search for: fvalu

...null, assuming dry_run")); + DEBUG(0,("Output is /dev/null, assuming dry_run\n")); dry_run = True; } tarhandle=-1; --- samba-2.2.7a/source/lib/snprintf.c Wed Jun 5 12:31:29 2002 +++ samba-2.2.7a-fixed/source/lib/snprintf.c Thu Dec 19 15:42:00 2002 @@ -341,6 +341,7 @@ fvalue = va_arg (args, LDOUBLE); else fvalue = va_arg (args, double); + fmtfp (buffer, &currlen, maxlen, fvalue, min, max, flags); break; case 'G': flags |= DP_F_UP; @@ -349,6 +350,7 @@ fvalue = va_arg (args, LDOUBLE); else fvalue = va_arg (args, d...

...initials parameters as indicated in the tutorials : optim(c(40,0.5), fn=LogLiketot) >Error in 1 - ms : 'ms' is missing But ms is 0.5 ... So I've tried this form : optimx(c(30,50),ms=c(0.4,0.5), fn=LogLiketot) with different values for the two parameters : par fvalues method fns grs itns conv KKT1 KKT2 xtimes >2 19.27583, 25.37964 2249.698 BFGS 12 8 NULL 0 TRUE TRUE 57.5 >1 29.6787861, 0.1580298 2248.972 Nelder-Mead 51 NA NULL 0 TRUE TRUE 66.3 The first line is not possible but as I've not constrained the optimiza...

...si(k-1) and also # (k - 1)(n - 1) becomes epsi(k - 1)(n - 1). The new # p-value for the F statistic is found with the # pf() function. For example, if we have # df1 = k - 1 = 3 - 1 = 2 # df2 = (k - 1)(n - 1) = (3 - 1)(10 - 1) = 18 # F = 40.719 # then the adjusted 2-tailed p-value is given by: # Fvalue <- 40.719 Pepsi <- 2 * (1 - pf(Fvalue, df1=epsi*(k-1), df2=epsi*(k-1)*(n-1) ) ) # Huynh-Feldt correction # The Greenhouse-Geisser epsilon tends to underestimate # epsilon when epsilon is greater than 0.70 (Stevens, 1990). # An estimated e=0.96 may be actually 1. Huynh-Feldt # correction is...

...ers with attribute ,,levels''. But unfortunately I am not been able to read the levels information. I am using: SEXP variable, levels; ... variable = VECTOR_ELT( data_frame, j); switch( TYPEOF( variable)) { case INTSXP: if( isFactor( VECTOR_ELT( data_frame, j))) { fvalue = INTEGER( variable)[i]; levels = STR_PTR( GET_LEVELS( variable)); .... but the variable levels is not an array of strings as I expected. I am probably doing something in the wrong way but how would it be correct? Thanks in advance, Ott Toomet -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-....

...d); static void fmtfloat(double value, char type, int ljust, int len, int precision, int pointflag); static void dostr(char *str, int cut); static void dopr_outch(int c); static char *output; static void dopr(char *buffer, const char *format, va_list args) { int ch; long_long value; double fvalue; int longlongflag = 0; int longflag = 0; int pointflag = 0; int maxwidth = 0; char *strvalue; int ljust; int len; int zpad; output = buffer; while ((ch = *format++)) { switch (ch) { case '%': ljust = len = zpad = maxwidth = 0; longflag = longlong...

...----------------- When I ran via OPTIMX, the results are that -------------------------------------------------------------------------------------------- > optimx(par=theta0, fn=obj.fy, method="BFGS", control=list(maxit=10000), > hessian=T) par fvalues method fns grs itns conv KKT1 KKT2 xtimes 1 0.6, 1.6, 0.6, 1.6, 0.7 -0.02764405 BFGS 1 1 NULL 0 TRUE NA 8.71 > -------------------------------------------------------------------------------------------- Whenever I used different initial values, the initial ones are...

...scores (no error) proc mixed data = dataSet; class treat group; model y = treat*time ; random intercept / subject=group(treat); PARMS (2.1) (1.2) / NOITER; contrast 'slopes' treat*time 1 -1 0,treat*time 1 0 -1; ods output contrasts=c; run; data dataSet; set c; alpha=0.05; ncparm=numdf*fvalue; fc=finv(1-alpha,numdf,dendf,0); power=1-probf(fc,numdf,dendf,ncparm); run; proc print;run; cheers, Wilfried Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

...) Analysis of Variance Table Response: A Df Sum Sq Mean Sq F value Pr(>F) factor1 15 85.99 5.73 3.1332 8.702e-05 *** factor1:factor2 42 131.75 3.14 1.7145 0.005654 ** Residuals 308 563.52 1.83 / and recalculate the Fvalue for factor1 as follows: /> ffactor1<-(85.99/15)/(131.75/42) > fpop [1] 1.827491 > 1-pf(ffactor1,15,42) [1] 0.06285258/ My question is: am I allowed to do the same with an Analysis of Deviance table, calculated on a glm model, as done hereafter: /> mod2<-glm(M~factor1/factor...

