On Fri, 05 Sep 2014 10:08:18 -0700, Ethan Furman <et...@stoneleaf.us> wrote:
>On 09/05/2014 09:48 AM, Seymore4Head wrote: >> I'm still doing practice problems. I haven't heard from the library >> on any of the books I have requested. >> >> http://www.practicepython.org/exercise/2014/04/16/11-check-primality-functions.html >> >> This is not a hard problem, but it got me to thinking a little. A >> prime number will divide by one and itself. When setting up this >> loop, if I start at 2 instead of 1, that automatically excludes one of >> the factors. Then, by default, Python goes "to" the chosen count and >> not "through" the count, so just the syntax causes Python to rule out >> the other factor (the number itself). >> >> So this works: >> while True: >> a=random.randrange(1,8) >> print (a) >> for x in range(2,a): >> if a%x==0: >> print ("Number is not prime") >> break >> wait = input (" "*40 + "Wait") >> >> But, what this instructions want printed is "This is a prime number" >> So how to I use this code logic NOT print (not prime) and have the >> logic print "This number is prime" > >Python's 'for' loop has a handy 'else' extension which is perfect for the >search-type of 'for' loop: > > while True: > a=random.randrange(1,8) > print (a) > for x in range(2,a): > if a%x==0: > print ("Number is not prime") > break > else: > print ("Number is prime") > wait = input (" "*40 + "Wait") > >Note the two lines I added after the 'break' and before the 'wait'. I had already tried this one. The solution I want should only print: "This number is prime" Adding else causes the program to also print "This number is not prime" I also tried the flag=True suggestion, but never got one that worked. I am unsure when to use flag=True and flag==True Then there is flag="True" and flag=="True" What I really wanted was something like: if !(a%x==0) BTW since I am getting no grade, I much prefer the answer than a hint. The best hint IMO is to tell me how you would do it. Thanks everyone -- https://mail.python.org/mailman/listinfo/python-list
