Simple - looking for a way to do an element exists check..

[rh0dium]

Hi all,

I have a simple list to which I want to append another tuple if
element 0 is not found anywhere in the list.

element = ('/smsc/chp/aztec/padlib/5VT.Cat',
'/smsc/chp/aztec/padlib',
'5VT.Cat', (33060))

element1 = ('/smsc/chp/aztec/padlib/5VT.Cat2',
'/smsc/chp/aztec/padlib',
'5VT.Cat2', (33060))

a = [ ('/smsc/chp/aztec/padlib/5VT.Cat',
'/smsc/chp/aztec/padlib',
'5VT.Cat', (33060)),
('/smsc/chp/aztec/padlib/padlib.TopCat%' ,
'/smsc/chp/aztec/padlib',
'padlib.TopCat% ', (33204)),
('/smsc/chp/aztec/padlib/Regulators.Cat% ',
'/smsc/chp/aztec/padlib',
'Regulators.Cat %', (33204))]

So my code would look something like this.

found = False
for item in a:
if item[0] == element[0]
found = True
break
if not found:
a.append(elemen t)

But this is just ugly - Is there a simpler way to interate over all
items in a without using a found flag?

Thanks

[Paul McGuire]

Re: Simple - looking for a way to do an element exists check..

On Feb 22, 11:20 am, rh0dium <steven.kl...@g mail.comwrote:

>
found = False
for item in a:
  if item[0] == element[0]
    found = True
    break
if not found:
  a.append(elemen t)
>
But this is just ugly - Is there a simpler way to interate over all
items in a without using a found flag?
>
Thanks

for item in a:
if item[0] == element[0]
break
else: # only called if we never 'break' out of the for loop
a.append(elemen t)


But what about a dict?

adict = dict((elem[0],elem) for elem in a)

if item[0] not in adict:
adict[item[0]] = item

# need the final list?
a = adict.values()

No list searching, and will scale well if a gets real long.

-- Paul

[Paul McGuire]

Re: Simple - looking for a way to do an element exists check..

On Feb 22, 11:20 am, rh0dium <steven.kl...@g mail.comwrote:

Hi all,
>
I have a simple list to which I want to append another tuple if
element 0 is not found anywhere in the list.
>
element =  ('/smsc/chp/aztec/padlib/5VT.Cat',
  '/smsc/chp/aztec/padlib',
  '5VT.Cat', (33060))
>
element1 =  ('/smsc/chp/aztec/padlib/5VT.Cat2',
  '/smsc/chp/aztec/padlib',
  '5VT.Cat2', (33060))
>
a =  [ ('/smsc/chp/aztec/padlib/5VT.Cat',
  '/smsc/chp/aztec/padlib',
  '5VT.Cat', (33060)),
 ('/smsc/chp/aztec/padlib/padlib.TopCat%' ,
  '/smsc/chp/aztec/padlib',
  'padlib.TopCat% ', (33204)),
 ('/smsc/chp/aztec/padlib/Regulators.Cat% ',
  '/smsc/chp/aztec/padlib',
  'Regulators.Cat %', (33204))]
>
So my code would look something like this.
>
found = False
for item in a:
  if item[0] == element[0]
    found = True
    break
if not found:
  a.append(elemen t)
>
But this is just ugly - Is there a simpler way to interate over all
items in a without using a found flag?
>
Thanks

Well, that's what I get for typing before thinking...

If the remaining items in each element tuple are the same for any
given element[0], then just use a set.

aset = set(a)
for element in list_of_new_ele ment_tuples:
aset.add(elemen t)

-- Paul

[Jason]

Re: Simple - looking for a way to do an element exists check..

On Feb 22, 10:20 am, rh0dium <steven.kl...@g mail.comwrote:

Hi all,
>
I have a simple list to which I want to append another tuple if
element 0 is not found anywhere in the list.
>
element = ('/smsc/chp/aztec/padlib/5VT.Cat',
'/smsc/chp/aztec/padlib',
'5VT.Cat', (33060))
>
element1 = ('/smsc/chp/aztec/padlib/5VT.Cat2',
'/smsc/chp/aztec/padlib',
'5VT.Cat2', (33060))
>
a = [ ('/smsc/chp/aztec/padlib/5VT.Cat',
'/smsc/chp/aztec/padlib',
'5VT.Cat', (33060)),
('/smsc/chp/aztec/padlib/padlib.TopCat%' ,
'/smsc/chp/aztec/padlib',
'padlib.TopCat% ', (33204)),
('/smsc/chp/aztec/padlib/Regulators.Cat% ',
'/smsc/chp/aztec/padlib',
'Regulators.Cat %', (33204))]
>
So my code would look something like this.
>
found = False
for item in a:
if item[0] == element[0]
found = True
break
if not found:
a.append(elemen t)
>
But this is just ugly - Is there a simpler way to interate over all
items in a without using a found flag?
>
Thanks

How-about using a generator expression and Python's built-in "in"
operator:

>>def example(myData, newData):

.... if newData[0] not in (x[0] for x in myData):
.... myData.append( newData )
....

