If-expression only runs the first expression?

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Sam
Sam 2014-12-20
编辑： Jan 2014-12-21
I created a for loop for a 5x5 matrix (it is a struct with data in it). But there are some values of the matrix I don't want to calculate. For example, I don't want to calculate cell (1,4). I also don't want to calculate the cell (3,4), (4,4),...
if ~((i_testen == 4) & (welke_pp == 1)) | ((i_testen == 4) & (welke_pp == 3)) | ((i_testen == 4) & (welke_pp == 4)) | ((i_testen == 4) & (welke_pp == 5)) %i_testen stands for the measurementnumber. welke_pp stand for the subjectnumber.
RASI = data_sts(welke_pp,i_testen).VideoSignals(:, strcmp('RASI', data_sts(welke_pp,i_testen).VideoSignals_headers)); %extract data
XY(2,1) = max(RASI) %maximum of RASI
XY(1,1) = 0; %mimimum is set to zero
Begin_Eind_sts.Begin(i_testen) = abs(XY(2,1)); %store data
Begin_Eind_sts.Eind(i_testen) = abs(XY(1,1));
close all %close all opened figures
else
continue %continue with the previous for-loop
end
The problem is that the program runs perfectly, and even when the values for i_testen = 4 and welke_pp = 1, the program goes to 'else' and continues the for loop. But when the next values for the if-expression comes up (being i_testen = 4 and welke_pp = 3), the program doens't jump to 'else'.
  2 个评论
John D'Errico
John D'Errico 2014-12-20
Please use the code formatting button to make your code readable.
Jan
Jan 2014-12-21
You have a "matrix", which is a "struct" and the elements are "cells"? These are contradictions and inconsquence confusing.

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采纳的回答

Jan
Jan 2014-12-21
编辑：Jan 2014-12-21
Of course your code does not enter the else part, when i_testen = 4 and welke_pp = 3.
if ~((i_testen == 4) & (welke_pp == 1)) | ...
((i_testen == 4) & (welke_pp == 3)) % shortend
The ~ operator matters the first condition only. So for i_testen = 4 and welke_pp = 3, the condition is true. Perhaps you want additional paranetheses?
if ~( ((i_testen == 4) && (welke_pp == 1)) || ...
((i_testen == 4) && (welke_pp == 3)) || ...
((i_testen == 4) && (welke_pp == 4)) || ...
((i_testen == 4) && (welke_pp == 5)) )
I'm used the && and operators, because the expressions are scalars, but the vector operators & and | are correct also.
Or shorter:
if i_testen ~= 4 || any(welke_pp ~= [1,3,4,5])

更多回答（1 个）

Star Strider
Star Strider 2014-12-20
The ‘else’ condition may not be necessary.
See if:
if ~((i_testen == 4) & (welke_pp == 1)) | ((i_testen == 4) & (welke_pp == 3)) | ((i_testen == 4) & (welke_pp == 4)) | ((i_testen == 4) & (welke_pp == 5))
. . . CODE . . .
close all %close all opened figures
end
works. The ‘else’ condition is likely implied.
  6 个评论
显示 4更早的评论
Sam
Sam 2014-12-21
I've ran your code, if I enter the values behind the if-expression, only the first one (i_testen = 4 and welke_pp = 1) works. The other ones doesn't work.
Star Strider
Star Strider 2014-12-21
编辑：Star Strider 2014-12-21
I have no idea what you want to do, so I’m not sure what to suggest. The code I wrote is designed to help you troubleshoot your code. You’ll have to experiment with your if logic to get the result you want. That’s the best I can do just now.
I would start with several large sheets of paper, then diagram on them what I want to do in various situations. (In the days when I did this for my FORTRAN programs on the back of 132-column fanfold line-printer paper, it was called a ‘flow chart’.) Write short programs to test your logic so you’re certain it will work as you want it to. Then write your program.
I’ll help as much as I can, but I can only do so much.

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