On Friday, October 24, 2014 12:36:23 PM UTC-7, Seymore4Head wrote:
> On Fri, 24 Oct 2014 12:25:33 -0700 (PDT), sohcahto...@gmail.com wrote:
> 
> >On Friday, October 24, 2014 12:12:10 PM UTC-7, Seymore4Head wrote:
> >> On Fri, 24 Oct 2014 11:57:12 -0700 (PDT), sohcahto...@gmail.com wrote:
> >> 
> >> >On Friday, October 24, 2014 11:17:53 AM UTC-7, Seymore4Head wrote:
> >> >> On Fri, 24 Oct 2014 11:52:15 -0600, Ian Kelly <ian.g.ke...@gmail.com>
> >> >> wrote:
> >> >> 
> >> >> >On Fri, Oct 24, 2014 at 11:03 AM, Seymore4Head
> >> >> ><Seymore4Head@hotmail.invalid> wrote:
> >> >> >> Actually I was a little frustrated when I added that line back in as
> >> >> >> the other lines all work.
> >> >> >> Using list(range(10)) Doesn't throw an error but it doesn't work.
> >> >> >>
> >> >> >> http://i.imgur.com/DTc5zoL.jpg
> >> >> >>
> >> >> >> The interpreter.   I don't know how to use that either.
> >> >> >
> >> >> >Try both of these in the interpreter, and observe the difference:
> >> >> >
> >> >> >7 in range(10)
> >> >> >
> >> >> >"7" in range(10)
> >> >> >
> >> >> >Do you understand what the difference between 7 and "7" is?
> >> >> 
> >> >> I do understand that.  7 is a number and "7" is a string.
> >> >> What my question was...and still is...is why
> >> >> Python 3 fails when I try using
> >> >> y=1 800 get charter
> >> >> 
> >> >> y in range str(range(10))
> >> >> should work because y is a string and str(range(10)) should be
> >> >> "y" in str(1) fails.
> >> >> It doesn't give an error it's just not True when y is a number.
> >> >> 
> >> >> These hints are just not working.  I am too thick for hints. :)
> >> >> If you could use it in the code, I might understand.
> >> >> The other work arounds that were posted work.
> >> >> I have used them.  str(range(10)) doesn't work.
> >> >> 
> >> >> import string
> >> >> def nametonumber(name):
> >> >>     lst=[]
> >> >>     nx=[]
> >> >>     digit=[]
> >> >>     digit="".join(str(i) for i in range(10))
> >> >>     for x in name:
> >> >>         lst.append(x)
> >> >>     for y in (lst):
> >> >>         if y in list(range(1,10)):
> >> >>         #if y in "1234567890":
> >> >>         #if y.isdigit():
> >> >>         #if y in digit:       
> >> >>         #if y in string.digits:
> >> >>             nx.append(y)
> >> >>         if y in " -()":
> >> >>             nx.append(y)
> >> >>         if y in "abc":
> >> >>             nx.append("2")
> >> >>         if y in "def":
> >> >>             nx.append("3")
> >> >>         if y in "ghi":
> >> >>             nx.append("4")
> >> >>         if y in "jkl":
> >> >>             nx.append("5")
> >> >>         if y in "mno":
> >> >>             nx.append("6")
> >> >>         if y in "pqrs":
> >> >>             nx.append("7")
> >> >>         if y in "tuv":
> >> >>             nx.append("8")
> >> >>         if y in "wxyz":
> >> >>             nx.append("9")
> >> >>     number="".join(e for e in nx)
> >> >>     return number
> >> >> a="1-800-getcharter"
> >> >> print (nametonumber(a))#1800 438 2427 837
> >> >> a="1-800-leo laporte"
> >> >> print (nametonumber(a))
> >> >> a="1 800 dialaho"
> >> >> print (nametonumber(a))
> >> >> 
> >> >> Please
> >> >
> >> >Your code here is actually pretty close to a correct answer.  Just a few 
> >> >things to consider...
> >> >
> >> >- Why are you converting your name string to a list?  It is unnecessary.  
> >> >When you do "for y in <some string>", then y will still be single 
> >> >characters on each iteration of the loop.
> >> >
> >> >- "if y in string.digits" should work fine.
> >> >
> >> >- "if y in list(range(1,10)" won't work for two reasons: First, it 
> >> >creates a list of numbers, not strings.  Second, even if it did, it would 
> >> >be missing the "0" digit.
> >> >
> >> >- At the end, when you convert your list to a string, you don't need to 
> >> >use list comprehension, since nx is already a list.  number = "".join(nx) 
> >> >should work fine.
> >> >
> >> >Also, in general, you need to stop and slow down and think like a 
> >> >programmer.  If you get an error, your instinct shouldn't be to just hack 
> >> >at it to make the error go away.  Look at the error and try to make sense 
> >> >of it.  Learn what the error means and try to fix the core problem.
> >> >
> >> >And for @#$%'s sake...stop saying "It isn't working" and not elaborating. 
> >> > You've been told by every other post in this thread to show us what you 
> >> >did and what the error was.  You've also been told to *NOT* retype what 
> >> >you see and to copy/paste your code and the error because when you make a 
> >> >typo when copying, we might see a problem that doesn't exist and then you 
> >> >just get more confused.
> >> 
> >> Ok  I think I may have the question you guys are looking for.
> >> I just posted it.
> >> See above.
> >> 
> >> But it's still broke.  :(
> >
> >str(range(10)) doesn't do what you think it does.
> >
> >Run 'print(str(range(10)))' and look at what you get.
> 
> Yeah, I know that.  My question is why?
> The answer was that Python 3 only stores the min and max values but
> you can still iterate over them.
> I don't think that means what I think it means.

"You can iterate over them" pretty much just means you can use them as the 
source of a 'for' loop.  But in your case, when you're calling 'for y in 
str(range(10))', you're not using 'range(10)' as the source of your loop, 
you're using the result of a str() function call, and you're calling str() on 
range(), which doesn't return a concrete value in Python 3.  If you try to 
print a range(), you're just getting a string containing your original call to 
range.

And that's why you're seeing the 1 and a in your output.  str(range(10)) 
returns the string 'range(10)'.

Like I said in a previous post, use 'string.digits'.  Try this test code and 
see what you get:

import string
name="123-xyz-abc" 
print("string.digits is", string.digits)
for x in name: 
    if x in range(10): 
        print ("Range",(x)) 
    if x in string.digits: 
        print ("string.digits",(x)) 
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