I need help using the strrep for this code.
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here are the three test cases. My question is how do I use strrep to replace the x of that function. The thing is it doesn't have to be x it can be anything but the basic layout is alway going to be functionName(varName) =
[val1] = ugaFunc('f(x) = 2*x^3', 5)
[val2] = ugaFunc('veloc(pos) = 2 / pos ^ 2 - 3', 0.5)
[val3] = ugaFunc('f(var_name) = 4 * (3 / var_name)^var_name-10*var_name',3.14)
function answer = hat(str, val)
% replace the input function with the value you want
a = strrep(str, ' ', 'val');
[b c] = strtok(a, '=');
[d e] = strtok(c, '0123456789 ');
% first number
[answer rest]= strtok(, '+-*/^');
answer = str2num(answer);
% a while loop to get a number and an operator each time
while (length(rest) ~= 0)
[operator rest]= strtok(rest, '0123456789 ');
[num rest] = strtok(rest, '+-*/^ ');
num = str2num(num);
% do the math
switch operator
case '+'
answer = answer + num;
case '-'
answer = answer - num;
case '*'
answer = answer .* num;
case '/'
answer = answer ./ num;
case '^'
answer = answer .^ num;
end
end
end
采纳的回答
You haven't said what you want to replace it with. The value specified in the second argument of your hat function?
In any case, to find the name of the variable, use regular expressions. For example, assuming a simple grammar where the function name is exclusively letters, the following would extract the variable name:
varname = regexp(str, '(?<=[A-Za-z]+\().*?(?=\))', 'match', 'once')
Note that there are many possible expression leading to the same result. The above looks for the shortest (lazy match) sequence of any characters (the .*?) immediately following (the (?<=....) one or more letters (the [A-Za-z]+) plus a bracket (the \(|) and immediately preceding (the |(?=...)) a closing bracket (the \)).
You can then replace the varname in the rest of the expression by the value with strrep or regexprep
newexpr = strrep(str, varname, num2str(val)); %is this what you want?
7 个评论
Kratos
2015-2-19
Guillaume
2015-2-19
I didn't get what I wanted. Then what is it you want?
I have no idea what it is you're asking now. Have you just copied the text of an assignment, and hoping I'll do it for you?
Kratos
2015-2-19
Hum. This is exactly what my answer does (minus a forgotten argument in strrep which I've now added).
str = 'function(varName) = 2*varName^2';
val = 3;
varname = regexp(str, '(?<=[A-Za-z]+\().*?(?=\))', 'match', 'once');
newexpr = strrep(str, varname, num2str(val))
Output:
newexpr =
function(3) = 2*3^2
Kratos
2015-2-19
Guillaume
2015-2-19
Oh, for Pete's sake, try the code. It makes no assumption on what's inside the brackets.
replacevar = @(expr, value) strrep(expr, regexp(expr, '(?<=[A-Za-z]+\().*?(?=\))', 'match', 'once'), num2str(value));
>>replacevar('f(x) = 2*x^3', 5)
ans =
f(3) = 2*3^3
>>replacevar('veloc(pos) = 2 / pos ^ 2 - 3', 0.5)
ans =
veloc(0.5) = 2 / 0.5 ^ 2 - 3
>>replacevar('f(var_name) = 4 * (3 / var_name)^var_name-10*var_name',3.14)
ans =
f(3.14) = 4 * (3 / 3.14)^3.14-10*3.14
Kratos
2015-2-19
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