Ramer-Douglas-Peucker line simplification: Difference between revisions
Ramer-Douglas-Peucker line simplification (view source)
Revision as of 19:18, 15 January 2024
, 4 months ago→{{header|REXX}}: add Version 2
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Walterpachl (talk | contribs) (→{{header|REXX}}: add Version 2) |
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=={{header|REXX}}==
===Version 1===
The computation for the ''perpendicular distance'' was taken from
the '''GO''' example.
Line 1,562 ⟶ 1,563:
end
return @.1 @.#</syntaxhighlight>
{{out|output|text= when using the default inputs:}}
<pre>
error threshold: 1
points specified: (0,0) (1,0.1) (2,-0.1) (3,5) (4,6) (5,7) (6,8.1) (7,9) (8,9) (9,9)
points simplified: (0,0) (2,-0.1) (3,5) (7,9) (9,9)
</pre>
===Version 2===
</syntaxhighlight>
<syntaxhighlight lang="rexx">
/*REXX program uses the Ramer-Douglas-Peucker (RDP) line simplification algorithm for*/
/*--------------------------- reducing the number of points used to define its shape. */
Parse Arg epsilon pl /*obtain optional arguments from the CL*/
If epsilon='' | epsilon=',' then epsilon= 1 /*Not specified? Then use the default.*/
If pl='' Then pl= '(0,0) (1,0.1) (2,-0.1) (3,5) (4,6) (5,7) (6,8.1) (7,9) (8,9) (9,9)'
Say ' error threshold:' epsilon
Say ' points specified:' pl
Say 'points simplified:' dlp(pl)
Exit
dlp: Procedure Expose epsilon
Parse Arg pl
plc=pl
Do i=1 By 1 While plc<>''
Parse Var plc '(' x ',' y ')' plc
p.i=x y
End
end=i-1
dmax=0
index=0
Do i=2 To end-1
d=distpg(p.i,p.1,p.end)
If d>dmax Then Do
index=i
dmax=d
End
End
If dmax>epsilon Then Do
rla=dlp(subword(pl,1,index))
rlb=dlp(subword(pl,index,end))
rl=subword(rla,1,words(rla)-1) rlb
End
Else
rl=word(pl,1) word(pl,end)
Return rl
distpg: Procedure
/**********************************************************************
* compute the distance of point P from the line giveb by A and B
**********************************************************************/
Parse Arg P,A,B
Parse Var P px py
Parse Var A ax ay
Parse Var B bx by
If ax=bx Then
res=px-ax
Else Do
k=(by-ay)/(bx-ax)
d=(ay-ax*k)
res=(py-k*px-d)/sqrt(1+k**2)
End
Return abs(res)
sqrt: Procedure
/* REXX ***************************************************************
* EXEC to calculate the square root of a = 2 with high precision
**********************************************************************/
Parse Arg x,prec
If prec<9 Then prec=9
prec1=2*prec
eps=10**(-prec1)
k = 1
Numeric Digits 3
r0= x
r = 1
Do i=1 By 1 Until r=r0 | (abs(r*r-x)<eps)
r0 = r
r = (r + x/r) / 2
k = min(prec1,2*k)
Numeric Digits (k + 5)
End
Numeric Digits prec
r=r+0
Return r
{{out|output|text= when using the default inputs:}}
<pre>
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