Calculate scores and probability -- Syntax issue

madhunk · #1 (**permalink**) 08-01-2006

Hi,

I am totally new to C programming on Sun Solaris environment. I am an active member on the UNIX forum and a good shell programmer.

I am trying to achieve some calculations in C programming. I have the pseudo code written down but don't know the syntax. I am reading a couple of books on C programming but still need some help.

Code:

#include <stdlib.h>
#include <stdio.h>

int transRiskLock,
    transRiskAdj,
    empirica;

double transRiskProb,
       transRiskNorm,
       fico;

   //Restrict range of transRiskScore from daily Trans-Union file PRM.JUNIPER.CSREJ.D*

   if (transRiskScore > 900)
      transRiskLock = 900;

   else
   if(transRiskScore < 150)
      transRiskLock = 150;

   else
      transRiskLock = transRiskScore;

   //Force score range between 150 and 900

   if (transRiskScore is missing)
      transRiskLock is missing;  //Handle missing values - missing transRiskScore should result in missing empirica


   for each record in recordset  //Loop through each record

      //Transrisk normalization Phase 1: Map transRiskScore to estimated probability

      if (transRiskLock >= 150 and < 257.14286)

         transRiskAdj = transRiskLock – 150;
         transRiskProb = 0.84415864 - (1.0488722e-005 * (transRiskAdj)2) + (3.2397817e-008 * (transRiskAdj)3);

      else
      if (transRiskLock >= 257.14286 and < 364.28571)

         transRiskAdj = transRiskLock – 257.14286;
         transRiskProb = 0.7636003 – (0.001131842 * transRiskAdj) – (2.0788434e-005 * (transRiskAdj)2) + (9.3295162e-008 * (transRiskAdj)3);

      else
      if (transRiskLock >= 364.28571 and < 471.42857)

         transRiskAdj = transRiskLock – 364.28571;
         transRiskProb = 0.51843775 – (0.0023735353 * transRiskAdj) – (6.0623834e-006 * (transRiskAdj)2) + (4.8571025e-008 * (transRiskAdj)3);

      else
         ;

      //Transrisk normalization Phase 2: Map estimated probability to EMPIRICASCORE

      if (transRiskProb >= 0 and < 0.00015241579)

         transRiskNorm = 820.36625 – (69591.494 * transRiskProb) – (1.312436e+008 * (transRiskProb)2) + (9.7484131e+011 * (transRiskProb)3);

      else
      if (transRiskProb >= 0.00015241579 and < 0.0024386526)

         transRiskNorm = 810.16217 – (41660.32 * (transRiskProb - 0.00015241579)) + (7018993.8 * (transRiskProb - 0.00015241579)2) +
         (1.8073426e+008 * (transRiskProb - 0.00015241579)3);

      else
      if (transRiskProb >= 0.0024386526) and <  0.012345679)

         transRiskNorm = 753.764 – (6732.1274 * (transRiskProb - 0.0024386526)) + (638606.42 * (transRiskProb - 0.0024386526)2) -
         (28380051 * (transRiskProb - 0.0024386526)3);

       
      else
         ;

   fico = round(transRiskNorm);  //Round to zero decimals
   empirica = (int) fico;  //cast to integer

   score_aligned = compress('+0000' || input(left(EMPIRICA),$8.))  // convert to string by adding "+" and four leading zeros, remove spaces from result

The input value is the transRiskScore here...and translate transRiskScore to Empirica score.

Please advise..
Thank You,
Madhu

madhunk · #2 (**permalink**) 08-01-2006

Any body who is an expert in C...Can you please help with the syntax here? If you can give me a heads up, then I can work the rest of the thing..

blowtorch · #3 (**permalink**) 08-01-2006

madhunk, please do not 'bump' up your posts. That goes against the rules of the forum.

nathan · #4 (**permalink**) 08-01-2006

This is the 1st thing that stands out:

Quote:

for each record in recordset //Loop through each record

C does not have a 'foreach'. You'll have to use a 'for' loop for this. If 'recordset' is defined as an array, it would be something like:

Code:

int i;
for( i = 0; recordset[i] ; i++ ) {
 /* stuff goes here */
}

This:

Quote:

if (transRiskLock >= 150 and < 257.14286)

becomes:

Code:

if (transRiskLock >= 150 && transRiskLock < 257.14286)

and this

Quote:

if (transRiskScore is missing)
transRiskLock is missing;

becomes

Code:

   if (transRiskScore == 0)
      transRiskLock = 0;

madhunk · #5 (**permalink**) 08-03-2006

Thank You Nathan....I appreciate your help. I got the idea and working on it...

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