Batch command to find highest number in set of strings? - batch-file

Question

For instance:
file.txt contains:
4.3 - random1
5.6 - random2
2.2 - random3
3 - random4
1.8 - random5
I need a command that will output the highest number only, not the preceding text.
Ie.
Output = 5.6

Answer 1

You can give SORTN.bat a try.
Here is the code for it as well.
#ECHO OFF
if "%~1"=="/?" (
echo.Sorts text by handling first number in line as number not text
echo.
echo.%~n0 [n]
echo.
echo. n Specifies the character number, n, to
echo. begin each comparison. 3 indicates that
echo. each comparison should begin at the 3rd
echo. character in each line. Lines with fewer
echo. than n characters collate before other lines.
echo. By default comparisons start at the first
echo. character in each line.
echo.
echo.Description:
echo. 'abc10def3' is bigger than 'abc9def4' because
echo. first number in first string is 10
echo. first number in second string is 9
echo. whereas normal text compare returns
echo. 'abc10def3' smaller than 'abc9def4'
echo.
echo.Example:
echo. To sort a directory pipe the output of the dir
echo. command into %~n0 like this:
echo. dir /b^|%~n0
echo.
echo.Source: http://www.dostips.com
goto:EOF
)
if "%~1" NEQ "~" (
for /f "tokens=1,* delims=," %%a in ('"%~f0 ~ %*|sort"') do echo.%%b
goto:EOF
)
SETLOCAL ENABLEDELAYEDEXPANSION
set /a n=%~2+0
for /f "tokens=1,* delims=]" %%A in ('"find /n /v """') do (
set f=,%%B
(
set f0=!f:~0,%n%!
set f0=!f0:~1!
rem call call set f=,%%%%f:*%%f0%%=%%%%
set f=,!f:~%n%!
)
for /f "delims=1234567890" %%b in ("!f!") do (
set f1=%%b
set f1=!f1:~1!
call set f=0%%f:*%%b=%%
)
for /f "delims=abcdefghijklmnopqrstuwwxyzABCDEFGHIJKLMNOPQRSTUWWXYZ~`##$*_-+=:;',.?/\ " %%b in ("!f!") do (
set f2=00000000000000000000%%b
set f2=!f2:~-20!
call set f=%%f:*%%b=%%
)
echo.!f1!!f2!!f!,%%B
rem echo.-!f0!*!f1!*!f2!*!f!*%%a>&2
)
I gave it a try using this input as an example
4.3 - random1
11.3 - random6
5.6 - random2
2.2 - random3
100.1 - random8
3 - random4
1.8 - random5
11.12 - random7
11.11 - random7
This is how I ran it but you should be able to capture the output as well using a FOR /F command just like Stephan showed you in his answer.
type sortme.txt |sortn.bat
Output
1.8 - random5
2.2 - random3
3 - random4
4.3 - random1
5.6 - random2
11.11 - random7
11.12 - random7
11.3 - random6
100.1 - random8

Answer 2

sort will sort in correct order (Attention: this is sorting strings, not numbers, but will work fine for your example. Note that with string comparison 5 or 6.3 are "bigger" than 15).
Put a for around to process the ouput (Standard tokens is 1 and Space is a standard delimiter, so the for /f gets only the first element - your desired number)
for /f %%a in ('sort t.txt') do set high=%%a
echo %high%
EDIT to also process numbers higher than 10. Note: there is no math involved - it's just clever string manipulation.
#echo off
setlocal enabledelayedexpansion
(
for /f "tokens=1,2 delims=. " %%a in (t.txt) do (
set a=0000%%a
if "%%b"=="-" (echo !a:~-4!) else (echo !a:~-4!.%%b)
)
)>temp.txt
type temp.txt
pause
for /f "tokens=1,2 delims=0" %%a in ('sort temp.txt') do set high=%%a
echo %high%

Answer 3

The following script does proper sorting of fractional numbers with eight digits before and after the decimal separator . at most:
#echo off
setlocal EnableExtensions EnableDelayedExpansion
rem // Define constants here:
set "_FILE=%~1" & rem // (input file; `%~1` means first command line argument)
set /A "_ZPAD=8" & rem // (number of zeroes to be used for temporary padding)
set "_ZERO=" & for /L %%K in (0,1,%_ZPAD%) do set "_ZERO=!_ZERO!0"
for /F "usebackq delims=- " %%L in ("%_FILE%") do (
for /F "tokens=1,2 delims=." %%I in ("%%L.") do (
set "INT=%_ZERO%%%I" & set "FRA=%%J%_ZERO%"
set "$NUM_!INT:~-%_ZPAD%!_!FRA:~,%_ZPAD%!=%%L"
)
)
set "MIN=" & set "MAX="
for /F "tokens=2 delims==" %%L in ('2^> nul set $NUM') do (
if not defined MIN set "MIN=%%L"
set "MAX=%%L"
)
echo Minimum: %MIN%
echo Maximum: %MAX%
endlocal
exit /B