how to make this situation return this result?
Ho Yeung Lee
jobmattcon at gmail.com
Sat Jul 1 06:44:30 EDT 2017
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.
More information about the Python-list
mailing list