how to insert a value into the array in a consistent manner.

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Yuki Koyama
Yuki Koyama 2019-11-20
评论： Bhaskar R 2019-11-20
If I want to get [0;1;1;1] when A=ones(3,1)=[1;1;1], I can get [0;1;1;1] by typing [0;A].
If I want to get [0 1 1 1] when A=ones(1,3)=[1 1 1], I can get [0 1 1 1] by typing [0 A].
How can I write them in the consistent way? Must I always use semicolon or space(or comma) dependign on whether A is a row vector or a column vector?
  1 个评论
Stephen23
Stephen23 2019-11-20
编辑：Stephen23 2019-11-20
"Must I always use semicolon or space(or comma) dependign on whether A is a row vector or a column vector? "
If you want to use the concatenation operator [] then a comma/space or semi-colon is required.
"How can I write them in the consistent way?"
What does "consistent" mean? In one case you are adding a new row, in the other case a new column: do you expect to be able to use the same syntax for both of these?

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Stephen23
Stephen23 2019-11-20
编辑：Stephen23 2019-11-20
>> A = ones(3,1);
>> A(end+1) = 0;
>> A([2:end,1]) = A
A =
0
1
1
1

更多回答（1 个）

Bhaskar R
Bhaskar R 2019-11-20
Are you asking for this ?
A = ones(3, 1);
if (size(A,2)>1)
A = [0 A];
else
A = [0;A];
end
  2 个评论
Steven Lord
Steven Lord 2019-11-20
To make this code a little clearer, you might want to use the isrow or iscolumn function instead of explicitly checking the size. These will also return false if the input is not a vector.
A = ones(3, 2);
if (size(A,2)>1)
A = [0 A]; % Error
else
A = [0;A];
end
With isrow and iscolumn:
A = ones(3, 2);
if isrow(A)
A = [0 A];
elseif iscolumn(A)
A = [0; A];
else
disp('A is not a vector')
end
Alternately, instead of using 0 use zeros to make a matrix with the appropriate number of rows or columns to be compatible with A. If you do that, you'd need to use some other criterion to determine whether to concatenate horizontally or vertically.
A = ones(3, 2);
B = [zeros(size(A, 1), 1), A]
C = [zeros(1, size(A, 2)); A]
Bhaskar R
Bhaskar R 2019-11-20
Yeah, I have been learning/growing with new commands, ways and techniques as a new contributer in this forum. Thanks for the feedback of the answer. Its my pleasure.!!

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