Piet van Oostrum writes:
>> John Machin (JM) wrote:
>
>>JM> On Apr 16, 8:14 am, Chris Rebert wrote:
def all_same(lst):
return len(set(lst)) == 1
def all_different(lst):
return len(set(lst)) == len(lst)
>
>>JM> @ OP: These are very reasonable interpret
> John Machin (JM) wrote:
>JM> On Apr 16, 8:14 am, Chris Rebert wrote:
>>>
>>> def all_same(lst):
>>> return len(set(lst)) == 1
>>>
>>> def all_different(lst):
>>> return len(set(lst)) == len(lst)
>JM> @ OP: These are very reasonable interpretations of "all same" and "all
>JM> dif
On Wed, Apr 15, 2009 at 8:48 PM, Paul Rubin wrote:
> I'd use:
>
> from operator import eq
> all_the_same = reduce(eq, mylist)
That won't work for a sequence of more than two items, since after the
first reduction, the reduced value that you're comparing to is the
boolean result:
>>> reduce(eq
John Posner writes:
> # get list of object-IDs
> ids = map(lambda x: id(x), mylist)
> # ALL THE SAME? ... test whether "average ID" matches "first ID"
> sum(ids)/len(ids) == ids[0]
I don't think you can rely on id's being the same if what
you want is that the values are the same:
>>>
On Apr 16, 8:14 am, Chris Rebert wrote:
> > On Wed, Apr 15, 2009 at 5:49 PM, Chris Rebert wrote:
>
> >> On Wed, Apr 15, 2009 at 2:36 PM, Gaurav Moghe wrote:
> > Thanks for the reply. But I am interested in analysing the contents of just
> > one list. For example,
>
> > list1=[1,2,3,4,5,6]
> > S
Scott David Daniels wrote:
Assuming you really are going for "is" comparison, how about:
max(id(x) for x in mylist) == min(id(x) for x in mylist)
It has the advantage that if [id(x) for x in mylist] = [2N, 1N, 3N],
you get the answer you desire.
Oops -- got me! -John
--
http://mail.python
John Posner wrote:
...This solution relies on the object ID -- no hashability required:
# get list of object-IDs
ids = map(lambda x: id(x), mylist)
# ALL THE SAME? ... test whether "average ID" matches "first ID"
sum(ids)/len(ids) == ids[0]
Assuming you really are going for "is" comparison,
On Wed, 15 Apr 2009 23:56:45 +0100, John Posner wrote:
Chris Rebert wrote:
Ah, okay. Then you want:
def all_same(lst):
return len(set(lst)) == 1
def all_different(lst):
return len(set(lst)) == len(lst)
Note that these require all the elements of the list to be hashable.
This solut
Chris Rebert wrote:
Ah, okay. Then you want:
def all_same(lst):
return len(set(lst)) == 1
def all_different(lst):
return len(set(lst)) == len(lst)
Note that these require all the elements of the list to be hashable.
This solution relies on the object ID -- no hashability required:
Thanks! That works!
On Wed, Apr 15, 2009 at 6:14 PM, Chris Rebert wrote:
> > On Wed, Apr 15, 2009 at 5:49 PM, Chris Rebert wrote:
> >>
> >> On Wed, Apr 15, 2009 at 2:36 PM, Gaurav Moghe wrote:
> >> > Hi,
> >> >
> >> > I am an amateur python user I wanted to know how do I know whether all
> >>
> On Wed, Apr 15, 2009 at 5:49 PM, Chris Rebert wrote:
>>
>> On Wed, Apr 15, 2009 at 2:36 PM, Gaurav Moghe wrote:
>> > Hi,
>> >
>> > I am an amateur python user I wanted to know how do I know whether all
>> > the
>> > contents of a list are all same or all different? Now, I could certainly
>> > w
Hi Chris,
Thanks for the reply. But I am interested in analysing the contents of just
one list. For example,
list1=[1,2,3,4,5,6]
So, the logical statement would probably be:
if list1==(contains all same values), print "Same">False
if list1==(contains all different values), print "Differen
On Wed, Apr 15, 2009 at 2:36 PM, Gaurav Moghe wrote:
> Hi,
>
> I am an amateur python user I wanted to know how do I know whether all the
> contents of a list are all same or all different? Now, I could certainly
> write a loop with a counter. But is there a ready command for that? Checked
> a lot
Hi,
I am an amateur python user I wanted to know how do I know whether all the
contents of a list are all same or all different? Now, I could certainly
write a loop with a counter. But is there a ready command for that? Checked
a lot of docs and this mailing list, but didnt get anything worthwhile
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