how to make this situation return this result?

[Ho Yeung Lee]

finally i searched dict.values()[index] solved this 


On Saturday, July 1, 2017 at 6:00:41 PM UTC+8, Peter Otten wrote:
> Ho Yeung Lee wrote:
> 
> > expect result as this first case
> > 
> > ii = 0
> > jj = 0
> > for ii in range(0,3):
> >     for jj in range(0,3):
> >         if ii < jj:
> >             print (ii, jj)
> > 
> > 
> > but below is different
> > as sometimes the situation is not range(0,3), but it a a list of tuple
> > 
> > iiii = 0
> 
> > for ii in range(0,3):
>       # jj starts with 0 on invocation of the inner loop
>       # to get that with an independent counter you have to
>       #initialise it explicitly:
>       jjjj = 0
> >     for jj in range(0,3):
> >         if iiii < jjjj:
> >             print (iiii, jjjj)
> >         jjjj = jjjj + 1
> >     iiii = iiii + 1
> > 
> > how to make this situation return result like the first case?
> 
> Because you don't reset jjjj the condition iiii < jjjj is always True the 
> second and third time the inner loop is executed.