Numerical Sorting (Difficulty: Difficult, Problem Solved) - China DOS Union Forum

If there is a column of values such as: 56 12 33 48 99 105 5 16
How to output it as: 5 12 16 33 48 56 99 105 through batch processing, and no temporary files are to be generated.
----------------------------------------------------------------------------------------------------------
Overall, my general code is as follows:


@echo off&setlocal enabledelayedexpansion
set "str=56 12 33 48 99 105 5 16"
for %%i in (%str%) do call,set var=!var! n%%in
for %%i in (%var%) do set a=%%i&call :lp
echo !var:n=!
pause>nul&goto :eof
:lp
for %%i in (%var%) do (
    set b=%%i&set b=!b:~1,-1!&set c=%a:~1,-1%
    if !b! lss !c! (
       set var=!var:%a% =!
       set var=!var:%%i=%%i %a%!
       ) else (
       set var=!var!
   )
)

Further expansion:


@echo off&setlocal enabledelayedexpansion
set "str=%random% %random% -%random% %random% %random% %random% -%random%"
for %%i in (%str%) do call,set var=!var! n%%in
for %%i in (%var%) do set a=%%i&call :lp
echo !var:n=!
pause>nul&goto :eof
:lp
for %%i in (%var%) do (
    set b=%%i&set b=!b:~1,-1!&set c=%a:~1,-1%
    if !b! lss !c! (
       set var=!var:%a% =!
       set var=!var:%%i=%%i %a%!
       ) else (
       set var=!var!
   )   
)

[ Last edited by zw19750516 on 2008-5-4 at 09:20 PM ]

Arrange them in order from small to large.

If it's purely P, I usually first compare the lengths, and if the lengths are the same, then compare the values.

Linux

$ cat TEST
56
12
33
48
99
105
5
16

$ sort -n <TEST
5
12
16
33
48
56
99
105

The moderator also comes to join in the fun...

cmd is not as powerful as shell.

There is still no perfect answer so far!

@echo off&setlocal enabledelayedexpansion
set/p str=Enter a number (separated by spaces)
for %%i in (%str%) do (
set str=0000000000%%i
set .!str:~-10! !random!=a
)
for /f "delims=.= " %%i in ('set .') do for /f "tokens=* delims=0" %%i in ("%%i") do if "%%i"=="" (set/p=0 <nul) else set/p=%%i <nul
pause>nul

```
@echo off&if not %1.==. goto%*
::plp@cn-dos.net|08-4-29
::Number sorting. Number range: (0-10^99) Numbers are saved in dat.txt, one number per line
if not exist dat.txt echo Missing dat.txt&pause&exit
for /f "delims= " %%a in ('"%0" :sor^|sort')do echo %%a
pause
:sor
setlocal enabledelayedexpansion
for /f %%a in (dat.txt)do (set d=%%a
call:byte&call:space
echo\!sp!!d!
)
exit/b
:byte
for /l %%z in (1 1 100)do if "!d:~%%z,1!"=="" set/a byte=100-%%z&goto:eof
:space
set "sp="
for /l %%z in (1 1 %byte%)do set "sp=!sp! "
goto:eof

Numbers are not one per line. First use this code to change it to one per line


@echo off
for /f "delims=" %%a in (test.txt)do call:dd %%a
exit
:dd
if %1.=. goto:eof
>>dat.txt echo %1
shift
goto:dd

[ Last edited by plp626 on 2008-4-30 at 01:41 PM ]
```

I wonder if brother s11ss's sorting algorithm can meet the building owner's needs

''''''''      Sorting      ''''''''
''''''''s11ss 2007-10-18''''''''

option explicit
'receive input:
dim s,r,n,i
s=inputbox(vbcrlf&vbcrlf&"Separate with spaces:","Please enter a group of numbers:","2007 10 18 21 15")
if s="" then wscript.quit
r=split(s," ")
n=ubound(r)
        'Convert string to Double subtype:
for i=0 to n
        r(i)=cdbl(r(i))
next

