解析正则表达式语法问题matlab /八度

Question

I can parse the regular expression in matlab / octave below:

A = 'Var Name 123.5'
[si ei xt mt] = regexp(A, '(\d)*(\.)?(\d)*$')
number = str2num(mt{1})
number =  123.50

But I get a syntax error below most likly caused due to the ]

A='[angle_deg = 75.01323334803705]'
[si ei xt mt] = regexp(A, '(\d)*(\.)?(\d)*$])

how can I fix this regular expression?

Answer 1

Your regular expression from the first method is fine assuming that you are looking for a number at the end of the string. Because you have an ending ] character in your new string, your regular expression won't work because your string does not end in a number . As such, simply removing the $ character should work, as you want to search for one number that may or may not be floating point. You have three capturing groups in your regex , where the first capture group grabs the integer part of the number, the second capture group optionally grabs a decimal point, and the last capture group grabs the floating point portion of your number.

You also did not close your string properly in your regex . It needs an ending single quotation. Therefore:

A='[angle_deg = 75.01323334803705]';
[si ei xt mt] = regexp(A, '(\d)*(\.)?(\d)*');

Displaying all of the output variables from regexp , this is what I get:

>> si

si =

    14

>> ei

ei =

    30

>> xt

xt = 

    [3x2 double]

>> mt

mt = 

    '75.01323334803705'

si denotes the starting index of where the match happened, which is index 14 in your string. ei denotes the ending index of where the match happened, which is index 30. xt shows you the starting and ending index that matched each token or capturing group of your regular expression. To display this, simply do:

>> xt{1}

ans =

    14    15
    16    16
    17    30

Therefore, the first capturing group begins at index 14 and ends at index 15, which is the 75 portion of your number. The second capturing group begins at index 16 and also ends there, which denotes the . character. Finally, index 17 to 30 denote the floating point portion of your number, which is 01323334803705 . To finish it all off, mt shows you the extracted string that matched the regular expression, which is the number at the end of this string. You can certainly convert this string into a number by using str2num .