Displaying 2 results from an estimated 2 matches for "error_3".
Did you mean:
error_r
2004 Sep 10
2
An assembly optimization and fix
...ed_asm.nasm 2002-09-17 16:19:08.000000000 +0200
@@ -76,107 +76,73 @@
push edi
sub esp, byte 16
; qword [esp] == temp space for loading FLAC__uint64s to FPU regs
- ; dword [esp] == last_error_0
- ; dword [esp + 4] == last_error_1
- ; dword [esp + 8] == last_error_2
- ; dword [esp + 12] == last_error_3
- ; eax == error
; ebx == &data[i]
; ecx == loop counter (i)
- ; edx == temp
- ; edi == save
; ebp == order
; mm0 == total_error_1:total_error_0
- ; mm1 == total_error_3:total_error_2
- ; mm2 == 0:total_error_4
- ; mm3/4 == 0:unpackarea
- ; mm5 == abs(error_1):abs(error_0)
- ; mm5 ==...
2010 Jul 21
0
Piecewise regression using lme()
...,5)
segment3=-40+0.5*1:15+rnorm(15,0,1)
group=c(rep(1,20),rep(2,30),rep(3,15))
y=c(segment1,segment2,segment3)
Data=data.frame(y,t=1:65,group=as.factor(group))
the model I'd like to fit is:
y_t=
(beta_01+beta_11*t+error_1)*(group==1)+(beta_02+beta_12*t+error_2)*(group==2)+(beta_03+beta_13*t+error_3)*(group==3)
It looks like a mixed effects model were the fixed effect are the piecewise
linear regression parts (beta_0i+beta_1i*t) and the random effects are the
variance components error_i.
The problem is that I can't find the way the write this model correctly
using the lme() function of...