'Call quick sort method:
quicksort r,0,n
'Call other sorting methods:
'insertsort r
'shellsort r
'bubblesort r
'selectsort r
'heapsort r

'Output result:
inputbox vbcrlf&vbcrlf&"In ascending order is:","Result",join(r," ")

'Customize various sorting subroutines:
'Direct insertion sort:
sub insertsort(r)
        dim i,n,t,j
        n=ubound(r)
        for i=1 to n'Insert r(1), r(2),..., r(n) in turn
                t=r(i)
                j=i-1
                do while t<r(j)'Find the insertion position of r(i)
                        r(j+1)=r(j)'Shift numbers larger than r(i) backward
                        j=j-1
                        if j=-1 then exit do
                loop
                r(j+1)=t'Insert r(i)
        next
end sub

'Shell sort:
sub shellsort(r)
                'Set increment sequence:
        dim i,d(),n,t,k,h,j
        n=ubound(r)
        i=0
        redim d(n)
        d(i)=fix(n/2)
        do until d(i)=1
                t=d(i)
                i=i+1
                d(i)=fix(t/2)
        loop
                'Sort:
        k=0
        do
                h=d(k)'Take the increment of this pass
                for i=h to n'r(h) to r(n) insert into the current ordered area
                        t=r(i)'Save the number to be inserted
                        j=i-h
                        do while t<r(j)'Find the correct insertion position
                                r(j+h)=r(j)'Shift backward
                                j=j-h'Get the position of the previous number
                                if j<0 then exit do
                        loop
                        r(j+h)=t'Insert r(i)
                next'Complete this pass of sorting
                k=k+1
        loop while h<>1
end sub

'Bubble sort:
sub bubblesort(r)
        dim i,n,noswap,j,t
        n=ubound(r)
        for i=0 to n-1'do n passes of sorting
                noswap=True'set unswapped flag
                for j=n-1 to i step -1'scan from bottom up
                        if r(j+1)<r(j) then'exchange
                                t=r(j)
                                r(j)=r(j+1)
                                r(j+1)=t
                                noswap=False
                        end if
                next
                if noswap then exit for'terminate the algorithm if no swap occurs in this pass of sorting
        next
end sub

'Quick sort:
        'Partition:
function partition(r,l,h)
        dim i,j,t
        i=l
        j=h
        t=r(i)'Initialize, t is the pivot
        do 
                while r(j)>=t and i<j
                        j=j-1'scan from right to left, find the first number less than t
                wend
                if i<j then 
                        r(i)=r(j)'exchange r(i) and r(j)
                        i=i+1
                end if
                while r(i)<=t and i<j
                        i=i+1'scan from left to right, find the first number greater than t
                wend
                if i<j then 
                        r(j)=r(i)'exchange r(i) and r(j)
                        j=j-1
                end if                
        loop while i<>j
        r(i)=t'the pivot t has been finally positioned
        partition=i
end function
        'Sort:
sub quicksort(r,s1,t1)
        dim i
        if s1<t1 then'no need to sort if there is one number or no number
                i=partition(r,s1,t1)'partition r(s1) to r(t1)
                quicksort r,s1,i-1'recursively process the left interval
                quicksort r,i+1,t1'recursively process the right interval
        end if
end sub

'Direct selection sort:
sub selectsort(r)
        dim i,n,k,j,t
        n=ubound(r)
        for i=0 to n-1'do n passes of sorting
                k=i
                for j=i+1 to n'select the smallest number r(k) in the current unsorted area
                        if r(j)<r(k) then k=j
                next
                if k<>i then
                        t=r(i)
                        r(i)=r(k)
                        r(k)=t
                end if
        next
end sub

