dimension mismatch with growing char matrix

Hello,

I have in a loop Name_Pairs growing

Name_Pairs(1,:) = sprintf('%s%d','p',10)

then

Name_Pairs(2,:) = sprintf('%s%d','p',1)
Subscripted assignment dimension mismatch.

Name_Pairs is first a 1x3 char, then I try to put in a 1x2 char.

Any ideas on how could this be arranged? thank you

Answer 1

[Walter Roberson]

Example:

Name_Paris(1:10, 1:3) = ' ';   %initialize matrix to blanks

for k = 1 : 10
  s = sprintf('%s%d','p',k);
  Name_Pairs(1,1:length(s)) = s;
end

This would end up with ['p1 '; 'p2 '; 'p3 '; 'p4 '; 'p5 '; 'p6 '; 'p7 '; 'p8 '; 'p9 '; 'p10']

If you have R2016b or later you could instead do

Name_Pairs = strings(10, 1);
for k = 1 : 10
  Name_Pairs(k) = sprintf('%s%d','p',k);
end

Or more compactly,

Name_Pairs = string('p') + (1:10).' ;

[Nicolas]

I have 2016a. Might need to update it !

Answer 2

[dpb]

>> num2str([10 1].','P%03d')
ans =
P010
P001
>>

numstr is vectorized.

To do with individual conversion again use a fixed-width field...

sprintf('P%03d',n)

on the RHS of the assignment. Or, use char to automagically pad the array...

 >> Name= sprintf('%s%d','p',10)
Name =
p10
>> Name=char(Name, sprintf('%s%d','p',1))
Name =
p10
p1 
>>

This has the effect of the names being truncated with no leading 0 in the field which has the side effect that wouldn't sort numerically as will the previous solution if that were to be wanted later for any reason, but each string is padded with blanks to the length of the longest which may also be or not be an issue in use.

The other way is to use cellstr which won't pad and has the other nicety that each string is an entity instead of an array so addressing is simpler...

>> clear Name
>> Name= {sprintf('%s%d','p',10)}
Name = 
  'p10'
>> Name=[Name; {sprintf('%s%d','p',1)}]
Name = 
  'p10'
  'p1'
>>

In the latter note the {} to convert to cell string and the '' around each in the display; that's the visual klew that what you have is a cellstr array, not a char array.