adam urbas said unto the world upon 05/23/2007 01:04 PM:
> Sorry, I don't think Hotmail has turn off HTML. If it does I
> havn't been able to find it. I think you're going to have to
> explain your little bit of text stuff down there at the bottom. I
> have no idea what most of that means. All my choice things are
> working now though. I think that is what you were trying to help
> me with. What I used wasif shape in["1","circle"]:and if shape ==
> "1" or shape =="circle":It works perfectly fine now.Ya that little
> bit o' code is really puzzling. I wish I knew more about this
> python deal. I understand the concept, but not the rules or the
> techniques and things of that sort. OK... I've got it... the
> data=raw_input('Feed Me!'). Ok I now understand that bit. Then it
> says Feed Me! and you put 42 (the ultimate answer to life the
> universe, everything). OK, it won't accept the <type 'str'> bit.
> it doesn't like the "<". Well, I just removed that bit and it
> said:Feed Me! and I put 42, and it said >>> (I guess it's
> satisfied now, with the whole feeding). Well if I understood what
> 'str' meant, then I could probably figure the rest out. Well I
> have to go do other things so I'll save the rest of this figuring
> out till later.I shall return,Adam> Date: Wed, 23 May 2007 12:12:16
> -0400> From: [EMAIL PROTECTED]> To: [EMAIL PROTECTED]> CC:
> tutor@python.org> Subject: Re: [Tutor] trouble with "if"> > adam
> urbas said unto the world upon 05/23/2007 11:57 AM:> > > > Hi all,>
> > > > I've been working with this new program that I wrote. I
> started out > > with it on a Ti-83, which is much easier to program
> than python. Now > > I'm trying to transfer the program to python
> but its proving to be quite > > difficult. I'm not sure what the
> whole indentation thing is for. And > > now I'm having trouble
> with the if statement things. > > > > #"Circle Data Calculation
> Program:"> > print "Welcome to the Circle Data Calcuation
> Program."> > print> > > > #"Menu 1:"> > print "Pick a shape:">
> > print "(NOTE: You must select the number of the shape and not the
> shape > > itself)"> > print "1 Circle"> > print "2 Square"> > print
> "3 Triangle"> > > > #"User's Choice:"> > shape=raw_input("> ")>
> > > > #"Select Given:"> > if shape == 1:> > print
> "Choose the given value:"> > print "1 radius"> >
> print "2 diameter"> > print "3 circumference"> >
> print "4 area"> > > > #"User's Choice:"> > given=raw_input("> ")> >
> > > if given == 1:> > radius=raw_input("Enter Radius:")> >
> diameter=(radius*2)> > circumference=(diameter*3.14)> >
> area=(radius**2*3.14)> > print "Diameter:", diameter> >
> print "Circumference:", circumference> > print "Area:",
> area> > > > if given == 2:> > diameter=raw_input("Enter
> Diameter:")> > radius=(diameter/2)> >
> circumference=(diameter*3.14)> > area=(radius**2*3.14)> >
> print "Radius:", radius> > print "Circumference:",
> circumference> > print "Area:", area> > > > if given == 3:>
> > circumference=raw_input("Enter Circumference:")> >
> radius=(circumference/3.14/2)> > diameter=(radius*2)> >
> area=(radius**2*3.14)> > print "Radius:", radius> >
> print "Diameter:", diameter> > print "Area:", area> > > >
> if given == 4:> > area=raw_input("Enter Area:")> >
> radius=(area/3.14)> > > > This is the whole program so
> far, because I haven't quite finished it > > yet. But I tried to
> get it to display another list of options after you > > select a
> shape but it just does this.> > > > Pick a shape:> > 1 Circle> > 2
> Square> > 3 Triangle> > >1> > >1> > >>>> > > > I'm not sure why
> it does that but I do know that it is skipping the > > second list
> of options.> > > > Another of my problems is that I can't figure
> out how to get it to > > accept two different inputs for a
> selection. Like I want it to accept > > both the number 1 and
> circle as circle then list the options for > > circle. It won't
> even accept words. I can only get it to accept > > numbers. It's
> quite frustrating actually.> > > > Any advice would be greatly
> appreciated.> > Thanks in advance,> > Adam> > > > > > > Adam,> >
> Could you send plain text email rather than html, please? At least
> for > me, your code's indentation is all messed up unless I take
> some steps > to rectify it.> > The problem is that raw_input
> returns a string, and you are testing > whether given is equal to
> integers. See if this helps make things clear:> > >>> data =
> raw_input('Feed me!')> Feed me!42> >>> type(data)> <type 'str'>>
> >>> data == 42> False> >>> int(data) == 42> True> >>>> > Best,> >
> Brian vdB
Adam,
As you can see from the above, the way hotmail is formatting things
makes the conversation a bit tricky :-) I'm only willing to spend so
much time trying to sort through it, so I hope what follows helps.
>>> data = raw_input("Feed me!")
Feed me!42
This calls the builtin function raw_input with a parameter setting the
prompt to "Feed me!" and assigns the result to data. Since I hit 42
and then enter,
>>> data
'42'
Notice the quotes around 42. They indicate that the value of data is a
string. That's what this tells us:
>>> type(data)
<type 'str'>
The string '42' is not the same as the integer 42:
>>> type(42)
<type 'int'>
>>> '42' == 42
False
So, when you had an if test that was something like:
if given == 1:
# Do stuff here
the equality comparison was never going to work---given was a string
returned by raw_input and no string is ever equal to an integer.
What I suggested was taking the string returned by raw_input and
feeding it to int() to transform it from a string to an integer, and
allow your if test to stand a chance:
>>> data = raw_input("Feed me!")
Feed me!42
>>> if data == 42:
... print "Matches!"
...
>>> data = int(raw_input("Feed me!"))
Feed me!42
>>> if data == 42:
... print "Matches!"
...
Matches!
>>>
There are other ways, for instance:
>>> data = raw_input("Feed me!")
Feed me!42
>>> if data == '42':
... print "Matches!"
...
Matches!
>>>
Here, instead of transforming data to an int and then testing for
equality with 42, I left data as a string and tested for equality with
the string '42'.
The way calling int() is a bit better, I think. If the user enters a
few spaces, then 42 then a few more spaces, that way will still work:
>>> data = int(raw_input("Feed me!"))
Feed me! 42
>>> if data == 42:
... print "Matches!"
...
Matches!
>>>
because
>>> int(' 42 ')
42
>>>
whereas
>>> ' 42 ' == '42'
False
I hope there is some help in there somewhere :-)
Brian vdB
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