finding and changing values in matrix that satisfies 2 conditions

Question

Rene Sebena 2017-2-12

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/324624-finding-and-changing-values-in-matrix-that-satisfies-2-conditions

评论： Star Strider 2017-2-12

Hi there, I am new in matlab and try to solve this problem. I have vector which contains various triggers and a lot of zeros: A=[0;0;0;0;0;0;0;0;2;0;0;1;0;0;0;0;0;0;0;0;3;0;0;0;1;0;0;0;0;0;0;0;4;0;1;0;0;0;0;0;0;0;0;5;1]; I need to find first non zero value (in this case the 9th value "2") and then from this point every second nonzero value and change it to other value. My code (see bellow) identifies the values but is not changing them. Any idea why? Thank you for your time and help.

idx = find(A > 0);

B=A;

for i = idx(1:2:20);

    if B(i)==2; B(i)=1;end 
    if B(i)==3; B(i)=2;end
    if B(i)==4; B(i)=3;end
    if B(i)==5; B(i)=4;end

end

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Star Strider 2017-2-12

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/324624-finding-and-changing-values-in-matrix-that-satisfies-2-conditions#answer_254467

在 MATLAB Online 中打开

See if this does what you want:

A=[0;0;0;0;0;0;0;0;2;0;0;1;0;0;0;0;0;0;0;0;3;0;0;0;1;0;0;0;0;0;0;0;4;0;1;0;0;0;0;0;0;0;0;5;1];
idx = find(A > 0);
Aidx = idx(1:2:end);
B=A;
B(Aidx) = B(Aidx)-1;                                % Desired Result
Compare = [A'; B']                                  % See Result & Compare (Delete)
Compare =
    Columns 1 through 16
       0     0     0     0     0     0     0     0     2     0     0     1     0     0     0     0
       0     0     0     0     0     0     0     0     1     0     0     1     0     0     0     0
    Columns 17 through 32
       0     0     0     0     3     0     0     0     1     0     0     0     0     0     0     0
       0     0     0     0     2     0     0     0     1     0     0     0     0     0     0     0
    Columns 33 through 45
       4     0     1     0     0     0     0     0     0     0     0     5     1
       3     0     1     0     0     0     0     0     0     0     0     4     1

2 个评论
显示无隐藏无

Rene Sebena 2017-2-12

Thank you. This is working, but I can not implement it. In some cases I canot use B(Aidx)-1, but need to change it to specific value..

Star Strider 2017-2-12

在 MATLAB Online 中打开

My pleasure.

Is there any pattern to the specific value you need to change the different values to?

In your example, there are 4 values that need to be changed, so if you want to change them respectively to 5, 9, 7 and 3 (chosen arbitrarily), this works:

A=[0;0;0;0;0;0;0;0;2;0;0;1;0;0;0;0;0;0;0;0;3;0;0;0;1;0;0;0;0;0;0;0;4;0;1;0;0;0;0;0;0;0;0;5;1];
idx = find(A > 0);
Aidx = idx(1:2:end);
B=A;
B(Aidx) = [5 9 7 3];                                % Desired Result
Compare = [A'; B']                                  % See Result & Compare (Delete)
Compare =
    Columns 1 through 16
       0     0     0     0     0     0     0     0     2     0     0     1     0     0     0     0
       0     0     0     0     0     0     0     0     5     0     0     1     0     0     0     0
    Columns 17 through 32
       0     0     0     0     3     0     0     0     1     0     0     0     0     0     0     0
       0     0     0     0     9     0     0     0     1     0     0     0     0     0     0     0
    Columns 33 through 45
       4     0     1     0     0     0     0     0     0     0     0     5     1
       7     0     1     0     0     0     0     0     0     0     0     3     1