'Heap sort:
        'Sift:
sub sift(r,i,m)'the complete binary tree with r(i) as the root forms a heap
        dim t,j
        t=r(i)
        j=2*i
        do while j<=m'j<=m, r(2*i) is the left child of r(i)
                if j<m then
                        if r(j)<r(j+1) then j=j+1'j points to the right child of r(i)
                end if
                if t<r(j) then'child node has a larger number
                        r(i)=r(j)'move r(j) to the parent position
                        i=j'revise the current adjusted node
                        j=2*i
                else
                        exit do'adjustment completed, exit loop
                end if
        loop
        r(i)=t'the initially adjusted node is placed in the correct position
end sub
        'Sort:
sub heapsort(r)
        dim i,n,t
        n=ubound(r)
        for i=fix(n/2) to 0 step -1'build initial heap
                sift r,i,n
        next
        for i=n to 0 step -1'do n+1 passes of sorting
                t=r(0)'swap the current heap top number with the last number
                r(0)=r(i)
                r(i)=t
                sift r,0,i-1'rebuild the heap from r(0) to r(i-1)
        next
end sub

Depressed, the single digits cannot have duplicates:

@echo off&setlocal enabledelayedexpansion
set "str=101 20 14 13 19 18 16 11 5"
for %%i in (%str%) do set a=%%i&call :lp
echo %str%
pause>nul&goto :eof
:lp
for %%i in (%str%) do (
    if %%i lss %a% (
       set str=!str:%a% =!
       set str=!str:%%i=%%i %a%!
    )
)

Is my post on floor 8 not meeting the requirements?
Change `set/p str=输入数字(用空格隔开)` to `set "str=101 20 14 13 19 18 16 11 5"`
It should be okay, right?

Originally posted by terse at 2008-4-30 00:11:
Is the code on my 8th floor not meeting the requirements?
Change "set/p str=Enter numbers (separated by spaces)" to "set "str=101 20 14 13 19 18 16 11 5"
It should be okay, right?

Terse's code on the 8th floor is really powerful. To be honest, I still don't understand how it works. Brother Terse, can you help explain it?

Originally posted by plp626 at 2008-4-30 12:04 AM:
Why use cmd when you can have the convenience, speed, and efficiency of a Linux shell?
Disclosure: I don't understand Linux.

This doesn't require understanding Linux; you just need to download GNU's sort.exe and call it in a batch script.

Originally posted by terse at 2008-4-29 11:56 PM:
@echo off&setlocal enabledelayedexpansion
set/p str=Enter a number (separated by spaces)
for %%i in (%str%) do (
set str=0000000000%%i
set .!str:~-10! !random!=a
)
for /f "delims=.= " ...

Really powerful! I didn't understand it at first...

@echo off&setlocal enabledelayedexpansion
set/p str=Enter a number (separated by spaces)
for %%i in (%str%) do (
set str=0000000000%%i
set .!str:~-10! !random!=a
)
set .
pause
for /f "delims=.= " %%a in ('set .') do for /f "tokens=* delims=0" %%i in ("%%a") do if "%%i"=="" (set/p=0 <nul) else set/p=%%i <nul
pause>nul

After changing the code like this, I got a bit enlightened.

I used this before, but I forgot where it came from.

@echo off
setlocal enabledelayedexpansion
    set numbers=%random% %random% %random% %random% %random% %random% %random% %random%
    echo The array before sorting is: %numbers%
    call :sort %numbers%
    echo The array after sorting is: %ret:~1%
pause
goto :eof

:sort
    set num=%1
    set str=%*
    for %%i in (%str%) do if !num! gtr %%i set num=%%i
    set ret=%ret% %num%
    if not "!str:%num%=!" == "" call :sort !str:%num%=!
goto :eof

