how to use if loop to determine to make a plot

Question

carly 2022-12-5

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1871852-how-to-use-if-loop-to-determine-to-make-a-plot

评论： Jan 2022-12-6

i cant post full code on here to keep my code private

if im making a function: function[output] = function(input1, input2, input3)

if input3 determines whether or not to make a plot, how can i go about it? in my code i cant directly ask the user to "input yes or no to make a plot" because i can only explain that yes mean to make one and no means not to make a plot in my syntax comments, so i dont think i can use strcmpi

this is what i have so far

if nargin == 3 && input3 == 'yes'
    input3 = true;
else 
    input3 = false;
end 
if nargin < 3 || isempty(input3)
    input3 = true;
end

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Answer 1

Jan 2022-12-5

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1871852-how-to-use-if-loop-to-determine-to-make-a-plot#answer_1120812

编辑：Jan 2022-12-5

在 MATLAB Online 中打开

function output = myfcn(input1, input2, input3)
if nargin < 3
   doPlot = true;  % Default if 3rd input is not provided
else
   doPlot = strcmpi(input3, 'yes');
end
if doPlot
    plot()
end
end

Alternative:

if nargin < 3
   input3 = 'yes';  % Default if 3rd input is not provided
end
doPlot = strcmpi(input3, 'yes');

The modern method is:

function out = myfcn(in1, in2, in3)
   arguments
       in1
       in2
       in3 char = 'yes'  % Default value
   end
doPlot = strcmpi(in3, 'yes');
if doPlot
    plot(in1, in2, 'ro');
end
end

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carly 2022-12-6

what if user inputs 'no'. wont it say "unrecognized function or variable 'no' "

Jan 2022-12-6

There is no try to access a variable or function called 'yes' or 'no'. So this error message will not appear. Simply try it by your own. You can run the code with different inputs.

If the 3rd input is 'no', strcmpi(in3, 'yes') replies false.

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