array: concatenate two columns of integers to one colum
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(Matlab novice)
I thought this would be so easy...
I have a 2-d array of integers and I want to combine two columns like this:
... 123 456 ...
... 123 456 ...
etc...
I want:
... 123456 ...
... 123456 ...
etc...
as single numbers (that is, I want (123 * 1000) + (456)
The number of rows will be variable from one run to the next.
Any help would be appreciated.
1 件のコメント
Jan
2012 年 9 月 13 日
The question is not clear. What is the wanted output for: [1,2,3,4; 5,6,7,8; 9,10,11,12] ?
採用された回答
Honglei Chen
2012 年 9 月 13 日
編集済み: Honglei Chen
2012 年 9 月 13 日
a = randi([1 100],[5 4]) % a 5x4 integer matrix
b = cellfun(@num2str,{a(:,2:3)},'UniformOutput',false)
c = cellfun(@(x) strrep(x,' ',''), cellstr(b{1}),'UniformOutput',false)
a(:,2) = str2double(c)
a(:,3) = []
2 件のコメント
Honglei Chen
2012 年 9 月 13 日
@Jan, thanks for the tip. That's a good one. I think you missed a b in your second expression.
その他の回答 (4 件)
Jan
2012 年 9 月 13 日
編集済み: Jan
2012 年 9 月 13 日
A = [1234, 567; 1, 234];
B = A(:, 1) .* nextpow10(A(:, 2)) + A(:, 2);
function Y = nextpow10(X)
Y = 10 .^ floor(log10(abs(n)) + 1); % [EDITED]
Please test this before using, I cannot run Matlab currently.
1 件のコメント
Andrei Bobrov
2012 年 9 月 13 日
編集済み: Andrei Bobrov
2012 年 9 月 13 日
A=randi(67,6,4)
n = ceil(log10(A(:,2:end)));
out = sum(A(:,1:end-1).*(10.^fliplr(cumsum(fliplr(n),2))),2)+ A(:,end)
William Sampson
2012 年 9 月 13 日
2 件のコメント
Honglei Chen
2012 年 9 月 13 日
It's the same. I updated my answer to show an example of combining 2nd and 3rd column of a matrix
Jan
2012 年 9 月 13 日
編集済み: Jan
2012 年 9 月 13 日
See my suggestion:
B = rawData(:, 9) .* nextpow10(rawData(:, 10)) + rawData(:, 10)
result = [rawData(:, 1:8), B, rawData(:, 11:end)]
I do not see, why the suggested method should not work.
Btw. What do you expect as result of two 16 digit numbers? You can store only 16 digits in variables of the type DOUBLE.
Ryan
2012 年 9 月 13 日
編集済み: Ryan
2012 年 9 月 13 日
A = [1234, 567; 1, 234]; %Sample Data
% Figure out how many spaces to shift the left column
C = ceil(log10(A(:,2)));
% Find areas where the value is a multiple of 10 or is 1
idx = A(:,2) == 1 | rem(A(:,2),10) == 0;
C(idx) = C(idx) + 1; % Handle the exception
% Determine how much to shift the left column
C = 10.^C;
% Add in the right column
A = A(:,1).*C + A(:,2);
0 件のコメント
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