Floor1　bat-zw	Posted 2008-04-29 22:22
金牌会员　Posts 1,276　Credits 3,105
If there is a column of values such as: 56 12 33 48 99 105 5 16 How to output it as: 5 12 16 33 48 56 99 105 through batch processing, and no temporary files are to be generated. ---------------------------------------------------------------------------------------------------------- Overall, my general code is as follows: @echo off&setlocal enabledelayedexpansion set "str=56 12 33 48 99 105 5 16" for %%i in (%str%) do call,set var=!var! n%%in for %%i in (%var%) do set a=%%i&call :lp echo !var:n=! pause>nul&goto :eof :lp for %%i in (%var%) do ( set b=%%i&set b=!b:~1,-1!&set c=%a:~1,-1% if !b! lss !c! ( set var=!var:%a% =! set var=!var:%%i=%%i %a%! ) else ( set var=!var! ) ) Further expansion: @echo off&setlocal enabledelayedexpansion set "str=%random% %random% -%random% %random% %random% %random% -%random%" for %%i in (%str%) do call,set var=!var! n%%in for %%i in (%var%) do set a=%%i&call :lp echo !var:n=! pause>nul&goto :eof :lp for %%i in (%var%) do ( set b=%%i&set b=!b:~1,-1!&set c=%a:~1,-1% if !b! lss !c! ( set var=!var:%a% =! set var=!var:%%i=%%i %a%! ) else ( set var=!var! ) ) [ Last edited by zw19750516 on 2008-5-4 at 09:20 PM ]

Floor2　qingfushuan	Posted 2008-04-29 22:40
高级用户　Posts 327　Credits 502
Arrange them in order from small to large.

Floor3　HAT	Posted 2008-04-29 22:44
版主　Posts 5,017　Credits 9,023
If it's purely P, I usually first compare the lengths, and if the lengths are the same, then compare the values.

Floor4　bjsh	Posted 2008-04-29 23:21
银牌会员　Posts 621　Credits 2,000
Linux $ cat TEST 56 12 33 48 99 105 5 16 $ sort -n <TEST 5 12 16 33 48 56 99 105

Floor5　plp626	Posted 2008-04-29 23:34
银牌会员　Posts 1,020　Credits 2,278
The moderator also comes to join in the fun... cmd is not as powerful as shell.

Floor6　bat-zw	Posted 2008-04-29 23:37
金牌会员　Posts 1,276　Credits 3,105
There is still no perfect answer so far!

Floor7　terse	Posted 2008-04-29 23:56
银牌会员　Posts 946　Credits 2,404
@echo off&setlocal enabledelayedexpansion set/p str=Enter a number (separated by spaces) for %%i in (%str%) do ( set str=0000000000%%i set .!str:~-10! !random!=a ) for /f "delims=.= " %%i in ('set .') do for /f "tokens=* delims=0" %%i in ("%%i") do if "%%i"=="" (set/p=0 <nul) else set/p=%%i <nul pause>nul

