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test for list equality

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  test for list equality noydb <jenn.duerr@gmail.com> - 2011-12-15 08:36 -0800
    Re: test for list equality noydb <jenn.duerr@gmail.com> - 2011-12-15 09:49 -0800
      Re: test for list equality John Gordon <gordon@panix.com> - 2011-12-15 17:57 +0000
      Re: test for list equality Miki Tebeka <miki.tebeka@gmail.com> - 2011-12-15 09:59 -0800
        Re: test for list equality MRAB <python@mrabarnett.plus.com> - 2011-12-15 18:12 +0000
        Re: test for list equality darnold <darnold992000@yahoo.com> - 2011-12-15 10:12 -0800
        Re: test for list equality Tim Chase <python.list@tim.thechases.com> - 2011-12-15 12:20 -0600
          Re: test for list equality Miki Tebeka <miki.tebeka@gmail.com> - 2011-12-15 10:25 -0800
          Re: test for list equality Miki Tebeka <miki.tebeka@gmail.com> - 2011-12-15 10:25 -0800
        Re: test for list equality Terry Reedy <tjreedy@udel.edu> - 2011-12-15 20:02 -0500
      Re: test for list equality Stefan Behnel <stefan_ml@behnel.de> - 2011-12-15 19:01 +0100
        Re: test for list equality noydb <jenn.duerr@gmail.com> - 2011-12-15 10:07 -0800
          Re: test for list equality Ian Kelly <ian.g.kelly@gmail.com> - 2011-12-15 11:28 -0700
            Re: test for list equality noydb <jenn.duerr@gmail.com> - 2011-12-15 10:32 -0800
      Re: test for list equality MRAB <python@mrabarnett.plus.com> - 2011-12-15 18:06 +0000
    Re: test for list equality MRAB <python@mrabarnett.plus.com> - 2011-12-15 17:54 +0000
    Re: test for list equality Alec Taylor <alec.taylor6@gmail.com> - 2011-12-16 17:30 +1100
    Re: test for list equality Ian Kelly <ian.g.kelly@gmail.com> - 2011-12-16 00:11 -0700
    Re: test for list equality Ian Kelly <ian.g.kelly@gmail.com> - 2011-12-16 00:32 -0700

#17284 — test for list equality

I want to test for equality between two lists.  For example, if I have
two lists that are equal in content but not in order, I want a return
of 'equal' -- dont care if they are not in the same order.  In order
to get that equality, would I have to sort both lists regardless?  if
yes, how (having issues with list.sort)?

Another way i tried, that I think is kind-of roundabout is like
x = [2, 5, 1, 88, 9]
y = [5, 2, 9, 1, 88]
inBoth = list(set(x) & set(y))

and then test that list.count is equal between inBoth and x and/or y.


Any better suggestions?

Thanks for any help!

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#17285

On Dec 15, 11:36 am, noydb <jenn.du...@gmail.com> wrote:
> I want to test for equality between two lists.  For example, if I have
> two lists that are equal in content but not in order, I want a return
> of 'equal' -- dont care if they are not in the same order.  In order
> to get that equality, would I have to sort both lists regardless?  if
> yes, how (having issues with list.sort)?
>
> Another way i tried, that I think is kind-of roundabout is like
> x = [2, 5, 1, 88, 9]
> y = [5, 2, 9, 1, 88]
> inBoth = list(set(x) & set(y))
>
> and then test that list.count is equal between inBoth and x and/or y.
>
> Any better suggestions?
>
> Thanks for any help!

My sort issue... as in this doesn't work
>>> if x.sort == y.sort:
... 	print 'equal'
... else:
... 	print 'not equal'
...
not equal


???

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#17287

In <61edc02c-4f86-45ef-82a1-61c7013003b4@t38g2000yqe.googlegroups.com> noydb <jenn.duerr@gmail.com> writes:

> My sort issue... as in this doesn't work
> >>> if x.sort =3D=3D y.sort:
> ... 	print 'equal'
> ... else:
> ... 	print 'not equal'
> ...
> not equal

> ???

Use x.sort() instead of x.sort .

-- 
John Gordon                   A is for Amy, who fell down the stairs
gordon@panix.com              B is for Basil, assaulted by bears
                                -- Edward Gorey, "The Gashlycrumb Tinies"

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#17288

> My sort issue... as in this doesn't work
> >>> if x.sort == y.sort:
You're missing the () to make it a function call.
Also list.sort() returns none, it mutates the original list.
You can either
    sorted(x) == sorted(y)
or
    set(x) == set(y)

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#17294

On 15/12/2011 17:59, Miki Tebeka wrote:
>>  My sort issue... as in this doesn't work
>>  >>>  if x.sort == y.sort:
> You're missing the () to make it a function call.
> Also list.sort() returns none, it mutates the original list.
> You can either
>      sorted(x) == sorted(y)
> or
>      set(x) == set(y)

But don't use sets if there may be duplicates.

