reshaping a vector of dates (2)
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Dear all, I have a cell column vector that contains the following elements
AAA={'AUGUST-SEPTEMBER 1999' ...
'OCTOBER-NOVEMBER 1999'...
'DECEMBER-JANUARY 2000'...
'FEBRUARY-MARCH 2000'...
'APRIL-MAY 2000'...
'JUNE-JULY 2000'}
I want to change that vector so as to have
AAA={'AS 1999'...
'ON 1999'...
'DJ 2000'...
'FM 2000'...
'AM 2000'...
'JJ 2000'}
Where AS for example stands for 'AUGUST-SEPTEMBER’
thanks
[EDITED, code formatted, Jan]
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채택된 답변
Andrei Bobrov
2012년 7월 20일
편집: Andrei Bobrov
2012년 7월 20일
g = regexp(AAA,'(^\w)|(-\w)|( \d*)','match');
AAA = strrep(cellstr(cell2mat(cat(1,g{:}))),'-','');
EDIT
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추가 답변 (1개)
Jan
2012년 7월 20일
편집: Jan
2012년 7월 20일
AAA={'AUGUST-SEPTEMBER 1999'; ...
'OCTOBER-NOVEMBER 1999'; ...
'DECEMBER-JANUARY 2000'; ...
'FEBRUARY-MARCH 2000'; ...
'APRIL-MAY 2000'; ...
'JUNE-JULY 2000'};
Replace = {'AUGUST-SEPTEMBER', 'AS'; ...
'OCTOBER-NOVEMBER', 'ON'; ...
'DECEMBER-JANUARY', 'DJ';, ...
'FEBRUARY-MARCH', 'FM'; ...
'APRIL-MAY', 'AM'; ...
'JUNE-JULY', 'JJ'};
for i = 1:size(Replace, 1)
AAA = strrep(AAA, Replace{i, 1}, Replace{i, 2});
end
[EDITED] faster:
...
for i = 1:size(Replace, 1)
key = Replace{i, 1};
match = strncmp(AAA, key, length(key));
AAA(match) = strrep(AAA(match), key, Replace{i, 2});
end
While the 1st method needs 3.93 seconds if AAA is a {1572864 x 1} cell string, the smarter 2nd methods needs 1.42 seconds.
댓글 수: 4
Jan
2012년 7월 20일
편집: Jan
2012년 7월 20일
I still do not understand, why the construction of "Replace" should take a lot of time and why this is influenced by the size of AAA. A vector of length 300 cannot be called "huge".
It seems to me like you did not post the complete problem. Unfortunately your decision to omit obviously necessary details leads to the fact, that the creation of my answer has wasted my time - and your time also. Please post all relevant details in future questions, especially the sizes of the real data.
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