reshaping a vector of dates (2)

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]

[Jan]

@antonet: Please learn how to format your code properly. The more readable the question is, the easier is it to answer. This is your 50th question and I asked you repeatedly to care for a proper formatting.

[antonet]

thanks simon. Sorry for this.

Answer 1

[Andrei Bobrov]

g = regexp(AAA,'(^\w)|(-\w)|( \d*)','match');
AAA = strrep(cellstr(cell2mat(cat(1,g{:}))),'-','');

EDIT

[Jan]

I assume "f1" is a typo and "f" is meant in the 2nd line.

The first line requires about 25 seconds on my Win7/Matlab 2009a/64 machine for a {1572864 x 1} cell string, while the 2nd line exhausted my small 2GB memory and made my machine unresponsive. Now, after a restart of the computer, I'm convinced, that the STRREP method is more efficient, especially for large inputs.

[antonet]

Hi Andrei. I get the following error mesage

??? Index exceeds matrix dimensions.

Error in ==> @(x)strcat(f1{x,1:2},',',f1{x,3:4})

thanks

[Andrei Bobrov]

Thank you Jan for your comment.

Answer 2

[Jan]

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.

[antonet]

thanks simon. The only problem is that AAA is a huge vector and it will be very time consuming to construct the "Replace" vector which contains the abbreviations. What I am looking for is something like

http://www.mathworks.de/matlabcentral/answers/43148-reshaping-2-vectors-of-dates

thanks again

[Jan]

Your original question looks, like you want to replace 6 strings only. Then the construction of the "Replace" cell will need micro-seconds only.

"Large" is relative. For e.g. a {1'572'864 x 1} cell string, the STRREP method is still faster than a REGEXP, because the later is much more powerful.

Please explain how large the actual data are and how many strings you want to replace. If the posted code is too slow for you, please measure the times using TIC/TOC and post the results.

[antonet]

thanks simon for your reply. The above AAA is a sample. My original vector is of size 300 by 1. So it will take a lot of time to construct the "Replace" cell. this is the only problem that I have and to be honest it is ok for me if the code is slow as long as I do not need to do something extra.

Thanks

[Jan]

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.