list comprehention

[Mathijs]

Hi,

Python beginner here and very much enjoying it. I'm looking for a
pythonic way to find how many listmembers are also present in a reference
list. Don't count duplicates (eg. if you already found a matching member
in the ref list, you can't use the ref member anymore).

Example1:
ref=[2, 2, 4, 1, 1]
list=[2, 3, 4, 5, 3]
solution: 2

Example2:
ref=[2, 2, 4, 1, 1]
list=[2, 2, 5, 2, 4]
solution: 3 (note that only the first two 2's count, the third 2 in the
list should not be counted)

Any suggestions or comments?

Thanks.
M.
#my failing effort:
sum([min(r.count(n)-l[:i].count(n),l.cou nt(n)) for i,n in enumerate(l)])

#test lists
import random
#reference list
r=[random.randint( 1,5) for n in range(5)]
#list
l=[random.randint( 1,5) for n in range(5)]

[Tim Chase]

> Python beginner here and very much enjoying it. I'm looking

for a pythonic way to find how many listmembers are also
present in a reference list. Don't count duplicates (eg. if
you already found a matching member in the ref list, you can't
use the ref member anymore).

Example1:
ref=[2, 2, 4, 1, 1]
list=[2, 3, 4, 5, 3]
solution: 2

Example2:
ref=[2, 2, 4, 1, 1]
list=[2, 2, 5, 2, 4]
solution: 3 (note that only the first two 2's count, the third
2 in the list should not be counted)

It sounds like you're looking for "set" operations: (using "ell"
for clarity)

from sets import Set
a = [2,2,4,1,1]
b = [2,3,4,5,3]
setA = Set(a)
setB = Set(b)
results = setA.intersecti on(setB)
results Set([2,4]) intersection = [x for x in results]
intersection

[2,4]
I'm a tad confused by the help, as it sounds like sets are
supposed to be first-class citizens, but in ver2.3.5 that I'm
running here (or rather "there", on a friend's box), I have to
"import sets" which I didn't see mentioned in the reference manual.

-one of many tims on the list
tim = Set(["bald", "vegetarian ", "loving husband"])

:)

[markscala]

def reference(alist 1,alist2):
counter = 0
for x in lis:
if x in ref:
ref.pop(ref.ind ex(x))
counter += 1
return counter

this works I think for your examples, but you should check it against
them and other cases.
good luck

[markscala]

revision of previous:

def reference(refli st,alist2):
counter = 0
for x in alist2:
if x in reflist:
reflist.pop(ref list.index(x))
counter += 1
return counter

[markscala]

another approach:

ref = [2,2,4,1,1]
lis = [2,2,5,2,4]

len([ref.pop(ref.ind ex(x)) for x in lis if x in ref])

[bonono]

Tim Chase wrote:

Python beginner here and very much enjoying it. I'm looking
> for a pythonic way to find how many listmembers are also
> present in a reference list. Don't count duplicates (eg. if
> you already found a matching member in the ref list, you can't
> use the ref member anymore).
>
> Example1:
> ref=[2, 2, 4, 1, 1]
> list=[2, 3, 4, 5, 3]
> solution: 2
>
> Example2:
> ref=[2, 2, 4, 1, 1]
> list=[2, 2, 5, 2, 4]
> solution: 3 (note that only the first two 2's count, the third
> 2 in the list should not be counted)

It sounds like you're looking for "set" operations: (using "ell"
for clarity)

>>> from sets import Set
>>> a = [2,2,4,1,1]
>>> b = [2,3,4,5,3]
>>> setA = Set(a)
>>> setB = Set(b)
>>> results = setA.intersecti on(setB)
>>> results Set([2,4]) >>> intersection = [x for x in results]
>>> intersection

[2,4]

won't set remove duplicates which he wants to preserve ? He is not just
looking for the 'values' that is in common, but the occurence as well,
if I understand the requirement correctly.

I would just build a dict with the value as the key and occurence as
the value then loop the list and lookup.

[Peter Otten]

Mathijs wrote:

Python beginner here and very much enjoying it. I'm looking for a
pythonic way to find how many listmembers are also present in a reference
list. Don't count duplicates (eg. if you already found a matching member
in the ref list, you can't use the ref member anymore).

Example1:
ref=[2, 2, 4, 1, 1]
list=[2, 3, 4, 5, 3]
solution: 2

Example2:
ref=[2, 2, 4, 1, 1]
list=[2, 2, 5, 2, 4]
solution: 3 (note that only the first two 2's count, the third 2 in the
list should not be counted)

Any suggestions or comments?

sum(min(list.co unt(n), ref.count(n)) for n in set(ref))

Is that it?

Peter

[Peter Otten]

Tim Chase wrote:

I'm a tad confused by the help, as it sounds like sets are
supposed to be first-class citizens, but in ver2.3.5 that I'm
running here (or rather "there", on a friend's box), I have to
"import sets" which I didn't see mentioned in the reference manual.

set and frozenset are a builtins starting with Python 2.4.

http://www.python.org/2.4/highlights.html

"""
New or upgraded built-ins

built-in sets - the sets module, introduced in 2.3, has now been implemented
in C, and the set and frozenset types are available as built-in types (PEP
218)
"""

Peter

[Matt]

Tim Chase wrote:
<snip>

I'm a tad confused by the help, as it sounds like sets are
supposed to be first-class citizens, but in ver2.3.5 that I'm
running here (or rather "there", on a friend's box), I have to
"import sets" which I didn't see mentioned in the reference manual.

-one of many tims on the list
tim = Set(["bald", "vegetarian ", "loving husband"])

:)

The builtin "set" was added in 2.4, I believe (note lowercase "s"):

http://docs.python.org/whatsnew/node2.html

print "vegetarian".re place("etari"," ")
:-)

[Tim Chase]

>> >Python beginner here and very much enjoying it. I'm looking

> for a pythonic way to find how many listmembers are also
> present in a reference list. Don't count duplicates (eg. if

[snipped]
won't set remove duplicates which he wants to preserve ?

My reading was that the solution shouldn't count duplicates
("Don't count duplicates"). However, once you mentioned it, and
I saw other folks' responses that looked very diff. from my own,
I re-read the OP's comments and found that I missed this key bit:

"note that only the first *two* 2's count, the third 2
in the list should not be counted"
(*emphasis* mine)

and that the desired result was a count (though a len(Set())
would return the right count if the OP wanted the true
intersection, but that's beside the point).

My error. Sorry, ladies and gentlemen :)

I'm partial to the elegance of markscala's suggestion of:

len([ref.pop(ref.ind ex(x)) for x in lis if x in ref])

though it might need to come with a caveat that it doesn't leave
"ref" in the same state is it was originally, so it should be
copied and then manipulated thusly:

r = ref[:]
len([r.pop(r.index(x )) for x in (lis if x in r])

which would then leave ref undisturbed. As tested:

import random
ref = [random.randint( 1,5) for n in range(5)]
for x in range(1,10):
lis = [random.randint( 1,5) for n in range(5)]
r = ref[:]
print repr((r,lis))
print len([r.pop(r.index(x )) for x in lis if x in r])

seems to give the results the OP describes.

-tim