Re: best way to compare contents of 2 lists?
On Apr 24, 4:41 pm, norseman norse...@hughes.net wrote: (How do I) ...explain these? [...] I can get the sashimi and take it home and BBQ it, I can roast it, I can steam it, I can wok it,..., but the other is what it is. (greasy) Besides, who says I like your cooking? :) Err... http://mail.python.org/pipermail/python-list/2008-September/679360.html -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
norseman wrote: Of course each method has its time and place of use and Python has some well oiled list search and list maintain routines of its own. List comparisons are most accurate when using presorted ones. (Some things don't lend themselves to sorting very well. Like paragraphs, books, chapters, computer programs, manuals, etc... These need the searchers (equivalent to the Unix diff) for checking equivalence.) HTH Steve Thanks Steve for bringing up various items to consider for these sort of tasks, very helpful indeed. Best, Esmail -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
Esmail: oh, I forgot to mention that each list may contain duplicates. Comparing the sorted lists is a possible O(n ln n) solution: a.sort() b.sort() a == b Another solution is to use frequency dicts, O(n): from itertools import defaultdict d1 = defaultdict(int) for el in a: d1[el] += 1 d2 = defaultdict(int) for el in b: d2[el] += 1 d1 == d2 As the arrays (Python lists) become large the second solution may become faster. Bye, bearophile -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
John Yeung gallium.arsen...@gmail.com (JY) wrote: JY It takes care of the duplicates, but so does your initial solution, JY which I like best: sorted(a)==sorted(b) JY This is concise, clear, and in my opinion, the most Pythonic. It may JY well even be the fastest. (If you didn't have to match up the numbers JY of duplicates, the set solution would be most Pythonic and probably JY fastest.) But it may fail if the list cannot be sorted because it contains incomparable elements: a = [1, 2, 3+4j] b = [2, 3+4j, 1] set(a) == set(b) True sorted(a)==sorted(b) Traceback (most recent call last): File stdin, line 1, in module TypeError: no ordering relation is defined for complex numbers In Python 3 there are even more incomparable objects pairs. -- Piet van Oostrum p...@cs.uu.nl URL: http://pietvanoostrum.com [PGP 8DAE142BE17999C4] Private email: p...@vanoostrum.org -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
On Thu, 23 Apr 2009 21:51:42 -0400, Esmail wrote: set(a) == set(b)# test if a and b have the same elements # check that each list has the same number of each element # i.e. [1,2,1,2] == [1,1,2,2], but [1,2,2,2] != [1,1,1,2] for elem in set(a): a.count(elem) == b.count(elem) Ah .. this part would take care of different number of duplicates in the lists. Cool. At significant cost of extra work. Counting the number of times a single element occurs in the list is O(N). Counting the number of times every element occurs in the list is O(N**2). Sorting is O(N*log N), so for large lists, sorting will probably be much cheaper. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
John Yeung wrote: so does your initial solution, which I like best: sorted(a)==sorted(b) This is concise, clear, and in my opinion, the most Pythonic. It may well even be the fastest. Great .. I can live with that :-) -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
MRAB wrote: You could use Raymond Hettinger's Counter class: http://code.activestate.com/recipes/576611/ on both lists and compare them for equality. thanks for the pointer, I'll study the code provided. Esmail -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
Arnaud Delobelle: Thanks to the power of negative numbers, you only need one dict: d = defaultdict(int) for x in a: d[x] += 1 for x in b: d[x] -= 1 # a and b are equal if d[x]==0 for all x in d: not any(d.itervalues()) Very nice, I'll keep this for future use. Someday I'll have to study this new new kind of numbers that can represent borrowed items too. Bye, bearophile -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
On Apr 24, 7:12 am, bearophileh...@lycos.com wrote: [...] Another solution is to use frequency dicts, O(n): from itertools import defaultdict d1 = defaultdict(int) for el in a: d1[el] += 1 d2 = defaultdict(int) for el in b: d2[el] += 1 d1 == d2 Thanks to the power of negative numbers, you only need one dict: d = defaultdict(int) for x in a: d[x] += 1 for x in b: d[x] -= 1 # a and b are equal if d[x]==0 for all x in d: not any(d.itervalues()) -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
Thanks all, after reading all the posting and suggestions for alternatives, I think I'll be going with sorted(a)==sorted(b) it seems fast, intuitive and clean and can deal with duplicates too. Best, Esmail -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
Piet van Oostrum wrote: John Yeung gallium.arsen...@gmail.com (JY) wrote: JY It takes care of the duplicates, but so does your initial solution, JY which I like best: sorted(a)==sorted(b) JY This is concise, clear, and in my opinion, the most Pythonic. It may JY well even be the fastest. (If you didn't have to match up the numbers JY of duplicates, the set solution would be most Pythonic and probably JY fastest.) But it may fail if the list cannot be sorted because it contains incomparable elements: Ah .. good point .. something I had not considered. Lucky for me in this case it's not an issue, but something I'll keep in mind for the future. -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
