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To: python-list@python.org
From: Dennis Lee Bieber <wlfraed@ix.netcom.com>
Subject: Re: Issue with my code
Date: Tue, 05 Feb 2013 17:20:09 -0500
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On Tue, 5 Feb 2013 10:38:55 -0800 (PST), maiden129
<sengokubasarafever@gmail.com> declaimed the following in
gmane.comp.python.general:

> Hi,
> 
> I'm trying to create this program that counts the occurrences of each digit in a string which the user have to enter.
> 
> Here is my code:
> 
> s=input("Enter a string, eg(4856w23874): ")
> s=list(s)
> 
> checkS=['0','1','2','3','4','5','6','7','8','9']
> 
> for i in s:
>     if i in checkS:
>         t=s.count(i)

	Let's see... for each character in the input, see if that character
is in the check list, and if it is, then reverse and count the
occurences in the input string. (Oh, that's after you converted the
input to a list of single characters)


>         if t>1:
>             for k in range(1,t):
>                 s=s.remove(i)
>                 print(i, "occurs", t,"times.")


	Oh Oh.... Besides the fact that .remove() does its work in-place
(and doesn't return a value -- hence the "s=" is binding None to s), you
are trying to modify the string that you are looping over... And that is
a no-no -- it will cause you to skip over items.

-=-=-=-=-=-
>>> s = list("123456789")
>>> for c in s:
... 	if c == "3":
... 		s.remove(c)
... 		print "c: %s\tREMOVED" % (c,)
... 	else:
... 		print "c: %s\ts: %s" % (c, s)
... 		
c: 1	s: ['1', '2', '3', '4', '5', '6', '7', '8', '9']
c: 2	s: ['1', '2', '3', '4', '5', '6', '7', '8', '9']
c: 3	REMOVED
c: 5	s: ['1', '2', '4', '5', '6', '7', '8', '9']
c: 6	s: ['1', '2', '4', '5', '6', '7', '8', '9']
c: 7	s: ['1', '2', '4', '5', '6', '7', '8', '9']
c: 8	s: ['1', '2', '4', '5', '6', '7', '8', '9']
c: 9	s: ['1', '2', '4', '5', '6', '7', '8', '9']
>>> 
-=-=-=-=-=-

	Note how there is no line for c=4. When you remove the "3", all the
rest shift left to fill in -- but the next round of the loop is going to
"increment" to the character after the position that "3" was at... That
is now the "5".

	Rather than working from the input list, why not turn it around...

-=-=-=-=-=-
>>> inpt = "4856w2304874"
>>> digits = "0123456789"
>>> for d in digits:
... 	o = inpt.count(d)
... 	if o:
... 		print "%s occurs %s times" % (d, o)
... 
0 occurs 1 times
2 occurs 1 times
3 occurs 1 times
4 occurs 3 times
5 occurs 1 times
6 occurs 1 times
7 occurs 1 times
8 occurs 2 times
>>> 
-=-=-=-=-=-
-- 
	Wulfraed                 Dennis Lee Bieber         AF6VN
        wlfraed@ix.netcom.com    HTTP://wlfraed.home.netcom.com/