为什么这个MATLAB'if'语句不起作用？

Question

I'm pretty new to MATLAB and I can't figure out why my "if" statement and condition don't work. I have a 100-by-4 matrix ( randQ ). Columns 1 through 3 are random integers, and column 4 contains either a 1 or 2 in each row.

I want to change all 1 's into 0 's and all 2 's into 1 's in column 4 in that 100-by-4 matrix. What I'm doing is:

if randQ(:,4) == 1
    randQ(:,4) = 0
elseif randQ(:,4) ==2
    randQ(:,4) = 1
end

It does not throw any errors but the matrix does not change and, in my mind, the conditions make perfect sense. I don't know why it doesn't work. Any explanation would be welcome.

Answer 1

Disclaimer

Some of the other answers contain incorrect/inaccurate claims:

• you are trying to compare a vector with a scalar, that would never be true.
• You are asking if your vector == 1 which is not the case.

As an example,

if [1, 1, 1] == 1
    disp("True")
end

will display "True" . See below for an explanation.

---

You write

I cant figure out why my "if" statement and condition does not work

What's happening is actually quite subtle. You need to develop an understanding of several things:

1. what the expression randQ(:,4) == 1 returns,
2. what the "if" statement does when presented with a predicate that is, not a scalar, but an array,
3. what randQ(:,4) = 0 does.

Furthermore, you need to start using logical indexing for this kind of operations.

What randQ(:,4) == 1 returns

Under the assumption that randQ is a 100-by-4 array, the expression randQ(:,4) == 1 returns a 100-by-1 logical array , ie full of (logical) zeros and ones:

• if the ith entry of this array is a (logical) 1 , it means that entry (i,4) of randQ is equal to 1 ;
• if the ith entry of this array is a (logical) 0 , it means that entry (i,4) of randQ is not equal to 1 .

Array as a predicate

Now that you know that you're using an array for the predicate of the "if" statement, let's see what happens. If the predicate of your "if" statement is an array, MATLAB will execute the "if" branch only if all entries of that array evaluate to logical 1 .

For instance,

if [1, 2; 3, 4]
    disp("True")
else
   disp("False")
end

will display "True" , because all entries of [1,2;3,4] get cast as a logical 1 , which causes the predicate to be evaluated as logical 1 (true). However,

if [1, 2; 3, 0]
    disp("True")
else
   disp("False")
end

will display "False" , because entry (2,2) of the [1,2;3,0] gets cast as a logical 0 , which causes the predicate to be evaluated as logical 0 (false). Therefore, if at least one entry in randQ(:,4) is zero, the if will not get executed.

Your assignment statements are incorrect anyway

The assignment statement

randQ(:,4) = 0

would overwrite all entries in the 4th column with 0 , which is not what you want.

A better approach: use logical indexing

You write

I want to change all 1's into 0's and all 2's into 1's in column 4 in that 100-by-4 matrix.

A more idiomatic approach for this, as pointed out by giuseppe , is to use logical indexing :

randQ(randQ(:,4) == 1, 4) = 0;
randQ(randQ(:,4) == 2, 4) = 1;

No need to use the find function, though, because randQ(:,4) == 1 already returns what you want: a 100-by-1 logical array indicating which entries of the 4th column of randQ are equal to 1 .

Answer 2

You are asking if your vector == 1 which is not the case.

There are more way to do it, but you can loop every value with a for.

Answer 3

The result of randQ(:,4) is a vector with a length of 100. You are then trying to compare it to a scalar, which will be true only if all the members of the vector (or matrix) are equal to the scalar.

> M=[1 1; 2 1; 3 2]
M =                                            

   1   1
   2   2
   3   1

> M(:,1)
ans =

   1
   2
   3

Also, the way you try to change the values isn't true, if you do randQ(:,4) = 1 then the whole 4 ^th column will get the value 1:

>M(:,2)=0
M =

   1   0
   2   0
   3   0

The simplest way to achieve what you want is by iterating over the matrix in a loop row by row, checking the value and then changing it.

Answer 4

As also answered by the others you are trying to compare a vector with a scalar, that would never be true.

Instead you can achieve your aim using the function find , this function in fact implements elements comparison.

randQ(find(randQ(:,4) == 1),4) = 0;
randQ(find(randQ(:,4) == 2),4) = 1;