validate string is valid maths

Matthew_WARREN at bnpparibas.com Matthew_WARREN at bnpparibas.com
Mon Jan 28 17:31:50 CET 2008


Ok, I was thinking along the same lines myself, replacing ++ etc.. until no
more replacements are made.

I hadnt considered creating a table of pairs and replacements though, or
realised that the same replace method would work in that case. handy :)

The dictionary is unordered; would differences in the order of replacements
being made effect the final outcome?

IE

*/*

replaced in order of */ then /* would give (using table below)

*/ => **
/* => **

But in the order /* then */

*/*


/* => */
*/ => *

I've tried testing, but I'm not certain wether repeated iterations over a
dict return different sequences of key,value pairs or wether I'll be
getting the same (but arbitrary) sequence each time even though they are
unordered, etc

So for testing, what could I do to guarantee the next iteration over the
dict will give keys/pairs in a different sequence to last time?


...actually, whenever I iterate over the dict say with for k,v in n.items()
I get exactly the same sequence each time.  For once I want to exploit the
dict's unorderedness and it's decided its going to be all ordered...

Matt.




                                                                                                                   
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                                                                        Re: validate string is valid maths         
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On Mon, 28 Jan 2008 15:10:54 +0000, Matthew_WARREN wrote:

> Hi pythoners.
>
> I am generating strings of length n, randomly from the symbols
>
> +-/*0123456789
>
> What would be the 'sensible' way of transforming the string, for example
> changing '3++++++8' into 3+8

That's easy: replace pairs of + into a single +, repeatedly until there's
nothing left to replace. And so forth. Here's an untested function. It's
probably not even close to optimal, but for playing around it is probably
fine.

def rationalise_signs(s):
    while "++" in s or "+-" in s or "-+" in s or "--" in s:
        s = s.replace("++", "+")
        s = s.replace("--", "+")
        s = s.replace("+-", "-")
        s = s.replace("-+", "-")
    return s




> or '3++--*-9' into '3+-9' such that  eval(string) will always return a
> number?
>
> in cases where multiple symbols conflict in meaning (as '3++--*-9' the
> earliest valid symbols in the sequence should be preserved

You have four symbols, so there are just 4*4=16 sets of two symbols.
Probably the easiest way is to just list them and their replacements out
in a table:

table = {"++": "+", "+-": "-", "+*": "+", "+/": "+",
    "-+": "-", "--": "+", "-*": "-", "-/": "-",
    "*+": "*", "**": "*", "*/": "*",
    "/+": "/", "/*": "/", "//": "/", }

Notice that both *- and /- don't get changed.



# Untested.
def rationalise_signs(s):
    prev = ''
    while s != prev:
        prev = s
        for key, value in table.items():
            s = s.replace(key, value)
    return s



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Steven
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