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#17295

On Dec 15, 11:59 am, Miki Tebeka <miki.teb...@gmail.com> wrote:
> > My sort issue... as in this doesn't work
> > >>> if x.sort == y.sort:
>
> You're missing the () to make it a function call.
> Also list.sort() returns none, it mutates the original list.
> You can either
>     sorted(x) == sorted(y)
> or
>     set(x) == set(y)

I'm pretty sure we don't want to use set() since it throws away
duplicates:

>>> x = [1,2,3,4]
>>> y = [1,1,2,2,3,3,4]
>>> sorted(x) == sorted(y)
False
>>> set(x) == set(y)
True

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#17297

On 12/15/11 11:59, Miki Tebeka wrote:
>> My sort issue... as in this doesn't work
>>>>> if x.sort == y.sort:
> You're missing the () to make it a function call.
> Also list.sort() returns none, it mutates the original list.
> You can either
>      sorted(x) == sorted(y)
> or
>      set(x) == set(y)

Duplicates cause issues in the set() version:

   a = [1,2,3,4]
   b = a + a
   print sorted(a) == sorted(b) # False
   print set(a) == set(b) # True

They mean different things, and the OP may want one or the other 
depending on how they want to consider duplicates.

-tkc

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#17299

> >      set(x) == set(y)
> 
> Duplicates cause issues in the set() version:
You're right, I stand corrected.

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#17303

> >      set(x) == set(y)
> 
> Duplicates cause issues in the set() version:
You're right, I stand corrected.

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#17323

On 12/15/2011 12:59 PM, Miki Tebeka wrote:
>> My sort issue... as in this doesn't work
>>>>> if x.sort == y.sort:
> You're missing the () to make it a function call.
> Also list.sort() returns none, it mutates the original list.
> You can either
>      sorted(x) == sorted(y)
> or
>      set(x) == set(y)

or x.sort(); y.sort(); x == y

-- 
Terry Jan Reedy

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#17289

noydb, 15.12.2011 18:49:
> On Dec 15, 11:36 am, noydb wrote:
>> I want to test for equality between two lists.  For example, if I have
>> two lists that are equal in content but not in order, I want a return
>> of 'equal' -- dont care if they are not in the same order.  In order
>> to get that equality, would I have to sort both lists regardless?  if
>> yes, how (having issues with list.sort)?
>>
>> Another way i tried, that I think is kind-of roundabout is like
>> x = [2, 5, 1, 88, 9]
>> y = [5, 2, 9, 1, 88]
>> inBoth = list(set(x)&  set(y))
>>
>> and then test that list.count is equal between inBoth and x and/or y.
>>
>> Any better suggestions?
>>
>> Thanks for any help!
>
> My sort issue... as in this doesn't work
> >>> if x.sort == y.sort:
> ... 	print 'equal'
> ... else:
> ... 	print 'not equal'
> ...
> not equal

alist.sort() is a method, so you have to call it in order to execute it. 
alist.sort will only give you a reference to the method. Comparing that to 
another method reference returns False, as expected.

Also, calling it does not return anything (useful), it modifies the list in 
place.

If you want to create a new list (which you don't want in this case, but 
anyway), you can use the sorted() builtin function.

Stefan

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#17293

Ahh, I see (on the sort issue), thanks All!

Still, any other slicker ways to do this?  Just for learning.

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#17300

On Thu, Dec 15, 2011 at 11:07 AM, noydb <jenn.duerr@gmail.com> wrote:
> Ahh, I see (on the sort issue), thanks All!
>
> Still, any other slicker ways to do this?  Just for learning.

MRAB's collections.Counter suggestion is what I would do.  Very tidy,
and also more efficient I think: O(n) instead of O(n log n).

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#17301

Thanks All, Thanks MRAB -- good note on the sort() vs sorted().

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#17292

On 15/12/2011 17:49, noydb wrote:
> On Dec 15, 11:36 am, noydb<jenn.du...@gmail.com>  wrote:
>>  I want to test for equality between two lists.  For example, if I have
>>  two lists that are equal in content but not in order, I want a return
>>  of 'equal' -- dont care if they are not in the same order.  In order
>>  to get that equality, would I have to sort both lists regardless?  if
>>  yes, how (having issues with list.sort)?
>>
>>  Another way i tried, that I think is kind-of roundabout is like
>>  x = [2, 5, 1, 88, 9]
>>  y = [5, 2, 9, 1, 88]
>>  inBoth = list(set(x)&  set(y))
>>
>>  and then test that list.count is equal between inBoth and x and/or y.
>>
>>  Any better suggestions?
>>
>>  Thanks for any help!
>
> My sort issue... as in this doesn't work
>>>>  if x.sort == y.sort:
> ... 	print 'equal'
> ... else:
> ... 	print 'not equal'
> ...
> not equal
>
>
> ???

