search for: e3

This patch replace list_entry with list_{next/prev}_entry as it makes the code more clear to read. Done using coccinelle: @@ expression e1; identifier e3; type t; @@ ( - list_entry(e1->e3.next,t,e3) + list_next_entry(e1,e3) | - list_entry(e1->e3.prev,t,e3) + list_prev_entry(e1,e3) ) Signed-off-by: Arushi Singhal <arushisinghal19971997 at gmail.com> --- drivers/gpu/drm/drm_lease.c | 2 +- drivers/gpu/drm/nouveau/nvkm/...

....fr> wrote: > > > On Mon, 19 Mar 2018, Arushi Singhal wrote: > > > This patch replace list_entry with list_{next/prev}_entry as it makes > > the code more clear to read. > > Done using coccinelle: > > > > @@ > > expression e1; > > identifier e3; > > type t; > > @@ > > ( > > - list_entry(e1->e3.next,t,e3) > > + list_next_entry(e1,e3) > > | > > - list_entry(e1->e3.prev,t,e3) > > + list_prev_entry(e1,e3) > > ) > > This looks like a rule that could be nice for the Linux kernel...

...es running Samba 4.2 (Debian Jessie) tdbdump gives a working password such as this: ]f2>lOR4NA~hbv\00 where the actual password is ]f2>lOR4NA~hbv On newer machines running Samba 4.5 (Debian Stretch) tdbdump gives an encrypted password such as this: \EE\A9\8D\EF\AD\AC\E2\A1\9D\E2\A0\8C\E3\96\8E\E7\B0\A8\EE\97\AA\E2\8E\9F\E2\A2\8F\EB\85\BF\EE\B7\8B\EA\A7\A9\EA\97\B8\D2\86\E6\83\AB\EE\82\AA\E3\A9\BB\E3\8A\8D\E2\86\9B\E2\8C\92\E6\8C\A6\EA\85\A5\E6\8F\82\EF\96\94\EF\9C\82\E7\8D\B3\E7\8F\93\E7\B8\AA\E7\A7\B7\EE\88\96\E2\A3\9B\EB\AA\B0\E6\B6\A7\EF\B6\B7\EA\A2\AD\EF\A8\88\EA\BB\B6\EE\A4\9A...

...length(sth)], k in sth_else : length(sth_else) ) - I'd like to have two independent conditions in the same loop "for". My secound question depend on program below. I'd like to write every result in matrix but I also want to call my function "odleg" using vector "e3" - exactly using values which are inside. Can anyone please guide me, how to do that? Thanks ## Function odleg <- function (Xa,Xb,Ya,Yb){ d <- ((Xa-Xb)^2+(Ya-Yb)^2)^(1/2) return (d) } # Database ma=matrix(c(0.51139630,-0.12937287, 0.19530080, 0.02273691,-0.43634186,-0.01717149...

Hi R-Users I need to estimate Lyapunov exponent of my time series. After reading description of all functions available I still don't know how to determine time delay. My time series length is 4200. Is it possible to determine time delay with other function's output or I can choose any random value? Thanks for your help! -- View this message in context:

...gives a working password such as this: > > > ]f2>lOR4NA~hbv\00 where the actual password is ]f2>lOR4NA~hbv > > > On newer machines running Samba 4.5 (Debian Stretch) tdbdump gives an encrypted password such as this: > > > \EE\A9\8D\EF\AD\AC\E2\A1\9D\E2\A0\8C\E3\96\8E\E7\B0\A8\EE\97\AA\E2\8E\9F\E2\A2\8F\EB\85\BF\EE\B7\8B\EA\A7\A9\EA\97\B8\D2\86\E6\83\AB\EE\82\AA\E3\A9\BB\E3\8A\8D\E2\86\9B\E2\8C\92\E6\8C\A6\EA\85\A5\E6\8F\82\EF\96\94\EF\9C\82\E7\8D\B3\E7\8F\93\E7\B8\AA\E7\A7\B7\EE\88\96\E2\A3\9B\EB\AA\B0\E6\B6\A7\EF\B6\B7\EA\A2\AD\EF\A8\88\EA\BB\B6\EE\A4\9A...

Dear all: I have this: A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 B3 C3 D3 E3 And I want this A1 E1 B1 E1 C1 E1 D1 E1 A2 E2 B2 E2 C2 E2 D2 E2 A3 E3 B3 E3 C3 E3 D3 E3 Example: m<- matrix(1:15,nrow=3,byrow=T) m v<- unlist(list(t(m[,1:4]))) u<- rep(c(5,10,15),c(4,4,4)) data.frame(v,u) This is the result I want but I would like to learn a simple...

