How to Write all of an if-statement in a Single Lline?
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Is there any way to write all of an if-statement in a single line?
if A == 1 B = 2 elseif B = 3 end
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Azzi Abdelmalek
2016년 8월 16일
편집: Azzi Abdelmalek
2016년 8월 16일
if A ==1 B = 2, else B = 3, end
But what is the aim of this?
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Steven Lord
2021년 4월 13일
Consider writing a "logme" function.
function logme(messageToLog)
verbose = true;
if verbose
fprintf('%s', messageToLog)
end
end
Now your code will contain calls to logme.
logme("Diagnostic message #1")
If you want to run without displaying these messages, change the definition of verbose inside logme. You could potentially also modify logme to log those messages not to the screen but to a log file without changing the callers of logme.
Angelo Yeo
2023년 7월 8일
편집: Angelo Yeo
2023년 7월 8일
If you want to set up if-else statement with a single line and make it into an inline funciton, you can think of such a thing.
ternary = @(varargin) varargin{end - varargin{1}};
ternary(true,'yes','no') % If the first argument is true, the result becomes 'yes'
ternary(false,'yes','no') % If the first argument is false, the result becomes 'no'
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KarlHoff
2021년 10월 20일
You can employ logicals as factors toggling between 0 and 1:
B = 2 + 1*(A~=1);
%or
%B = 3 + (-1)*(A==1) ; % just roles reversed - depending on taste
This approach works whenever your if-statement just discriminates between two versions of a single assignment to one and the same variable as your example does.
I personally find this particular useful if there is a base value (here 2) and some delta added conditionally (here, +1 to obtain 3). You can also extend for further conditions.
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Reno Filla
2021년 12월 9일
The problem is if you want to get NaN on the case of, for example A==1
2+NaN*(A==1) doesnt work because 0*NaN return NaN just as 1*NaN does.
Also employing the NaN() function like
2+NaN(A==1) doesnt help since NaN(0) returns [] and canot be added to a scalar.
Any ideas?
KarlHoff
2021년 12월 10일
How about
0/(A-1) + 3 %returns different numeric values for different A ~= 1
or
0/( (A==1)-1 ) + 3 % returns 3 for all A ~= 1
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