[Tutor] Algorithm

[Wayne]

On Tue, Aug 25, 2009 at 2:14 PM, kreglet <kreglet at gmail.com> wrote:

>
> Wayne,
>  I appreciate your patience with me. I still can't get this to work:
>
>
> from operator import itemgetter
> class testwords:
>         def __init__(self):
>                                self.lettercount={}
>                                 self.inword=False
>                                self.mainword=""
>                                self.cmpword=""
>
> def countletters(word):
>        lc.lettercount = {}
>        for letter in word:
>                lc.lettercount[letter] =lc.lettercount.get(letter,0) + 1
>        print sorted(lc.lettercount.iteritems(), key=itemgetter(1))
>
> def comparewords(cmpword, mainword):
>        for letter in cmpword:
>                if mainword.get(letter):
>                        print letter, cmpword[letter], mainword[letter]
>                         if cmpword[letter] >mainword[letter]:
>                                 lc.inword=False
>
>                else:
>                         if cmpword[letter] <=mainword[letter]:
>                                 lc.inword=True
>
>
> lc=testwords()
>
> lc.mainword="batty"
> lc.cmpword="bat"
>
> countletters(lc.mainword)
> mainword = lc.lettercount
>
> countletters(lc.cmpword)
> cmpword = lc.lettercount
>
> comparewords(cmpword, mainword)
>
> if lc.inword==True:
>        print lc.cmpword + " IS in: " + lc.mainword
> if lc.inword==False:
>        print lc.cmpword + " IS NOT in: " + lc.mainword


This is a bit redundant -  since lc.inword returns True or False you can
simply test:

if lc.inword:
  #do stuff
else:
  #do other stuff

That's really the proper way to do it. Also, AFAIK two "if" statements take
longer than an if/else statement. Of course we're talking about ms or less,
but it still reads better to have an if/else.

It appears that I've not been coding enough lately - and my explanation has
been a bit of a failure.

I just did a quick test and this code seems to work correctly:
def countletters(word):
    lettercount = {}
    for letter in word:
        lettercount[letter] = lettercount.get(letter, 0) + 1

    return lettercount

def comparewords(cmpword, mainword):
    for letter in cmpword:
        if mainword.get(letter):
            if cmpword[letter] > mainword[letter]:
                return False
        else:
            return False
    return True

word1 = countletters('python')
word2 = countletters('pyz')

print comparewords(word2, word1)


at least "py" was in "python", so was "pyn", but 'pyz' was not.

HTH,
Wayne
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