Forcing matrices to match in size

Let's say I have two single-row matrices A and B.
I want to check if A and B have the same number of columns. If they do, do nothing.
If not, I want to remove the last character in B.
For example, if we have
A = [30 25 78]
B = [17 91 44 83]
Then I want to remove the last column of B, leaving
A = [30 25 78]
B = [17 91 44]
But if we originally had
A = [30 25 78]
B = [17 91 44]
then do nothing. What scheme could accomplish this?

Antworten (2)

Arturo Mendoza Quispe
Arturo Mendoza Quispe am 8 Sep. 2019

0 Stimmen

B = B(1:numel(A));
This will set B to have the size of A, hence there is no need for an if statement.

3 Kommentare

Not correct according to the problem statement. If A is longer than B, then the number of columns do not match and it is required to remove the last column in B, even though that would make B even shorter than A. Also, if A were (say) 3 shorter than B, then according to the problem statement, only the last column of B is to be removed, not the last two columns.
The logic is:
is length(A) the same as length(B)
if not then remove the last column of B
It is not clear what is intended to happen if B is already empty but A is not empty.
g
g am 8 Sep. 2019
Bearbeitet: g am 8 Sep. 2019
Based on what I'm working with, the number of columns in B is always greater than or equal to the number of columns in A. I should have been more precise.
Walter Roberson
Walter Roberson am 8 Sep. 2019
But you still only want to remove one column of B in that case, right? Because your problem title talks about making them match. Arturo's solution is great to make them match, but is not the correct solution if exactly one column is to be removed from B.

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Walter Roberson
Walter Roberson am 8 Sep. 2019

0 Stimmen

B = B(1:end-(length(A) ~= length(B)));
Leaves B the same length if A and B are the same length, and otherwise removes exactly 1 element from B.

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Gefragt:

g
g
am 7 Sep. 2019

Beantwortet:

am 8 Sep. 2019

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