Displaying 2 results from an estimated 2 matches for "_pi".
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2011 Nov 23
1
Measure of separation for survival data
...h there is no R code for this, the author informs me that it should be easy:
1. Calculate the linear predictor xb from the proportional hazards model or other model of interest.
2. Calculate the ranks r1,...,rn of xb. In case of ties, do not average the ranks.
3. Calculate scaled rankits zi = sqrt(_pi/8) * invnormal((ri - 3/8)/(n +1/4)) where invnormal() is the inverse normal distribution function, and where n is the sample size and _pi = 3.141592654 ...
4. In case of ties in the original xb, substitute average zi's over each of the tied sets.
5. Perform regression (e.g. Cox) on z = z1,...,z...
2010 Jun 07
0
No subject
...at *) malloc(sizeof(*x) * n2);
=A0=A0 memcpy(x, buffer, sizeof(*x) * n2);
=A0=A0 for (i=3D0; i < n; ++i) {
=A0=A0=A0=A0=A0 float acc =3D 0;
=A0=A0=A0=A0=A0 for (j=3D0; j < n2; ++j)
=A0=A0=A0=A0=A0=A0=A0=A0 // formula from paper:
=A0=A0=A0=A0=A0=A0=A0=A0 //acc +=3D n/4.0f * x[j] * (float) cos(M_PI / 2 / =
n * (2 * i + 1 + n/2.0)*(2*j+1));
=A0=A0=A0=A0=A0=A0=A0=A0 // formula from wikipedia
=A0=A0=A0=A0=A0=A0=A0=A0 //acc +=3D 2.0f / n2 * x[j] * (float) cos(M_PI/n2 =
* (i + 0.5 + n2/2)*(j + 0.5));
=A0=A0=A0=A0=A0=A0=A0=A0 // these are equivalent, except the formula from t=
he paper inverts the...