does an "if does not contain operator" exist
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I am trying to write a function where the input must be entered in radians. My idea was to make the function return if the input does not contain "pi". Is there a way to do this?
ex. if you wanted to enter 90 you have to enter pi/2
Thanks
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Réponse acceptée
Jan
le 16 Mar 2013
Modifié(e) : Jan
le 17 Mar 2013
Trust the users of your program:
d = input('Input angle in radians [0 <= x <= 2*pi]: ')
Then even "pi/2" is a valid answer.
Testing the input to contain the string 'pi' would be not sufficient: E.g. the string '32 pineappletrees' should not be accepted, but '3.14159265358979 / 2' should.
3 commentaires
Jan
le 18 Mar 2013
@Steve: Yes, of course. You can call input() within a function also. But I'd prefer inputdlg.
Plus de réponses (3)
Matt Kindig
le 16 Mar 2013
The "if does not contain" operation can be done with ismember, using something like:
notPi = ~ismember( user_input, pi)
However, due to numerical precision, you probably want to use a tolerance to determine how close the user is to pi, e.g.
notPi = abs(user_input-pi) > 1e-7
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Azzi Abdelmalek
le 16 Mar 2013
If the input does not contain pi does not mean that the input is not in radian.
Now if you want the input to contain pi
s=[]
while isempty(s)
u=input('Enter value (like 2*pi)','s')
s=strfind(u,'pi')
end
u=str2num(u)
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Image Analyst
le 16 Mar 2013
Modifié(e) : Image Analyst
le 16 Mar 2013
Every number contains pi - some factor times it. So what you want is to have the user enter a string and see if "pi" is in the string and alert them if it's not in the string. Try this:
% Ask user to enter some string containing pi.
% Keep asking until they do.
defaultValue = 42;
titleBar = 'Enter a value';
userPrompt = 'Enter the value containing pi:';
while true
% Ask user for a number containing pi.
caUserInput = inputdlg(userPrompt, titleBar, 1, {num2str(defaultValue)});
if isempty(caUserInput),break,end; % Bail out if they clicked Cancel.
% Convert to lower case, then see if "pi" is in there.
userString = char(lower(caUserInput{1}));
piLocation = strfind(userString, 'pi');
if piLocation >= 1
% String contains pi so we can exit the loop now.
break;
end
end
1 commentaire
Greg Heath
le 18 Mar 2013
Why not change the input to a normalized angle in [ 0 1) translated as radianangle/2/pi?
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