.... If I specify the legend positions in characters like "topright"..etc, it appears, if i specify it in terms of coordinates like "-1, 1" .. etc, it does not appear. Can anyone help me with this? script - x.date <- as.Date(paste(year, month, day, sep="-")) ts1.n.e3 <- ts(data.nemr.e3[,3]) z1.n.e3 <- zoo(ts1.n.e3, x.date) plot(z1.n.e3, ylim = c(min(data.nemr.e3[,3]), max(data.nemr.e3[,5])), col="orange", main = "Monthly variations of SST in El-Nino3", xlab = "Year", ylab="SST (deg C)") lines(z2.n.e3, lty = 2, col=&...

On Mon, 19 Mar 2018, Arushi Singhal wrote: > This patch replace list_entry with list_{next/prev}_entry as it makes > the code more clear to read. > Done using coccinelle: > > @@ > expression e1; > identifier e3; > type t; > @@ > ( > - list_entry(e1->e3.next,t,e3) > + list_next_entry(e1,e3) > | > - list_entry(e1->e3.prev,t,e3) > + list_prev_entry(e1,e3) > ) This looks like a rule that could be nice for the Linux kernel in general, because the code really is much simpler....

...AccountCode: 11 Context: ABC Exten: 3002 Priority: 8 Uniqueid: 1521665055.4920 Linkedid: 1521665055.4920 Result: %F7%EF%F0%F4z%7C%EE%EF%F6%FCkWDDLO%5Cm%EF%7B__%60h%FE%E0%D4%D5%D9%DA%DD%E2%E8%E1%DC%DE%E0%F3%EC%EF%7C%ED%ED%EC%FAx~oov%F9%F2%F6%EB%ED%FA%FD%F9%FEvjXGFMKWg%FF%7Cdidew%E8%D9%D9%D8%D6%DA%DF%E3%E3%E2%E5%E9%EA%E1%E6%EF%E9%F8%F8%F6%FD%F8%7C~%7D%FC%F9%F7%ED%F4%F6%F9oh%60JFMKP%5Do%FFgtjch%7B%E3%DF%DB%D6%D8%DB%E6%EC%E7%E6%E5%DF%DD%E0%EA%F6%ED%EC%F7%E8%E4%EB%F2%F8%F9%7Cu%FC%F8%FAwhXIIIHQ%5Dksw%7Fgfl%7F%EA%E2%DA%D8%DC%DE%DE%E2%E4%DF%E0%E3%E9%E4%E8%EC%E6%E3%E1%E4%E0%E3%EE%F4%FAxy%FCj%5BLIJFLU%5Bd...

Hi all, I have a data.frame e3. I use a function on it: n3 <- normalize.qspline(x = e3, samples = 0.07, na.rm = TRUE); The resulting data.frame has lost it's row-names and column names. This is the natural behaviour of the function I am calling, and I can't alter that at the moment. However, the data *is* in the c...

Hi Please check out this product http://www.sangoma.com/products/hardware_products/data_networking/a301.html Does it work on Asterisk or Freeswitch ? Do Telcos provide an E3 connection ? One of our customers had an inquiry for terminating 6000 calls simultaneously. I want to do some homework before taking it further with him. If I use E1 lines, I will need 6000 / 30 = 200 E1 lines, which does not look feasible ? Thanks for any input you may provide. regards, Anita...

Dear all:   I have this:  A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 B3 C3 D3 E3   And I want this A1 E1 B1 E1 C1 E1 D1 E1 A2 E2 B2 E2 C2 E2 D2 E2 A3 E3 B3 E3 C3 E3 D3 E3   Example:   m<- matrix(1:15,nrow=3,byrow=T) m v<- unlist(list(t(m[,1:4]))) u<- rep(c(5,10,15),...

...ty: 8 > > Uniqueid: 1521665055.4920 > > Linkedid: 1521665055.4920 > > Result: > %F7%EF%F0%F4z%7C%EE%EF%F6%FCkWDDLO%5Cm%EF%7B__%60h%FE%E0%D4%D5%D9%DA%D > D%E2%E8%E1%DC%DE%E0%F3%EC%EF%7C%ED%ED%EC%FAx~oov%F9%F2%F6%EB%ED%FA%FD% > F9%FEvjXGFMKWg%FF%7Cdidew%E8%D9%D9%D8%D6%DA%DF%E3%E3%E2%E5%E9%EA%E1%E6 > %EF%E9%F8%F8%F6%FD%F8%7C~%7D%FC%F9%F7%ED%F4%F6%F9oh%60JFMKP%5Do%FFgtjc > h%7B%E3%DF%DB%D6%D8%DB%E6%EC%E7%E6%E5%DF%DD%E0%EA%F6%ED%EC%F7%E8%E4%EB > %F2%F8%F9%7Cu%FC%F8%FAwhXIIIHQ%5Dksw%7Fgfl%7F%EA%E2%DA%D8%DC%DE%DE%E2% > E4%DF%E0%E3%E9%E4%E8%EC%E6%E3%E1%E4%E0%E3%EE%...

