Some usages of Matlab's find() function (quickly find qualified values)

Sometimes , There's a set of data , And the task , It requires us to find the position of data that conforms to certain rules in this group of data , such as Greater than a number , Or is it Equal to one Count , Or a multiple of a certain number, and so on . Such problems sometimes make us feel headache . Now? , Let's think of a way , So that this problem can be well solved .

For this kind of problem , One of the simplest , Write a Loop traversal , Find eligible data , Make sure that Location ( Index value ), Then save the location data . you 're right , It's very simple , Next, we will implement this method .

The problem background ： One set of data is [1,2,3,4,0,1,0,10,12,16], Of course , Maybe more , Find something that can be 2 Divisible data , And record its location .

clc;clear;M=[1,2,3,4,0,1,0,10,12,16];n=length(M);List=[];j=0;for i=1:n    if mod(M(i),2)==0  % Determine whether it can be 2 to be divisible by         j=j+1;        List(j)=i;    endenddisp(List)% Where to output qualified data RE=M(List);% Get qualified data disp(RE)% Output qualified data 

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 result ： 2     4     5     7     8     9    10  % Location of eligible data  2     4     0     0    10    12    16  % Eligible data 

Of course , This can solve the problem well . however , But it's a little complicated , Is there a simpler way ？ ok , It does .

Form 1 ：find( Judge the condition )​​​​​​

 M=[1,2,3,4,0,1,0,10,12,16]; r1=find(M==0); % Data found 0 Position of elements  disp(r1); r2=find(M~=0);% Find the non 0 Position of elements  disp(r2); r3=find(M==1);% Find a value equal to ( Here for 1) The location of the elements of  disp(r3); r4=find(M>10);% Find a value greater than ( Here for 10) The location of all the elements of  disp(r4); r5=find(mod(M,2)==0);% Find something that can be 2 The position of all elements divisible  disp(r5);  result ：     5     7
     1     2     3     4     6     8     9    10
     1     6
     9    10
     2     4     5     7     8     9    10

Of course , This judgment condition can be many, many , You can set it according to your actual needs .

Form 2 ( Main operation and 0 of ):

1   index=find(M)

Find the data M Position of all non-zero elements in ( Index value ), The return value is the location label .M It can be a vector , It can also be a matrix .

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>> M=[1,2,3,4,6,0,0,9,0];>> find(M)ans =     1     2     3     4     5     8%-------------------------------------->> M=[1,2;0,8];>> find(M)ans =     1     3     4 notes ：matlab The elements of the matrix are stored by column 

2   index=find(M,n)

Return to the former n The position of a non-zero element ,M It can be a vector , It can also be a matrix .​​​​​​​

>> M=[1,2,3,4,1,6,0,0,9,0];>> find(M,1)ans =     1>> find(M,2)ans =     1     2>> find(M,0) Misuse  find The second argument must be a positive scalar integer .

3   index=find(M,1,'last')

Returns the position of the last non-zero element ,M It can be a vector , It can also be a matrix .​​​​​​​

>> M=[1,2,3,4,1,6,0,0,9,0];>> find(M,1,'last')ans =     9

notes ：matlab The elements of the matrix are stored by column

Form 3 ( Mainly for matrix )

Through the operation in form 2 , For matrices , We got the position , However, this position , Computers can recognize , But for us, we have to do some calculations . such as , One 2 Order matrix . Start with the first in the first column , Store... In columns , The serial numbers are 1,2,3,4. In fact, if you use two numbers to determine , Then the corresponding relationship is like this

1-----------（1,1）

2-----------（2,1）

3-----------（1,2）

4-----------（2,2）

[r,c]=find( Judge the condition )

r: Row index

c: Column index

example ：​​​​​​​

M=[1,2,3;4,0,8;0,0,9];[r,c]=find(M~=0);disp(r);disp(c);disp([r,c]); result ：     1     2     1     1     2     3
     1     1     2     3     3     3
     1     1     2     1     1     2     1     3     2     3     3     3

obviously , Use find() Function can make our programming very simple , It will greatly reduce our programming time , So as to leave us more time to think about more complex problems .