[Tutor] trouble with "if"

[Brian van den Broek]

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: broek at cc.umanitoba.ca> To: adamurbas at hotmail.com> CC:
> tutor at 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