Hi, I am translating some Gauss code to R. Gauss has an interesting way of handling constraints. Observe the following code snipplet: e1 = x[.,23] .eq 0; @ remove obs with Regular Hours = 0 @ e2 = x[.,12] .gt 1; @ remove obs with non-regular work status @ e3 = x[.,4] .lt 15; @ remove obs with agricultural and mining industry code (< 15)@ esum = e1 + e2 + e3; e = esum .gt 0; @ remove obs that fail all three of the above tests @ x = delif(x,e); I'm hoping that the above is self explanatory. Currently I am using the "subset&quo...

...m0,(%rsp) 9: e8 00 00 00 00 callq e <_Z4fct1Dv4_f+0xe> e: c5 f8 29 44 24 30 vmovaps %xmm0,0x30(%rsp) 14: c5 fa 16 04 24 vmovshdup (%rsp),%xmm0 19: e8 00 00 00 00 callq 1e <_Z4fct1Dv4_f+0x1e> 1e: c5 f8 29 44 24 20 vmovaps %xmm0,0x20(%rsp) 24: c4 e3 79 05 04 24 01 vpermilpd $0x1,(%rsp),%xmm0 2b: e8 00 00 00 00 callq 30 <_Z4fct1Dv4_f+0x30> 30: c5 f9 29 44 24 10 vmovapd %xmm0,0x10(%rsp) 36: c4 e3 79 04 04 24 e7 vpermilps $0xe7,(%rsp),%xmm0 3d: e8 00 00 00 00 callq 42 <_Z4fct1Dv4_f+0x42> 42: c5 f8 28 4c 24...

...:16])) betao3<-t(t(theta[17:18])) betao4<-t(t(theta[19:20])) gamma<-theta[21:24] pd<-theta[25] po<-theta[26] pw<-theta[27] E1<- Y-alpha[1]*l%*%t(l)-X%*%betad1%*%t(l)-l%*%t(betao1)%*%t(X)-gamma[1]*G E2<- WY-alpha[2]*l%*%t(l)-X%*%betad2%*%t(l)-l%*%t(betao2)%*%t(X)-gamma[2]*G E3<- YW-alpha[3]*l%*%t(l)-X%*%betad3%*%t(l)-l%*%t(betao3)%*%t(X)-gamma[3]*G E4<-WYW-alpha[4]*l%*%t(l)-X%*%betad4%*%t(l)-l%*%t(betao4)%*%t(X)-gamma[4]*G Q<-matrix(0,4,4) Q[1,1]<-sum(E1*t(E1)) Q[1,2]<-sum(E1*t(E2)) Q[1,3]<-sum(E1*t(E3)) Q[1,4]<-sum(E1*t(E4)) Q[2,1]<-sum(E2*t(E1...

...difficult for me to run my analysis. The lab outputs the results where each parameter is its own row and it’s not consistently in the same order (and not each sample is tested for the same suite of variables). e.g. >dataset site parameter value a e2 3 a e1 1 a e3 5 b e3 1 b e1 2 c e5 4 c e4 5 c e2 2 d e2 4 d e4 3 e e1 2 How would I go about transforming the above table to the one below, where each parameter has its own column and there is only 1 row of data for eac...

Hi, At the moment I am controlling my LAN client access to the Inet by their MAC address. Currently I am putting their MAC address in the rules file - now the number of the PC that I want to manage is getting more and more and it is not practicle to do this way anymore. My question is, how can I have their MAC address in other separate file? Regards http://www.debian.org/consultants/#Malaysia

...I am aware that this list uses GNU mailman for the list management. I have my options set that it sends a email digest. One thing I find is that the digest consists of emails that ordered temporarlly. For eg lets say there are two threads t1 and t2 and the emails arrive as e1 of t1, e2 of t2, e3 of t3 . The digest lists them as e1,e2 and then e3. Is it possible to somehow configure it as T1 : e1,e3 and then T2 : e2 ? This is the digest format that google groups uses which is incredibly helpful as you can read all the messages in a thread. Additionally, it also helpfully includes a hea...