Converting strings to operators

11 次查看（过去 30 天）
v k
v k 2020-11-26
评论： Walter Roberson 2020-11-26
I think this is a naive query. Yet, let me pose it:
If '(' '2' '*' '3' ')' are elements of a string vector S, then what function f results in the following:
f(S) = (2*3) = 6
str2num for '2' and '3' is not an option, since they can occur anywhere in S, and there is no apriori information as to where they will occur.
For the same reasons, keeping the bracket structure intact is also important, even though it seems that we can do away with the brackets in this example.

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2020-11-26
Ran in:
S = '(2*3)+Z*14*(3*5)+(7)*(9)-(11)*13'
S = '(2*3)+Z*14*(3*5)+(7)*(9)-(11)*13'
while true
newS = regexprep(S, {'\((\d+)\)\*\((\d+)\)', '\((\d+)\)\*(\d+)', '(\d+)\*\((\d+)\)', '(\d+)\*(\d+)'}, {'${string(times(str2double($1),str2double($2)))}'})
if strcmp(newS, S); break; end
S = newS;
end
newS = '(6)+Z*14*(15)+63-143'
newS = '(6)+Z*210+63-143'
newS = '(6)+Z*210+63-143'
Note that the logic gets more complex if you have operators that are higher priority, such as ^ . For example '2*3^Z' should not have the 2*3 multiplied out.
Also, you could reasonably argue that I did not preserve brackets in some of the cases such as (11)*13 -- it would not be unreasonable to claim that the result of that should be (143) instead of 143 . You could take care of those cases by having four different replacement strings instead of the single replacement I have there.
I did try to unify all of the cases into one case, but stripping off the brackets without using nested regexp or using an auxillary function would have been a bit messy. It would probably be cleanest to use some auxillary functions like
{'${s_times($1,$2)}'}
with s_times responsible for stripping off brackets and converting to numeric and doing the multiplication and converting the result to character. It would probably be a good idea to use something like that, and then the cases could probably be reduced to a single case.
  5 个评论
显示 3更早的评论
v k
v k 2020-11-26
Shunting Yard --- How did you find this??!! Thank you for directing to that algorithm! Come to think of it - there has to have been an algorithm for the strings to function as operators. They won't become operators until such an algorithm is in play.
Also, thanks for the logical true and false - that works.
Walter Roberson
Walter Roberson 2020-11-26
Shunting Yard got mentioned by other people a couple of times recently... they found it, not me :)

请先登录，再进行评论。

更多回答（1 个）

Matt J
Matt J 2020-11-26
编辑：Matt J 2020-11-26
Ran in:
I can't tell if you really intend S to be a string vector because that dpesn't agree with the example you posted. If I assume it is what you meant, though, then you could do,
S=["(" "2" "*" "3" ")"];
whos S
Name Size Bytes Class Attributes S 1x5 366 string
eval(cell2mat(cellstr(S)))
ans = 6
  1 个评论
Walter Roberson
Walter Roberson 2020-11-26
If this is user input then evalc is dangerous as the user could have coded a call to delete files

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by