Floor8　plp626	Posted 2008-04-29 23:59
银牌会员　Posts 1,020　Credits 2,278
``` @echo off&if not %1.==. goto%* ::plp@cn-dos.net\|08-4-29 ::Number sorting. Number range: (0-10^99) Numbers are saved in dat.txt, one number per line if not exist dat.txt echo Missing dat.txt&pause&exit for /f "delims= " %%a in ('"%0" :sor^\|sort')do echo %%a pause :sor setlocal enabledelayedexpansion for /f %%a in (dat.txt)do (set d=%%a call:byte&call:space echo\!sp!!d! ) exit/b :byte for /l %%z in (1 1 100)do if "!d:~%%z,1!"=="" set/a byte=100-%%z&goto:eof :space set "sp=" for /l %%z in (1 1 %byte%)do set "sp=!sp! " goto:eof Numbers are not one per line. First use this code to change it to one per line @echo off for /f "delims=" %%a in (test.txt)do call:dd %%a exit :dd if %1.=. goto:eof >>dat.txt echo %1 shift goto:dd [ Last edited by plp626 on 2008-4-30 at 01:41 PM ] ```

Floor9　HAT	Posted 2008-04-30 00:01
版主　Posts 5,017　Credits 9,023
I wonder if brother s11ss's sorting algorithm can meet the building owner's needs '''''''' Sorting '''''''' ''''''''s11ss 2007-10-18'''''''' option explicit 'receive input: dim s,r,n,i s=inputbox(vbcrlf&vbcrlf&"Separate with spaces:","Please enter a group of numbers:","2007 10 18 21 15") if s="" then wscript.quit r=split(s," ") n=ubound(r) 'Convert string to Double subtype: for i=0 to n r(i)=cdbl(r(i)) next 'Call quick sort method: quicksort r,0,n 'Call other sorting methods: 'insertsort r 'shellsort r 'bubblesort r 'selectsort r 'heapsort r 'Output result: inputbox vbcrlf&vbcrlf&"In ascending order is:","Result",join(r," ") 'Customize various sorting subroutines: 'Direct insertion sort: sub insertsort(r) dim i,n,t,j n=ubound(r) for i=1 to n'Insert r(1), r(2),..., r(n) in turn t=r(i) j=i-1 do while t<r(j)'Find the insertion position of r(i) r(j+1)=r(j)'Shift numbers larger than r(i) backward j=j-1 if j=-1 then exit do loop r(j+1)=t'Insert r(i) next end sub 'Shell sort: sub shellsort(r) 'Set increment sequence: dim i,d(),n,t,k,h,j n=ubound(r) i=0 redim d(n) d(i)=fix(n/2) do until d(i)=1 t=d(i) i=i+1 d(i)=fix(t/2) loop 'Sort: k=0 do h=d(k)'Take the increment of this pass for i=h to n'r(h) to r(n) insert into the current ordered area t=r(i)'Save the number to be inserted j=i-h do while t<r(j)'Find the correct insertion position r(j+h)=r(j)'Shift backward j=j-h'Get the position of the previous number if j<0 then exit do loop r(j+h)=t'Insert r(i) next'Complete this pass of sorting k=k+1 loop while h<>1 end sub 'Bubble sort: sub bubblesort(r) dim i,n,noswap,j,t n=ubound(r) for i=0 to n-1'do n passes of sorting noswap=True'set unswapped flag for j=n-1 to i step -1'scan from bottom up if r(j+1)<r(j) then'exchange t=r(j) r(j)=r(j+1) r(j+1)=t noswap=False end if next if noswap then exit for'terminate the algorithm if no swap occurs in this pass of sorting next end sub 'Quick sort: 'Partition: function partition(r,l,h) dim i,j,t i=l j=h t=r(i)'Initialize, t is the pivot do while r(j)>=t and i<j j=j-1'scan from right to left, find the first number less than t wend if i<j then r(i)=r(j)'exchange r(i) and r(j) i=i+1 end if while r(i)<=t and i<j i=i+1'scan from left to right, find the first number greater than t wend if i<j then r(j)=r(i)'exchange r(i) and r(j) j=j-1 end if loop while i<>j r(i)=t'the pivot t has been finally positioned partition=i end function 'Sort: sub quicksort(r,s1,t1) dim i if s1<t1 then'no need to sort if there is one number or no number i=partition(r,s1,t1)'partition r(s1) to r(t1) quicksort r,s1,i-1'recursively process the left interval quicksort r,i+1,t1'recursively process the right interval end if end sub 'Direct selection sort: sub selectsort(r) dim i,n,k,j,t n=ubound(r) for i=0 