Search for elements in a string
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I am trying to find whether a certain string of letters and special characters ( like /) and then output true if it does and false if it doesn't. I know I should convert the characters to numbers using double() but after that I am not sure what to do. An example would be to see if 'add?/adfd' has any elements from ascii value 32-47.
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Cedric
2015-1-28
编辑:Cedric
2015-1-28
Could give an example of text and string that you need to find? There are several tools available, e.g. STRFIND and REGEXP. the former is easy to understand and use, but limited, and the latter is powerful but we would need to know more about the type of strings that you need to find before we can propose a solution.
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Chad Greene
2015-1-28
For multiple special characters, use a few calls of regexp and any:
s = 'abcd\potatoes\eddie\murphy?martin/lawrence';
specialChars = [regexp(s,'\.') regexp(s,'&') regexp(s,'/') regexp(s,'^')];
if any(specialChars)
disp('Yeah, they''re some special characters alright.')
end
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Guillaume
2015-1-28
And even, if it is, a single regexp still suffice:
s = 'abcd\potatoes\eddie\murphy?martin/lawrence';
[pos, match] = regexp(s, '[\\?/]', 'start', 'match');
[characters, ~, subs] = unique(match);
positions = accumarray(subs, pos, [], @(p) {p})';
[characters; positions]
Chad Greene
2015-1-28
Use regexp
s = 'abcd\potatoes\eddie\murphy';
indicesOfSlash = regexp(s,'\')
indicesOfSlash =
5 14 20
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Image Analyst
2015-1-28
Here's a method that (in my opinion) is a little more intuitive, straightforward, and less cryptic than regexp() and bsxfun():
s = 'add?/adfd' % The starting string
d = s - ' ' + 32 % Convert to ASCII number array
% Get indexes of where d is between 32 and 47
specialCharIndices = find(d >= 32 & d <= 47)
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Stephen23
2015-2-1
"a little more intuitive, straightforward, and less cryptic", that is until you want to compare more than one contiguous group of characters:
(s>=32 & s<=47) | (s>=58 & s<=60) | (s>=62 & s<=64) | (s>=133 & s<=140) % etc
bsxfun(@eq,'add?/adfd',[32:47,58:60,62:64,133:140].')
The bsxfun solution is much more robust in comparing any two input vectors, regardless of the contiguity or later decisions about what characters to include/exclude.
per isakson
2015-1-28
编辑:per isakson
2015-1-29
A code based on regexp
str = 'add\?/adfd';
chr = char(32:47);
xpr = ['[',regexptranslate('escape',chr),']'];
pos = regexp( str, xpr, 'once' );
has = not(isempty(pos));
This code may be squeezed into one obscure line.
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