851731
3,22851731
ok, thx, but is there no way to get the two lists of output while comparing the sets only once??
– Anthony
Nov 14 '18 at 23:46
Is my updated answer gives you what you asked in your comment? Please let me know.
– TeeKea
Nov 14 '18 at 23:48
no, unfortunately that's not what i meant. what i'm trying to get at it is thatdif_1_from_2compares the sets once, thendif_2_from_1compares the sets for a 2nd time. is there no way to compareL1andL2once, but get two different lists as the output? does that make sense?
– Anthony
Nov 14 '18 at 23:50
Not sure, actually. I believe you would really need two operations to do that at the end.
– TeeKea
Nov 15 '18 at 0:21
ok, interesting. thanks anyways!!
– Anthony
Nov 15 '18 at 0:50
add a comment |
ok, thx, but is there no way to get the two lists of output while comparing the sets only once??
– Anthony
Nov 14 '18 at 23:46
Is my updated answer gives you what you asked in your comment? Please let me know.
– TeeKea
Nov 14 '18 at 23:48
no, unfortunately that's not what i meant. what i'm trying to get at it is thatdif_1_from_2compares the sets once, thendif_2_from_1compares the sets for a 2nd time. is there no way to compareL1andL2once, but get two different lists as the output? does that make sense?
– Anthony
Nov 14 '18 at 23:50
Not sure, actually. I believe you would really need two operations to do that at the end.
– TeeKea
Nov 15 '18 at 0:21
ok, interesting. thanks anyways!!
– Anthony
Nov 15 '18 at 0:50
ok, thx, but is there no way to get the two lists of output while comparing the sets only once??
– Anthony
Nov 14 '18 at 23:46
ok, thx, but is there no way to get the two lists of output while comparing the sets only once??
– Anthony
Nov 14 '18 at 23:46
Is my updated answer gives you what you asked in your comment? Please let me know.
– TeeKea
Nov 14 '18 at 23:48
Is my updated answer gives you what you asked in your comment? Please let me know.
– TeeKea
Nov 14 '18 at 23:48
no, unfortunately that's not what i meant. what i'm trying to get at it is that
dif_1_from_2 compares the sets once, then dif_2_from_1 compares the sets for a 2nd time. is there no way to compare L1 and L2 once, but get two different lists as the output? does that make sense?– Anthony
Nov 14 '18 at 23:50
no, unfortunately that's not what i meant. what i'm trying to get at it is that
dif_1_from_2 compares the sets once, then dif_2_from_1 compares the sets for a 2nd time. is there no way to compare L1 and L2 once, but get two different lists as the output? does that make sense?– Anthony
Nov 14 '18 at 23:50
Not sure, actually. I believe you would really need two operations to do that at the end.
– TeeKea
Nov 15 '18 at 0:21
Not sure, actually. I believe you would really need two operations to do that at the end.
– TeeKea
Nov 15 '18 at 0:21
ok, interesting. thanks anyways!!
– Anthony
Nov 15 '18 at 0:50
ok, interesting. thanks anyways!!
– Anthony
Nov 15 '18 at 0:50
add a comment |
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