is it possible to expand [0 {9}, 45 {2}, 0 {9}]

I have a matrix like [0 {9}, 45{2}, 0{9}] i want to expand it like

 [0 0 0 0 0 0 0 0 0 45 45 0 0 0 0 0 0 0 0 0]? and it works like loop, [ 0{k}, 45{k}, 0{k}]

please help.

[Image Analyst]

What is [0 {9}, 45{2}, 0{9}]? Is it a cell array? Please give code for us to get [0 {9}, 45{2}, 0{9}] so we can then expand it.

[Triveni]

Actually, I have to solve large number of data i am irritate to copy and paste it from lyx.

when i copy and paste this gives me [-45_{2}, 120, 90, 60, 120, 90, -45, 90, -45, 60, 90_{2}, 45, 120, 90, 120, 60, 30, 45, 120, 45, 90, 60, 30, 45, 30_{3}, 120, 60_{2}, 30] this type of format.

Answer 1

[Image Analyst]

If you're stuck starting with a string in that special format, then this code will parse it and produce the output vector

str = '[-45_{2}, 120, 90, 60, 120, 90, -45, 90, -45, 60, 90_{2}, 45, 120, 90, 120, 60, 30, 45, 120, 45, 90, 60, 30, 45, 30_{3}, 120, 60_{2}, 30]'
% Get rid of brackets
str(str== '[') = [];
str(str== ']') = [];
words = strsplit(str, ',')
output = [];
for k = 1 : length(words)
  % See if there is a brace
  thisWord = words{k}
  braceLocation = strfind(thisWord, '{');
  if isempty(braceLocation)
    % There no brace there.
    theNumber = str2double(thisWord);
    theRepeat = 1;
  else
    % There is a brace there.
    theNumber = str2double(thisWord(1:braceLocation-2));
    theRepeat = str2double(thisWord(braceLocation+1:end-1));
  end
  fprintf('Repeat %d for %d times.\n', theNumber, theRepeat);
  % Repeat this number the required number of times and append to the output.
  output = [output, theNumber * ones(1, theRepeat)];
end
% Echo to command window:
output

[Triveni]

 str = '[-45_{2} 0, -45_{2}, 0_{7}, 45, 0_{4}, 45_{3}]'
% Get rid of brackets
str(str== '[') = [];
str(str== ']') = [];
words = strsplit(str, ',')
output = [];
for k = 1 : length(words)
  % See if there is a brace
  thisWord = words{k}
  braceLocation = strfind(thisWord, '{');
  if isempty(braceLocation)
    % There no brace there.
    theNumber = str2double(thisWord);
    theRepeat = 1;
  else
    % There is a brace there.
    theNumber = str2double(thisWord(1:braceLocation-2));
    theRepeat = str2double(thisWord(braceLocation+1:end-1));
  end
  fprintf('Repeat %d for %d times.\n', theNumber, theRepeat);
  % Repeat this number the required number of times and append to the output.
  output = [output, theNumber * ones(1, theRepeat)];
end
% Echo to command window:
output
%Not working at all. By the way thank you so much.

[Triveni]

Thank you so much "Image Analyst " It's works very fine, "," comma was missed in the data. Can you fix this bug.

[Image Analyst]

Just put in the comma:

str = '[-45_{2}, 0, -45_{2}, 0_{7}, 45, 0_{4}, 45_{3}]'

Or you can use strrep to replace "brace space" with "brace comma":

str = strrep(str, '} ', '},');

before you call strsplit().

[Triveni]

str = '[45 0, 45, 0_{5}]';

 str = strrep(str, '} ', '},');
 % Get rid of brackets
 str(str== '[') = [];
 str(str== ']') = [];
 words = strsplit(str, ',');
 output = [];
 for k = 1 : length(words)
  % See if there is a brace
  thisWord = words{k};
  braceLocation = strfind(thisWord, '{');
  if isempty(braceLocation)
    % There no brace there.
    theNumber = str2double(thisWord);
    theRepeat = 1;
  else
    % There is a brace there.
    theNumber =    str2double(thisWord(1:braceLocation-2));
    theRepeat =  str2double(thisWord(braceLocation+1:end-1));
  end
 %   fprintf('Repeat %d for %d times.\n', theNumber, theRepeat);
  % Repeat this number the required number of times and append to the output.
  output = [output, theNumber * ones(1, theRepeat)];
  end
 output
 output =

   NaN    45     0     0     0     0     0

[Image Analyst]

You keep changing the format. Why is there no comma after the first 45? Or is it that the first one never has a comma but all the others do? Where are you getting these strings from anyway?

[Triveni]

Thanks, between Some sequences don't have comma. I'll change it.

Answer 2

[Steven Lord]

If you write your problem slightly differently, you can use repelem.

v = repelem([0 45 0], [9 2 9])

[Triveni]

then tell me program to convert "[0 {9}, 45{2}, 0{9}]" to "repelem([0 45 0], [9 2 9])", because my problem is remains same.

[Triveni]

can i print ([0 45 0], [9 2 9]) to "[0 {9}, 45{2}, 0{9}]"

Answer 3

[Sean de Wolski]

This should do it:

str = '[0 {9}, 45{2}, 0{9}]';
expr = '(?<num>\d*)'; % match digits

nums = cellfun(@str2double,regexp(str,expr,'match')); % extract and convert

repelem(nums(1:2:end),nums(2:2:end)) % repelem

[Walter Roberson]

Not every entry has a repeat; also the repeats have an underscore between the number and the {

[Triveni]

Thanks, but unable to count negatives,

str = '[[0_{9}, 45_{2}, -45_{2}, 0_{7}]]';
expr = '(?<num>\d*)'; % match digits

 nums =  cellfun(@str2double,regexp(str,expr,'match')); 
 repelem(nums(1:2:end),nums(2:2:end)) % repelem

Answer 4

[Walter Roberson]

Paste it into a string, and then you could use string processing. For example you could use regexprep with dynamic regular expressions and the ${} output replacement construct.