How do I locate all integer values within a matrix (of string and integer values in the same cell), then replace all those integer values with a 1 or 0 thus forming a new matrix with the replaced integers?

Question

Michael Tross 2019-11-14

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/490977-how-do-i-locate-all-integer-values-within-a-matrix-of-string-and-integer-values-in-the-same-cell

关闭： MATLAB Answer Bot 2021-8-20

"( x(2) | x(1) )"

"x(3)"

"( x(4) | x(6) | x(5) )"

"( x(7) | x(8) )"

""

"x(9)"

"( x(10) & x(11) & x(12) )"

"x(13)"

"( x(15) | x(14) | x(1) )"

The text above are in a column vector, but I need to replace the integers with a zero or one. Eg. All integer values not equal to two should be zeros (x(0)), whereas those equal to two should be '1' (x(1)

4 个评论
显示 2更早的评论隐藏 2更早的评论

Daniel M 2019-11-14

I know there's a way to do this in one line, but I'm not great with regular expressions. Here's what I came up with, perhaps you can make it more elegant.

s = ["( x(2) | x(1) )", "x(3)", "( x(4) | x(6) | x(5) )", "( x(7) | x(8) )", "( x(7) | x(8) )", "", "x(9)", "( x(10) & x(11) & x(12) )", "x(13)", "( x(15) | x(14) | x(1) )"];
r = regexprep(s,'\(2\)','(NaN)');
r = regexprep(r,'\(\d*\)','(0)');
r = regexprep(r,'NaN','1');

Michael Tross 2019-11-14

编辑：Michael Tross 2019-11-14

Thank you Daniel! That did the trick!

请先登录，再回答此问题。

Answer 1

Stephen23 2019-11-14

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/490977-how-do-i-locate-all-integer-values-within-a-matrix-of-string-and-integer-values-in-the-same-cell#answer_401439

编辑：Stephen23 2019-11-14

Method one: multiple regular expressions in one regexprep call:

>> c = {'( x(2) | x(1) )', 'x(3)', '( x(4) | x(6) | x(5) )', '( x(7) | x(8) )', '( x(7) | x(8) )', '', 'x(9)', '( x(10) & x(11) & x(12) )', 'x(13)', '( x(15) | x(14) | x(1) )'};
>> d = regexprep(c,{'\(2\)','\d+','%%%'},{'(%%%)','0','1'});
>> d{:}
ans =
( x(1) | x(0) )
ans =
x(0)
ans =
( x(0) | x(0) | x(0) )
ans =
( x(0) | x(0) )
ans =
( x(0) | x(0) )
ans =
     ''
ans =
x(0)
ans =
( x(0) & x(0) & x(0) )
ans =
x(0)
ans =
( x(0) | x(0) | x(0) )

Method two: dynamic replacement expression:

>> fun = @(x)num2str(isequal(str2double(x),2));
>> d = regexprep(c,'\d+','${fun($&)}');
>> d{:}
ans =
( x(1) | x(0) )
ans =
x(0)
ans =
( x(0) | x(0) | x(0) )
ans =
( x(0) | x(0) )
ans =
( x(0) | x(0) )
ans =
     ''
ans =
x(0)
ans =
( x(0) & x(0) & x(0) )
ans =
x(0)
ans =
( x(0) | x(0) | x(0) )