Problem using sscanf in picking out numbers from string.

So, I have the string:

 str1=D993920x2ExlsSuited20.xls

And I want to pick out the numbers in the string, to give me a vector like

 res=[993920 2 20]

Allowing me to pick up the last number for use in naming a txt document.

However, I cannot get sscanf to give me this.

I tried using.

 res = sscanf(str1, ['%d'])

Shouldn't this give me the numbers in the string at least? All I get is a space sign...

All help on this would be greatly appreciated.

Best regards, Nicki Eriksen.

Answer 1

[Laura Proctor]

The following code should give you results close to what you want:

res = sscanf(str1,'%*[^0123456789]%d')

The idea is that sscanf will try to match all the strings specified, and if it doesn't find a match, then it returns what is found. So, when you use '%d', since the first character in the string is a letter and not a number, it doesn't find a match, and returns nothing.

In the code above, the first element is looking for anything that doesn't match the numbers 0-9. The carat in the beginning indicates a non-match. The asterisk indicates that you don't want to save that information in the output. So, it looks for non-numbers then numbers alternating until the end of the string and only saves the numbers. The documentation for textscan explains it better.

[Nicki]

Yes, thank you, that helped me. I would suggest more, or maybe more diverse, examples in the documentation. I will look in to the documentation for textscan. Thank you for your help.

One (two actually) final question(s), to make sure I got it right,

%[ ] The symbols in these mark what I am looking for, the carat, *, inverts this, so that I am looking for everything not in here.

Why is the ^ there?

[Stephen23]

"...the carat, *, inverts this, so that I am looking for everything not in here. Why is the ^ there?"

You have mixed up the names of the symbols. The caret ^ is used to invert the match inside the square brackets. The asterisk * excludes that data from the output.

[Nicki]

Yes. I realize that now, thank you for clearing it up though.

Answer 2

[Walter Roberson]

 str1 = 'D993920x2ExlsSuited20.xls';
 
 %method 1
 res = str2double(regexp(str1, '\d+', 'match'))

res = 1×3

993920 2 20

 %method 2
 res = double(extract(string(str1), digitsPattern))

res = 3×1

993920 2 20

[Nicki]

Hi. Thanks for the solution. My question is 10 years old, but his was the exact solution I was looking for. Can't remember what I ended up using though.