Array in loop - Index exceeds the number of array elements (3) - problem

R = string(data{:});
conf_val = [R(1) "2" "3" "4" "5" "6" "7" "8" "9" "10" "11" "12" "13" "14" "15" "16" "17" "18" "19" "20" "21" "22" "23" "24" "25" "26" "27"]; 
for ii=1:27
    if isempty(R(ii))
        x = "nill";
    else
        x = R(ii);
end
   text_str{ii} = ['Confidence: ' convertStringsToChars(x),'%s' '%'];
end

data = textscan(fileID,'%s');
fclose(fileID);
R = vertcat(data{:});
for ii=1:27
  text_str{ii} = R{ii};
end

text string should have 27 elements.

I need a character vector of 27 strings read from a file.

if there are not 27 elements in my file, I would like to display 'nil' for that element at the missing index

Please can someone help

[Saud Alfalasi]

@Image Analyst Please can you help

[Saud Alfalasi]

@Jan Hi please can you help

[Mario Malic]

With this new feature of tagging users in post, I believe some have already turned off such notifications as they are misused as in this question. There are a lot of identical questions, you're not even posting the exact error message.

Most likely, width of R is 3, but you want to access the index that doesn't exist. You can remove semicolon to see the output, or inspect the variable in Workspace.

R = string(data{:});

Please spend few hours in MATLAB Onramp, it helps a lot.

[Saud Alfalasi]

@Mario Malic Thank you for your input. And please don't make asking questions an unpleseant experience - questions should be encouraged, this is how people learn.

[Saud Alfalasi]

@Mario Malic I understand the width of R is 3. However I would like it to be 27 - so if some lines of string from text are missing, It flags up yet my code still runs. The exact error message is in the title

[Jan]

@Saud Alfalasi: "And please don't make asking questions an unpleseant experience - questions should be encouraged, this is how people learn."

I agree with you. When you mention this explicitly I'd like to add: Please keep it pleasent for others to answer questions. Imagine what would happen, if all users catch the attraction of the top ten answerers explicitly for all of their questions. This would be a reason for me to leave the forum to avoid stress.

I answer a question, if I find the time to read it and have an idea about a possible solution. A further pushing decreases your chance to get answers. Therefore Mario's hint is valuable and you can take it as a supoort.

Answer 1

[Walter Roberson]

data = textscan(fileID,'%s');
fclose(fileID);
text_str = vertcat(data{:});
if length(text_str) > 27; text_str = text_str(1:27); end  %in case it was too long
text_str(end+1:27) = {'nil'};   %if it was too short, put in nil

[Saud Alfalasi]

Thank you Walter, I think it would be better if i showed an example of the output I'm after:

say my txt file was

apple
pear
bannana

juice

I would want the new array to be:

apple
pear
bannana 
nil
nil
juice
nil
nil
nil 
ect (until index 27 is reached)

[Walter Roberson]

data = textscan(fileID,'%s');
fclose(fileID);
text_str = data{1};
mask = cellfun(@isempty, text_str);
text_str(mask) = {'nil'}; 

[Jan]

Or because it is faster:

mask = cellfun('isempty', text_str);

[Saud Alfalasi]

Records = 'record.txt';
fileID2 = fopen(Records);
data = textscan(fileID2,'%s');
fclose(fileID2);
text_str = data{1:4};
mask = cellfun(@isempty, text_str);
text_str(mask) = {'nil'}; 

Hi guys, array element 4 doesn't exist,

The output I was expecting is

apple
pear
cheese
nil 

However this is the output:

mask =

  3×1 logical array

   0
   0
   0

[Walter Roberson]

You need to switch your method of reading entirely. I suggest that you use readlines() if you have a new enough matlab.