How can I add leading zeros in a column vector?

assuming X is my vertical vector of numbers, I've tried

X_with_zeros=sprintf('%04d\n',X)

It prints out what a want but saves it all combined in a 1x1 char array

I tried this loop:

j=length(length(X));
for i=1:j
    X_with_zeros=zeros(j,1);
    X_with_zeros(i,1)=X(sprintf('%04d',X(i)),1);
end

It's telling me that my assignment doesn't match the dimensions of the subscript, I've tried different ways. I realize I probably need different rules for a list of strings or char array.

[Azzi Abdelmalek]

Your question is confusing, you start with leading zeros and end with date problem. Can you make your question clear. Post an example and explain what you want.

[James Condie]

thanks, i limited this to one question. Here's an example: I have a list of hours and minutes in a column vector like this

0
1
2
3
...
59
100
101
..
2359
0

etc. Id like to add padding so all entries have four digits padded with leading zeros so it looks like this

0000
0001
...
0059
0100
..
etc.

Answer 1

[Azzi Abdelmalek]

v=[0 1 2 3 59 100 101]'
w=arrayfun(@(x) sprintf('%04d',x),v,'un',0)

[Daniel Pare]

Thanks, It works great and fast on table of more than 200k records.

Answer 2

[Andrei Bobrov]

sprintf(['%0',sprintf('%d',max(ceil(log10(X)))),'d\n'],X)

[James Condie]

this prints out what I want, but when I assign this to variable it still saves it as a big character in a 1x1 variable, maybe I need to assign it to something different?