>>l = []
>>example( l, ('a', 'apple', 'aviary') )
>>l

[('a', 'apple', 'aviary')]

>>example( l, ('s', 'spam', 'silly') )
>>l

[('a', 'apple', 'aviary'), ('s', 'spam', 'silly')]

>>example( l, ('s', 'suck-tastic') )
>>l

[('a', 'apple', 'aviary'), ('s', 'spam', 'silly')]

>>>

[Paul Rubin]

Re: Simple - looking for a way to do an element exists check..

Paul Rubin <http://phr.cx@NOSPAM.i nvalidwrites:

if any(x==element[0] for x in a):
a.append(elemen t)

Should say:

if any(x[0]==element[0] for x in a):
a.append(elemen t)

[Paul McGuire]

Re: Simple - looking for a way to do an element exists check..

On Feb 22, 12:54 pm, Paul Rubin <http://phr...@NOSPAM.i nvalidwrote:

Paul Rubin <http://phr...@NOSPAM.i nvalidwrites:

    if any(x==element[0] for x in a):
      a.append(elemen t)

>
Should say:
>
     if any(x[0]==element[0] for x in a):
        a.append(elemen t)

I think you have this backwards. Should be:

if not any(x[0]==element[0] for x in a):
a.append(elemen t)

or

if all(x[0]!=element[0] for x in a):
a.append(elemen t)

-- Paul

[Paul Rubin]

Re: Simple - looking for a way to do an element exists check..

Paul McGuire <ptmcg@austin.r r.comwrites:

I think you have this backwards. Should be:
>
if not any(x[0]==element[0] for x in a):
a.append(elemen t)

I think you are right, it was too early for me to be reading code when
I posted that ;-)

[Boris Ozegovic]

Re: Simple - looking for a way to do an element exists check..

Paul Rubin wrote:

if any(x[0]==element[0] for x in a):

How come this list comprehension isn't in [] brackets?

[TeroV]

Re: Simple - looking for a way to do an element exists check..

Boris Ozegovic wrote:

Paul Rubin wrote:
>

> if any(x[0]==element[0] for x in a):

>
How come this list comprehension isn't in [] brackets?

It isn't list comprehension, it is generator expression

http://en.wikipedia.org/wiki/Python_syntax_and_semantics#Generator_expressions

--
Tero

[Boris Ozegovic]

Re: Simple - looking for a way to do an element exists check..

TeroV wrote:

It isn't list comprehension, it is generator expression
http://en.wikipedia.org/wiki/Python_...or_expressions

Nice. :)

[Paul Hankin]

Re: Simple - looking for a way to do an element exists check..

On Feb 22, 7:01 pm, Paul McGuire <pt...@austin.r r.comwrote:

On Feb 22, 12:54 pm, Paul Rubin <http://phr...@NOSPAM.i nvalidwrote:
>

Paul Rubin <http://phr...@NOSPAM.i nvalidwrites:

    if any(x==element[0] for x in a):
      a.append(elemen t)

>

Should say:

>

     if any(x[0]==element[0] for x in a):
        a.append(elemen t)

>
I think you have this backwards.  Should be:
>
     if not any(x[0]==element[0] for x in a):
        a.append(elemen t)

IMO Jason's solution of testing containment in a generator is better
(more readable).
if element[0] not in (x[0] for x in a):
a.append(elemen t)

--
Paul Hankin

[Marc 'BlackJack' Rintsch]

Re: Simple - looking for a way to do an element exists check..

On Sat, 23 Feb 2008 17:19:47 -0800, thebjorn wrote:

On Feb 23, 6:18 pm, Paul Hankin <paul.han...@gm ail.comwrote:
>

>IMO Jason's solution of testing containment in a generator is better
>(more readable).
> if element[0] not in (x[0] for x in a):
> a.append(elemen t)

>
It may be more readable (although that's debatable), but it always
traverses the entire list.

The ``not in`` stops if the element is found.

Ciao,
Marc 'BlackJack' Rintsch