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ommiting blanks in a string

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lowcalorie on 11 Feb 2013
how would i write a code to print each line of the string array G omitting any trailing blanks and blanks in-between letters

Answers (4)

Thorsten on 11 Feb 2013
s = [' hallo a b c'; 'asdsdf v vvvv '];
for i = 1:size(s, 1)
disp(s(i, setdiff(1:length(s(i,:)), findstr(' ', s(i,:)))))

José-Luis on 11 Feb 2013
Edited: José-Luis on 11 Feb 2013
your_array = char(randi([32 127],100,10));
for ii = your_array'
sub_set = ii(ii ~= char(32))'

Image Analyst
Image Analyst on 11 Feb 2013
Edited: Image Analyst on 11 Feb 2013
Use strtrim(). From the help:
S = strtrim(str) returns a copy of string str with all leading and trailing white-space characters removed. A white-space character is one for which the isspace function returns logical 1 (true).
for k = 1:size(yourString, 1)
fprintf('%s', strtrim(yourString(k,:)));

Jan on 11 Feb 2013
This means, that you want to conserve the leading spaces, correctly?
Deleting all spaces would be:
s = strrep(s, ' ', ''); % Works for cell strings also
Trailing spaces:
s = deblank(s);
Conserve leading spaces - assuming that you have a cell string:
for iC = 1:numel(C)
S = C{iC};
first = find(isletter(S), 1);
C{iC} = [S(1:first - 1), strrep(S, ' ', '')];
And an equivalent approach work for a string also.


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