How to evaluate numeric expression in a string, which contains "12u" (for 12e-6) and "0.1m" (for 0.1e-3) formated numbers (standing for micro and milli)?

Hi,

I'd like to have a "supercharged" eval function, which would accept expressions where numbers can be given with 'm' (mili) or 'u' (micro extension.

Here is my naive approach is to define function seval

seval("1m")     % works fine

ans = 1.0000e-03

seval(['10*(0.2m+20*3.1u)'])   % works fine

ans = 0.0026

seval('max(1u,2m)')  % errors, since 'm' in 'max' is also replaced

Error using solution>seval
Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error. To construct matrices, use brackets instead of parentheses.

function rez = seval(s)
    s = strrep(s, 'u', 'e-6');  % micro
    s = strrep(s, 'm', 'e-3');  % mili
    s = strrep(s, 's', 'e-0');  % :-)
    rez = eval(s);
end

This works fine for simple expressions, but obviously not for

seval('max(1u,2m)')

So, I'd like to replace 'm' with 'e-3', but only when 'm' follows a well formated number. Seems like a job for regexp, but I couldn't find an elegant way to solve this.

P.S. Bonus question:

I'd also like to use 'variables'. A way to acieve this is to replace e.g. '1+a' with '1+var(a)', but only when 'a' is "standalone factor" in the expression. Any ideas?

seval2('1m + a')      % this would translate to ...
seval('1m + var(a)')  

function rez = var(s) % an example of 'variable' function
switch s
    case 'a'
        rez =1;
    case('b')
        rez = 2;
    otherwise
        rez = 0;
end
end

Answer 1

[Star Strider]

Here is my regexprep attempt at ‘seval’ —

str = 'max(1u,2m)';
out = seval(str)

out = 'max(1e-6,2e-3)'

str = 'max(1.5u,2.01m)';
out = seval(str)

out = 'max(1.5e-6,2.01e-3)'

function res = seval(s)
res = s;
res = regexprep(res, '(?<=\d)m','e-3');
res = regexprep(res, '(?<=\d)u','e-6');
res = regexprep(res, '(?<=\d)s','e-0');
end

The ‘Bonus question’ eludes me, likely because I am not confident that I understand it.

.

[Star Strider]

With respect to @Steven Lord’s quite valid point, instead of eval (the use of which is generally discouraged and for good reason), one option could be to use str2func with the replaced numeric values and evaluate the resulting anonymous function —

out = 'max(1.5e-6,2.01e-3)';
outfcn = str2func(['@(x)' out])

outfcn = function_handle with value:
    @(x)max(1.5e-6,2.01e-3)

outfcn(1)

ans = 0.0020

The argument itself in this example is irrelevant. It just has to be something that the function can use.

In other situations with other functions, if the function actually uses the argument, the argument would have to be appropriate.

.

[Tomazz]

Thank you, your first solution is perfect for what I needed. I knew there had to be a nice hack for that.

It seems I didn't state the bonus question clearly enough, but it is less difficult to solve, anyway.

Of course you are both right that my approach is not the best in a general case, but for limited and specialized requirements, your first solution works great.

Also thank you for pointing out that str2func can be used this way. I was not aware of that.

[Star Strider]

As always, my pleasure!

Answer 2

[Steven Lord]

I'd like to have a "supercharged" eval function

This doesn't sound like a good idea.

which would accept expressions where numbers can be given with 'm' (mili) or 'u' (micro extension.

I strongly recommend that instead of making a "supercharged" version of eval that you think smaller. A function that accepts a string or char vector containing a number followed by a suffix then:

1. splits the text into the number part and the suffix part
2. converts the number part to an actual double value then
3. multiplies the double value by a scaling factor determined from the suffix

would be more focused and less likely to get you into trouble. The splitting may be a bit tricky, since the "number part" may contain digits, a decimal point, perhaps an arithmetic symbol, and potentially the letters 'd', 'e', 'i', and/or 'j'.

x = 1.23e-1+2i

x = 0.1230 + 2.0000i

[Tomazz]

Thank you for comments. I know my hack is a bad practice and won't work in a general case.

However, my requirements are more specialized and limited, and I don't need to process every kind of allowed number format, just simple “1.2m”, “exp(-3m*10)”, so Star's solution works perfectly for me.