fprintf %E Leading zero

Hi there,

I am saving a text file using the fprintf command. I have specified the following format string: % 13.6E and the following number is printed: 2.166250E+02.

Can anyone suggest an easy change to the format string such that the number is printed with a leading zero and the exponential increases by one? I.E the number becomes 0.2166250E+03.

Many thanks for the help.

Answer 1

[Sven]

Interesting question Peter...

Here's a solution I found, which pretty much involves brute force shifting of characters around the ".", and incrementing characters after the "E"

sprintfPrettyE = @(num)regexprep(sprintf('%13.6E', num), '(\d)\.(\d+)E((+|-)\d+)', '0.$1$2E${sprintf(''%02d'',str2num($3)+1)}')

>> sprintfPrettyE(2.166250E+02)

ans =

   0.2166250E03

This is pretty close to what you're looking for... You'll notice that the +/- sign after the "E" is gone though... It will only show up when it is a "-" sign. Do you need it in for the plus sign too?

If so, you can wrap the whole thing in another regexprep to add a "+" sign:

sprintfPrettyE = @(num)regexprep( regexprep(sprintf('%13.6E', num), '(\d)\.(\d+)E((+|-)\d+)', '0.$1$2E${sprintf(''%02d'',str2num($3)+1)}'), 'E(\d)', 'E+$1')

>> sprintfPrettyE(2.166250E+02)

ans =

    0.2166250E+03

Thanks, Sven.

Answer 2

[Star Strider]

This came up in another question, Normalizing the E descriptor a few months ago. I sketched out a solution for it.

Answer 3

[Matt Fig]

Here is another:

F = @(x)regexprep(sprintf('%13.6E\n',x*10),'(\d)(\.)(.*)','0$2$1$3')

Answer 4

[Dr. Seis]

Here is a potentially lame way of doing it:

temp_num = 216.6250;
temp_str = sprintf('%13.11e',temp_num*10);
temp_str = sprintf('0.%s%s',temp_str(1:end~=2))

temp_str =
0.216625000000e+03