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?

edited Nov 15 '18 at 0:05

asked Nov 14 '18 at 23:31

Anthony

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

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?

edited Nov 15 '18 at 0:05

asked Nov 14 '18 at 23:31

Anthony

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

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?

edited Nov 15 '18 at 0:05

asked Nov 14 '18 at 23:31

Anthony

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

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

edited Nov 15 '18 at 0:05

asked Nov 14 '18 at 23:31

Anthony

1,375527

edited Nov 15 '18 at 0:05

asked Nov 14 '18 at 23:31

Anthony

1,375527

edited Nov 15 '18 at 0:05

asked Nov 14 '18 at 23:31

Anthony

1,375527

asked Nov 14 '18 at 23:31

Anthony

1,375527

asked Nov 14 '18 at 23:31

Anthony

1,375527

add a comment |

1 Answer
1

active

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.

edited Nov 14 '18 at 23:46

answered Nov 14 '18 at 23:41

TeeKea

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 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

ok, interesting. thanks anyways!!

– Anthony
Nov 15 '18 at 0:50

add a comment |

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1 Answer
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active

oldest

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1 Answer
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active

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.

edited Nov 14 '18 at 23:46

answered Nov 14 '18 at 23:41

TeeKea

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 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

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.

edited Nov 14 '18 at 23:46

answered Nov 14 '18 at 23:41

TeeKea

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 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

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.

edited Nov 14 '18 at 23:46

answered Nov 14 '18 at 23:41

TeeKea

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.

edited Nov 14 '18 at 23:46

answered Nov 14 '18 at 23:41

TeeKea

3,22851731

edited Nov 14 '18 at 23:46

answered Nov 14 '18 at 23:41

TeeKea

3,22851731

answered Nov 14 '18 at 23:41

TeeKea

3,22851731

answered Nov 14 '18 at 23:41

TeeKea

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 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

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 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

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

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

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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