Re: Help for for loop

On 12/9/2010 12:24 PM, Cal Leeming [Simplicity Media Ltd] wrote: > Uh, you *might* be able to use: > > {% for x in mylist %} > {% if x % 2 %} > yay: {{x}} > {% else %} > nay: {{x}} > {% endif %} > {% endfor %} > Blerch! This is a really good indication of why it's

Re: Help for for loop

Ahh thanks Bruno, that gave the correct solution: >>> l ['pos 0', 'pos 1', 'pos 2', 'pos 3'] >>> l[1::2] ['pos 1', 'pos 3'] My slice-fu is clearly weak. Cheers Tom On Thu, Dec 9, 2010 at 12:03 PM, bruno desthuilliers wrote: > > > On 9 déc, 12:24, Tom Evans

Re: Help for for loop

Thanx for help. I solved the problem using divisibleby -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-us...@googlegroups.com. To unsubscribe from this group, send email to

Re: Help for for loop

On 9 déc, 12:24, Tom Evans wrote: > Hmm, those are the values that are odd, he wanted the values from odd > indices Well spotted ;) There's a builtin "slice" filter that should have done the trick but I just couldn't manage to make it work with a for loop :-/ The

Re: Help for for loop

Because that looks at the values of the array and returns the items which have a value that is divisible by two, not the position of the value within the array. It is simpler to comprehend if you don't put numbers in the list: >>> l = [ 'pos 0', 'pos 1', 'pos 2', 'pos 3' ] >>> filter(lambda x:

Re: Help for for loop

Can't you use the % operator within if statements in templates? I was almost sure I'd done this before :| On 09/12/2010 11:27, bruno desthuilliers wrote: On 9 déc, 12:18, Phani Chand wrote: Can i use filter(lambda x: x%2, mylist) directly in my html page s/html

Re: Help for for loop

I see no need to get itertools involved ;) Why not just use this: filter(lambda x: not x%2, a) On 09/12/2010 11:24, Tom Evans wrote: Hmm, those are the values that are odd, he wanted the values from odd indices, eg: a=[2,3,4,5,6,7] filter(lambda x: x%2, a) [3, 5, 7] [ val for val, i in

Re: Help for for loop

On 9 déc, 12:18, Phani Chand wrote: > Can i use filter(lambda x: x%2, mylist) directly in my html page s/html page/template/ And no, you cannot use Python code in a template. You can either 1/ filter the list in the view 2/ use the builtin "divisibleby" filter

Re: Help for for loop

Hmm, those are the values that are odd, he wanted the values from odd indices, eg: >>> a=[2,3,4,5,6,7] >>> filter(lambda x: x%2, a) [3, 5, 7] >>> [ val for val, i in itertools.izip(a, itertools.count()) if not i % 2 ] [2, 4, 6] That could probably be written a bit nicer.. On Thu, Dec 9, 2010 at

Re: Help for for loop

Uh, you *might* be able to use: {% for x in mylist %} {% if x % 2 %} yay: {{x}} {% else %} nay: {{x}} {% endif %} {% endfor %} On 09/12/2010 11:18, Phani Chand wrote: Can i use filter(lambda x: x%2, mylist) directly in my html page -- You received this message

Re: Help for for loop

Can i use filter(lambda x: x%2, mylist) directly in my html page -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-us...@googlegroups.com. To unsubscribe from this group, send email to

Re: Help for for loop

>>> mylist = [1,2,3,4,5,6,7,8] >>> filter(lambda x: x%2, mylist) [1, 3, 5, 7] >>> This what you need? On 09/12/2010 10:34, Phani Chand wrote: i am passing though a list{1,2,3,4} but i want only the numbers with odd index in template written in html -- You received this message because you are

Help for for loop

i am passing though a list{1,2,3,4} but i want only the numbers with odd index in template written in html -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-us...@googlegroups.com. To unsubscribe from