Steven D'Aprano wrote: On Thu, 23 Apr 2009 21:51:42 -0400, Esmail wrote: set(a) == set(b)# test if a and b have the same elements # check that each list has the same number of each element # i.e. [1,2,1,2] == [1,1,2,2], but [1,2,2,2] != [1,1,1,2] for elem in set(a): a.count(elem) == b.count(elem) Ah .. this part would take care of different number of duplicates in the lists. Cool. At significant cost of extra work. Counting the number of times a single element occurs in the list is O(N). Counting the number of times every element occurs in the list is O(N**2). A frequency dict should be O(n) also, and hence faster than sorting. Sorting is O(N*log N), so for large lists, sorting will probably be much cheaper. -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
Esmail wrote: What is the best way to compare the *contents* of two different lists regardless of their respective order? The lists will have the same number of items, and be of the same type. E.g. a trivial example (my lists will be larger), a=[1, 2, 3] b=[2, 3, 1] should yield true if a==b I suppose I could sort them and then compare them. I.e., sorted(a)==sorted(b) I am wondering if there is a more efficient/preferred way to do so. Thanks, Esmail -- http://mail.python.org/mailman/listinfo/python-list = Technically, == is reserved for identical, as in byte for byte same -If sorted(listA) == sorted(listB): - etc Is the preferred. While there are other ways, the ones I know of are much more computer intensive. Things like: get (next) line from file1 if checking it against each line of file2 yields a found loop else ... are a real I/O killers. Indexes are good but introduce a separate maintenance issue. By the way - does original order of original files count? If not, sorting AND KEEPING can speed up future things. Of course each method has its time and place of use and Python has some well oiled list search and list maintain routines of its own. List comparisons are most accurate when using presorted ones. (Some things don't lend themselves to sorting very well. Like paragraphs, books, chapters, computer programs, manuals, etc... These need the searchers (equivalent to the Unix diff) for checking equivalence.) HTH Steve -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
On Fri, 24 Apr 2009 10:39:39 -0700, norseman wrote:
Technically, == is reserved for identical, as in byte for byte same
Really? Then how do you explain these?
u'abc' == 'abc'
True
1 == 1.0
True
2L == 2
True
import decimal
decimal.Decimal('42') == 42
True
Here's one to think about:
d1 = {-1: None, -2: None}
d2 = {-2: None, -1: None}
print d1, d2
{-2: None, -1: None} {-1: None, -2: None}
d1 == d2
True
If d1 and d2 are equal and both are dictionaries with the same keys and
same values, why do they print in different orders?
--
Steven
--
http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
Steven D'Aprano wrote:
On Fri, 24 Apr 2009 10:39:39 -0700, norseman wrote:
Technically, == is reserved for identical, as in byte for byte same
Really? Then how do you explain these?
u'abc' == 'abc'
True
1 == 1.0
True
2L == 2
True
import decimal
decimal.Decimal('42') == 42
True
Here's one to think about:
d1 = {-1: None, -2: None}
d2 = {-2: None, -1: None}
print d1, d2
{-2: None, -1: None} {-1: None, -2: None}
d1 == d2
True
If d1 and d2 are equal and both are dictionaries with the same keys and
same values, why do they print in different orders?
=
(How do I) ...explain these?
Python cooks the test function.
That explains a few things I thought were kinda weird.
...one to think about:
Either someone has a programming glitch or else likes stack use.
LoadA push, read-LILO, readB, compare, loop to read-LILO.
And they decided to just pre-load as a matter of course, but maybe
forgot to read-LILO in print... ie.. didn't finish cooking print.
(or left it out as a reminder or)
Technically, == is reserved for identical, as in byte for byte same
(when == is used to check identical, uncooked)
As a matter of self - I don't favor cooking. I prefer 'cooked' things
have a different name so I can get what I want, when I want.
Choice on wall:
fish, deep fried in light beer batter
sashimi
in case you didn't notice - cooking takes longer :)
I can get the sashimi and take it home and BBQ it, I can roast it, I can
steam it, I can wok it,..., but the other is what it is. (greasy)
Besides, who says I like your cooking? :)
Steve
--
http://mail.python.org/mailman/listinfo/python-list
best way to compare contents of 2 lists?
What is the best way to compare the *contents* of two different lists regardless of their respective order? The lists will have the same number of items, and be of the same type. E.g. a trivial example (my lists will be larger), a=[1, 2, 3] b=[2, 3, 1] should yield true if a==b I suppose I could sort them and then compare them. I.e., sorted(a)==sorted(b) I am wondering if there is a more efficient/preferred way to do so. Thanks, Esmail -- http://mail.python.org/mailman/listinfo/python-list
RE: best way to compare contents of 2 lists?
