From your reply, it happens that one SPACE should be present, in the search regex, between the opening round bracket and the block [+-] !

Well, I just realized that your numbers could be separated by more than one space characters ! Moreover, your file may, also, contain some tabulation characters. In that case, prefer the search regex below ( The replacement regex is unchanged ):

(\h+[+-]?\d+(\.(\d+)?)?[Ee][+-]\d\d){3}

Notes :

• The \h regex syntax matches one horizontal blank character, that is to say :

  • The normal space character ( \x20 )

  • The tabulation character ( \x09 )

  • The No-Break space ( abbreviated NBSP ) ( \xA0 )

• As usual, the + quantifier means one or more times the previous character

So given the test example below, where I put, between numbers, several spaces, in the first line, several tabulations in the second line and a mix of these two, in the third line :

    ospht421 4.52E-02     -1.20E-02 2.98E-03          3.19E-02             -1.93E-02 -6.87E-02 4.16E-02 1.71E-02 -2082.95E-02

ospht000	+9.9999998E+03	+1.23e-07	-5.4625e+04	-6.258E-00	5.789E+02	1.008E-06	+9.E+03	+1.E-07	-5.e+04

        ospht999         		-6.E-00 5.E+02			        1.e-06 +9E+03 +1e-07 -5E+04    		    	-6e-00 5E+02 1E-06

After clicking on the Replace All button, we get the following text :

        ospht421
 4.52E-02     -1.20E-02 2.98E-03
          3.19E-02             -1.93E-02 -6.87E-02
 4.16E-02 1.71E-02 -2082.95E-02
    
    ospht000
	+9.9999998E+03	+1.23e-07	-5.4625e+04
	-6.258E-00	5.789E+02	1.008E-06
	+9.E+03	+1.E-07	-5.e+04
    
            ospht999
         		-6.E-00 5.E+02			        1.e-06
 +9E+03 +1e-07 -5E+04
    		    	-6e-00 5E+02 1E-06

Now, we have to get rid of all the possible blanks characters, at the beginning of the lines and just keep one space character between the different numbers. The following S/R will achieve it, nicely :

Find what : ^\h+|(\h+)

Replace with : (?1 ) , with a space after the digit 1

Notes :

• Due to the alternative regex symbol |, this search regex looks :

  • for any non null range of blank characters, that begins the line ( the ^ symbol represents the zero length assertion beginning of line )

  • for any subsequent range of these blank characters, between the numbers

• The second part (\h+), ( the subsequent blanks ), is enclosed in round brackets, and represents the group 1

• In replacement, we use the conditional form (?n....) :

  • which replace the searched regex with the string located after digit n , till the ending round bracket, if the group n exists

  • which does NO replacement if the group n does NOT exist

So :

• When the first part, of the search regex, matches possible blank characters, at beginning of a line, group 1, does not exist. Therefore, there’s NO replacement

• When the second part, of the search regex, matches blank characters, before the numbers, group 1 exists. Then they are replaced with the single space, located between the digit 1 and the ending round bracket, of the replacement regex

This time, after performing this second S/R, we obtain the final text below :

ospht421
4.52E-02 -1.20E-02 2.98E-03
3.19E-02 -1.93E-02 -6.87E-02
4.16E-02 1.71E-02 -2082.95E-02

ospht000
+9.9999998E+03 +1.23e-07 -5.4625e+04
-6.258E-00 5.789E+02 1.008E-06
+9.E+03 +1.E-07 -5.e+04

ospht999
-6.E-00 5.E+02 1.e-06
+9E+03 +1e-07 -5E+04
-6e-00 5E+02 1E-06

Cheers,

guy038

I have never used Notepad++ before, so I am not familiar with how to use it. I have tried your suggestion in the Replace box but it say there are 0 replacements.

Find what: ([±]?\d+(.(\d+)?)?[Ee][±]\d\d){3}
Replace with: \r\n$0
And the Regular Expression box is ticked.

I am sure it is just a silly mistake on my part, but I can’t seem to get it working.

Thanks!
Kim

Once more, this kind of modification of text can be easily achieved with regular expressions !

At first sight, your text seems to contain a list of numbers, in the classical E scientific notation ! Am I right ?

I supposed so. Therefore, I, first, tried to build a regex, in order to match such numbers. Taking in account, all the different syntaxes for these numbers, with an example below :

+9.9999998E+03
+1.23e-07
-5.4625e+04
-6.258E-00
5.789E+02
1.008E-06

+9.E+03
+1.E-07
-5.e+04
-6.E-00
5.E+02
1.e-06

+9E+03
+1e-07
-5E+04
-6e-00
5E+02
1E-06

I got this regex : [+-]?\d+(\.(\d+)?)?[Ee][+-]\d\d

• The part [+-]? matches an optional sign ( remain that the regex quantifier ? means 0 or 1 time )

• The part \d+ matches the mantissa part, before the dot ( The regex quantifier + means 1 or more time(s) )

• The part (\.(\d+)?)? represents the optional fractional part of a number, itself composed of a dot \. ( which must be escaped as it’s a meta character in regexes ) and an optional digits (\d+)?, after the dot

• The part [Ee] matches the E notation letter, whatever the case

• The part [+-] matches the mandatory sign

• The part \d\d represents the two digits of the exponent

Right. Now, from your text, below :

ospht421 4.52E-02 -1.20E-02 2.98E-03 3.19E-02 -1.93E-02 -6.87E-02 4.16E-02 1.71E-02 -2082.95E-02

We notice that any E notation number is preceded by a space. So, as you expect to split on several lines, every 3 numbers, we can build the final search regex ( [+-]?\d+(\.(\d+)?)?[Ee][+-]\d\d){3}, which will matches 3 numbers

• Enclosed in round brackets, it begins with a space and ends with the {3} quantifier

• The replacement part \r\n$0 will replace this range of numbers by an Windows End of Line characters \r\n, followed by the entire regex matched $0

• Don’t forget, in the Replace dialog, to set the Regular expression search mode !

So , from the test example, below

ospht421 4.52E-02 -1.20E-02 2.98E-03 3.19E-02 -1.93E-02 -6.87E-02 4.16E-02 1.71E-02 -2082.95E-02

ospht000 +9.9999998E+03 +1.23e-07 -5.4625e+04 -6.258E-00 5.789E+02 1.008E-06 +9.E+03 +1.E-07 -5.e+04

ospht999 -6.E-00 5.E+02 1.e-06 +9E+03 +1e-07 -5E+04 -6e-00 5E+02 1E-06

After clicking on the Replace All button, we get the new text :

ospht421
 4.52E-02 -1.20E-02 2.98E-03
 3.19E-02 -1.93E-02 -6.87E-02
 4.16E-02 1.71E-02 -2082.95E-02

ospht000
 +9.9999998E+03 +1.23e-07 -5.4625e+04
 -6.258E-00 5.789E+02 1.008E-06
 +9.E+03 +1.E-07 -5.e+04

ospht999
 -6.E-00 5.E+02 1.e-06
 +9E+03 +1e-07 -5E+04
 -6e-00 5E+02 1E-06

Possibly, to get rid of the leading space(s), just use the simple search regex ^ + and leave the replacement zone empty

Et voilà !

Best Regards,

guy038