to n-1'do n passes of sorting k=i for j=i+1 to n'select the smallest number r(k) in the current unsorted area if r(j)<r(k) then k=j next if k<>i then t=r(i) r(i)=r(k) r(k)=t end if next end sub 'Heap sort: 'Sift: sub sift(r,i,m)'the complete binary tree with r(i) as the root forms a heap dim t,j t=r(i) j=2i do while j<=m'j<=m, r(2i) is the left child of r(i) if j<m then if r(j)<r(j+1) then j=j+1'j points to the right child of r(i) end if if t<r(j) then'child node has a larger number r(i)=r(j)'move r(j) to the parent position i=j'revise the current adjusted node j=2*i else exit do'adjustment completed, exit loop end if loop r(i)=t'the initially adjusted node is placed in the correct position end sub 'Sort: sub heapsort(r) dim i,n,t n=ubound(r) for i=fix(n/2) to 0 step -1'build initial heap sift r,i,n next for i=n to 0 step -1'do n+1 passes of sorting t=r(0)'swap the current heap top number with the last number r(0)=r(i) r(i)=t sift r,0,i-1'rebuild the heap from r(0) to r(i-1) next end sub

Floor10　bat-zw	Posted 2008-04-30 00:04
金牌会员　Posts 1,276　Credits 3,105
Depressed, the single digits cannot have duplicates: @echo off&setlocal enabledelayedexpansion set "str=101 20 14 13 19 18 16 11 5" for %%i in (%str%) do set a=%%i&call :lp echo %str% pause>nul&goto :eof :lp for %%i in (%str%) do ( if %%i lss %a% ( set str=!str:%a% =! set str=!str:%%i=%%i %a%! ) )

Floor11　terse	Posted 2008-04-30 00:11
银牌会员　Posts 946　Credits 2,404
Is my post on floor 8 not meeting the requirements? Change `set/p str=输入数字(用空格隔开)` to `set "str=101 20 14 13 19 18 16 11 5"` It should be okay, right?

Floor12　bat-zw	Posted 2008-04-30 00:18
金牌会员　Posts 1,276　Credits 3,105
Originally posted by terse at 2008-4-30 00:11: Is the code on my 8th floor not meeting the requirements? Change "set/p str=Enter numbers (separated by spaces)" to "set "str=101 20 14 13 19 18 16 11 5" It should be okay, right? Terse's code on the 8th floor is really powerful. To be honest, I still don't understand how it works. Brother Terse, can you help explain it?

Floor13　HAT	Posted 2008-04-30 00:32
版主　Posts 5,017　Credits 9,023
Originally posted by plp626 at 2008-4-30 12:04 AM: Why use cmd when you can have the convenience, speed, and efficiency of a Linux shell? Disclosure: I don't understand Linux. This doesn't require understanding Linux; you just need to download GNU's sort.exe and call it in a batch script.

Floor14　pusofalse	Posted 2008-04-30 00:39
银牌会员　Posts 646　Credits 1,604
Originally posted by terse at 2008-4-29 11:56 PM: @echo off&setlocal enabledelayedexpansion set/p str=Enter a number (separated by spaces) for %%i in (%str%) do ( set str=0000000000%%i set .!str:~-10! !random!=a ) for /f "delims=.= " ... Really powerful! I didn't understand it at first... @echo off&setlocal enabledelayedexpansion set/p str=Enter a number (separated by spaces) for %%i in (%str%) do ( set str=0000000000%%i set .!str:~-10! !random!=a ) set . pause for /f "delims=.= " %%a in ('set .') do for /f "tokens=* delims=0" %%i in ("%%a") do if "%%i"=="" (set/p=0 <nul) else set/p=%%i <nul pause>nul After changing the code like this, I got a bit enlightened.

Floor15　zh159	Posted 2008-04-30 01:51
金牌会员　Posts 1,467　Credits 3,687
I used this before, but I forgot where it came from. @echo off setlocal enabledelayedexpansion set numbers=%random% %random% %random% %random% %random% %random% %random% %random% echo The array before sorting is: %numbers% call :sort %numbers% echo The array after sorting is: %ret:~1% pause goto :eof :sort set num=%1 set str=%* for %%i in (%str%) do if !num! gtr %%i set num=%%i set ret=%ret% %num% if not "!str:%num%=!" == "" call :sort !str:%num%=! goto :eof