'set' comes to mind, though I'm not sure if there are performance inplications with large amopunts of data. a=[1, 2, 3] b=[2, 3, 1] set(a) == set(b) True Cheers, Hans -Original Message- From: python-list-bounces+hans.dushanthakumar=hcn.com...@python.org [mailto:python-list-bounces+hans.dushanthakumar=hcn.com...@python.org] On Behalf Of Esmail Sent: Friday, 24 April 2009 11:31 AM To: python-list@python.org Subject: best way to compare contents of 2 lists? What is the best way to compare the *contents* of two different lists regardless of their respective order? The lists will have the same number of items, and be of the same type. E.g. a trivial example (my lists will be larger), a=[1, 2, 3] b=[2, 3, 1] should yield true if a==b I suppose I could sort them and then compare them. I.e., sorted(a)==sorted(b) I am wondering if there is a more efficient/preferred way to do so. Thanks, Esmail -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
On Thu, Apr 23, 2009 at 9:31 PM, Esmail ebo...@hotmail.com wrote: What is the best way to compare the *contents* of two different lists regardless of their respective order? The lists will have the same number of items, and be of the same type. E.g. a trivial example (my lists will be larger), a=[1, 2, 3] b=[2, 3, 1] should yield true if a==b I suppose I could sort them and then compare them. I.e., sorted(a)==sorted(b) I am wondering if there is a more efficient/preferred way to do so. Thanks, Esmail -- http://mail.python.org/mailman/listinfo/python-list set(a) == set(b)# test if a and b have the same elements # check that each list has the same number of each element # i.e.[1,2,1,2] == [1,1,2,2], but [1,2,2,2] != [1,1,1,2] for elem in set(a): a.count(elem) == b.count(elem) -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
Esmail wrote: What is the best way to compare the *contents* of two different lists regardless of their respective order? The lists will have the same number of items, and be of the same type. E.g. a trivial example (my lists will be larger), a=[1, 2, 3] b=[2, 3, 1] should yield true if a==b I suppose I could sort them and then compare them. I.e., sorted(a)==sorted(b) I am wondering if there is a more efficient/preferred way to do so. oh, I forgot to mention that each list may contain duplicates. So I suppose this means I would have to sort the two lists and compare them afterall, unless there's another way? -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
David Robinow wrote: On Thu, Apr 23, 2009 at 9:31 PM, Esmail ebo...@hotmail.com wrote: What is the best way to compare the *contents* of two different lists regardless of their respective order? The lists will have the same number of items, and be of the same type. E.g. a trivial example (my lists will be larger), a=[1, 2, 3] b=[2, 3, 1] should yield true if a==b I suppose I could sort them and then compare them. I.e., sorted(a)==sorted(b) I am wondering if there is a more efficient/preferred way to do so. Thanks, Esmail -- http://mail.python.org/mailman/listinfo/python-list set(a) == set(b)# test if a and b have the same elements # check that each list has the same number of each element # i.e.[1,2,1,2] == [1,1,2,2], but [1,2,2,2] != [1,1,1,2] for elem in set(a): a.count(elem) == b.count(elem) Ah .. this part would take care of different number of duplicates in the lists. Cool. thanks, Esmail -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
Esmail wrote: Esmail wrote: What is the best way to compare the *contents* of two different lists regardless of their respective order? The lists will have the same number of items, and be of the same type. E.g. a trivial example (my lists will be larger), a=[1, 2, 3] b=[2, 3, 1] should yield true if a==b I suppose I could sort them and then compare them. I.e., sorted(a)==sorted(b) I am wondering if there is a more efficient/preferred way to do so. oh, I forgot to mention that each list may contain duplicates. So I suppose this means I would have to sort the two lists and compare them afterall, unless there's another way? You could use Raymond Hettinger's Counter class: http://code.activestate.com/recipes/576611/ on both lists and compare them for equality. -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
Hans DushanthaKumar wrote: 'set' comes to mind, Yes, I thought about that, but I wasn't sure how to potentially deal with different number of duplicates in the lists. The post by David seems to address that issue nicely. though I'm not sure if there are performance inplications with large amopunts of data. Yes, I wonder about these sort of things too, not knowing too much about how Python does things internally. Thanks for the suggestion, Esmail -- http://mail.python.org/mailman/listinfo/python-list
Re: best way to compare contents of 2 lists?
Esmail ebo...@hotmail.com wrote: What is the best way to compare the *contents* of two different lists regardless of their respective order? The lists will have the same number of items, and be of the same type. Best can mean different things. Fastest? Shortest code? Most readable? David Robinow wrote: set(a) == set(b) # test if a and b have the same elements # check that each list has the same number of each element # i.e. [1,2,1,2] == [1,1,2,2], but [1,2,2,2] != [1,1,1,2] for elem in set(a): a.count(elem) == b.count(elem) Ah .. this part would take care of different number of duplicates in the lists. Cool. It takes care of the duplicates, but so does your initial solution, which I like best: sorted(a)==sorted(b) This is concise, clear, and in my opinion, the most Pythonic. It may well even be the fastest. (If you didn't have to match up the numbers of duplicates, the set solution would be most Pythonic and probably fastest.) John -- http://mail.python.org/mailman/listinfo/python-list