.sort is a method which sorts the list in-place and returns None. You
must provide the () if you want to call it, otherwise you just get a
reference to the method:

 >>> x
[2, 5, 1, 88, 9]
 >>> x.sort
<built-in method sort of list object at 0x00A58508>
 >>> x.sort()
 >>> x
[1, 2, 5, 9, 88]

There's also a function "sorted" which returns its argument as a sorted
list. The argument itself isn't altered:

 >>> y = [5, 2, 9, 1, 88]
 >>> sorted(y)
[1, 2, 5, 9, 88]
 >>> y
[5, 2, 9, 1, 88]

It's all in the documentation! :-)

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#17286

On 15/12/2011 16:36, noydb wrote:
> I want to test for equality between two lists.  For example, if I have
> two lists that are equal in content but not in order, I want a return
> of 'equal' -- dont care if they are not in the same order.  In order
> to get that equality, would I have to sort both lists regardless?  if
> yes, how (having issues with list.sort)?
>
> Another way i tried, that I think is kind-of roundabout is like
> x = [2, 5, 1, 88, 9]
> y = [5, 2, 9, 1, 88]
> inBoth = list(set(x)&  set(y))
>
> and then test that list.count is equal between inBoth and x and/or y.
>
>
> Any better suggestions?
>
> Thanks for any help!

You could count the number of times each item occurs using the Counter
class in the collections module:

 >>> x = [2, 5, 1, 88, 9]
 >>> y = [5, 2, 9, 1, 88]
 >>> from collections import Counter
 >>> cx = Counter(x)
 >>> cy = Counter(y)
 >>> cx
Counter({88: 1, 1: 1, 2: 1, 5: 1, 9: 1})
 >>> cx == cy
True

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#17333

Just for fun, use the Hungarian Algorithm

(Python implementation: http://software.clapper.org/munkres/)

On Fri, Dec 16, 2011 at 3:36 AM, noydb <jenn.duerr@gmail.com> wrote:
> I want to test for equality between two lists.  For example, if I have
> two lists that are equal in content but not in order, I want a return
> of 'equal' -- dont care if they are not in the same order.  In order
> to get that equality, would I have to sort both lists regardless?  if
> yes, how (having issues with list.sort)?
>
> Another way i tried, that I think is kind-of roundabout is like
> x = [2, 5, 1, 88, 9]
> y = [5, 2, 9, 1, 88]
> inBoth = list(set(x) & set(y))
>
> and then test that list.count is equal between inBoth and x and/or y.
>
>
> Any better suggestions?
>
> Thanks for any help!
> --
> http://mail.python.org/mailman/listinfo/python-list

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#17335

On Thu, Dec 15, 2011 at 11:30 PM, Alec Taylor <alec.taylor6@gmail.com> wrote:
> Just for fun, use the Hungarian Algorithm
>
> (Python implementation: http://software.clapper.org/munkres/)

That's a pretty silly approach, but okay:

def listequals(a, b):
    if len(a) != len(b):
        return False
    matrix = [[int(item_a != item_b) for item_b in b] for item_a in a]
    path = Munkres().compute(matrix)
    return sum(matrix[r][c] for (r, c) in path) == 0

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#17336

On Fri, Dec 16, 2011 at 12:11 AM, Ian Kelly <ian.g.kelly@gmail.com> wrote:
> On Thu, Dec 15, 2011 at 11:30 PM, Alec Taylor <alec.taylor6@gmail.com> wrote:
>> Just for fun, use the Hungarian Algorithm
>>
>> (Python implementation: http://software.clapper.org/munkres/)
>
> That's a pretty silly approach, but okay:
>
> def listequals(a, b):
>    if len(a) != len(b):
>        return False
>    matrix = [[int(item_a != item_b) for item_b in b] for item_a in a]
>    path = Munkres().compute(matrix)
>    return sum(matrix[r][c] for (r, c) in path) == 0

Amendment -- it seems that Hungarian implementation fails on an empty matrix:

def listequals(a, b):
    if len(a) == len(b) == 0:
        return True
    if len(a) != len(b):
        return False
    matrix = [[int(item_a != item_b) for item_b in b] for item_a in a]
    path = Munkres().compute(matrix)
    return sum(matrix[r][c] for (r, c) in path) == 0

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