validate string is valid maths

Matthew_WARREN at Matthew_WARREN at
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?



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...


           steve at                                                                    
           Sent by:                                                                                             cc 
                                                                        Re: validate string is valid maths         
           28/01/2008 15:43                                                                                        

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

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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