Hi It is probably simple but how to force in plot command number format: 20000, 30000, 40000 etc. instead of 2e+04, 3e+04 etc. example of command: plot(x=data$along, y=data$rzedna,type="l", xlim=c(0,max(data$along)), ylim=c(200,1200), main="", xlab="length", ylab="height", cex.main=2, cex.lab=1.5, cex.axis=1.2, lwd=2) thanks Jarek

...nt function line by line. However, I could not find anything wrong. When I remove the gradient, I've got ----------------------------------------------------------------- > optimx(par=theta0, fn=obj.fy, method="BFGS") par fvalues method fns grs itns conv KKT1 KKT2 xtimes 1 0.4423958, 0.9665069, 0.7920856, 1.1952092, 0.3083377 -0.01733672 BFGS 35 22 NULL 0 TRUE FALSE 76.02 ----------------------------------------------------------------- where the true theta is (0.5, 1.0, 0.8, 1.2, 0.6). However, I've go...

...test tol = 0.1018794 KKT1 result = TRUE Hessian eigenvalues: [1] 7.138974e+02 9.931285e-04 KKT2 result = TRUE KKT results: gmax= 0.004032794 evratio= 1.391136e-06 KKT1 & 2: TRUE TRUE [1] 7.138974e+02 9.931285e-04 Save results from method BFGS > zz method par fvalues fns grs hes rs conv KKT1 KKT2 2 BFGS 0.2832468, 52.6096161 100.8794 10 1 0 -2 3 TRUE TRUE 1 NM 0.2827563, 54.3551491 100.8779 53 0 0 1 0 TRUE TRUE mtilt xtimes meths 2 NA 0.11 BFGS 1 0.0005130808 0.57 NM There were 50 or more warnings (use warning...

Hi All, I meant to take the min row by row. But the result is apparently not what I want. Changing min to pmin solve the problem. > df=data.frame(X=1:10, Y=1:10) > transform(df, Z=min(X,10-Y)) X Y Z 1 1 1 0 2 2 2 0 3 3 3 0 4 4 4 0 5 5 5 0 6 6 6 0 7 7 7 0 8 8 8 0 9 9 9 0 10 10 10 0 I try to look at the source code to understand what transform() does. I know

...atic void fmtint(char *buffer, size_t *currlen, size_t maxlen, - long value, int base, int min, int max, int flags); + LLONG value, int base, int min, int max, int flags); static void fmtfp(char *buffer, size_t *currlen, size_t maxlen, LDOUBLE fvalue, int min, int max, int flags); static void dopr_outch(char *buffer, size_t *currlen, size_t maxlen, char c); @@ -468,10 +468,10 @@ /* Have to handle DP_F_NUM (ie 0x and 0 alternates) */ static void fmtint(char *buffer, size_t *currlen, size_t maxlen, - long value, int base, i...

Let's say we have the following scheme using C++ and virtual functions: class DSP { public: DSP() {} virtual ~DSP() {} virtual int Compute(int count, float** in, float** out) = 0; }; class CONCRETE_DSP : public DSP { public: CONCRETE_DSP():fValue() {} virtual ~CONCRETE_DSP() {} virtual int Compute(int count, float** in, float** out) { DoSomeProcess(); } }; class SEQ_DDSP : public DSP { private: DSP* fArg1; DSP* fArg2; public: SEQ_DDSP(DSP* a1, DSP* a2):fArg1(a1), fArg2(a2) {} virtual~SEQ_DDSP() {delete fArg1...

Hello! I am trying to create an R optimization routine for a task that's currently being done using Excel (lots of tables, formulas, and Solver). However, otpim seems to be finding a local minimum. Example data, functions, and comparison with the solution found in Excel are below. I am not experienced in optimizations so thanks a lot for your advice! Dimitri ### 2 Inputs:

Hi all, I was intending on trying out gjournal on a new disk i've added in my desktop. I had a look to see what the most recent patch provided by Pawel and found http://people.freebsd.org/~pjd/patches/gjournal6_20061024.patch I created the directories as per Pawel's original post (http://lists.freebsd.org/pipermail/freebsd-fs/2006-June/001962.html) and the patch succeeded with no failed