search for: m_k

...185 9 2000-04-09 133 186 10 2000-04-16 NA 186 # ? 1,236 more rows # ? Use `print(n = ...)` to see more rows > Here's my attempts. You can see some of the things I've tried in the commented out sections: weights %>% group_by(year(Date)) %>% summarize( m_K = mean(K, na.rm = TRUE), m_J = mean(J, na.rm = TRUE), ) %>% ggplot(aes(x = `year(Date)`)) + geom_point(aes(y = m_K, color = "red")) + geom_smooth(aes(y = m_K, color = "red")) + geom_point(aes(y = m_J, color = "blue")) + geo...

...parms) { with(as.list(c(init, parms)),{ m_Na <- function(v) 1 / (1 + 15 * exp(-0.769 * (v+34)) * ((1 - exp(-0.05(v+6)/0.11 * v + 0.66)))) dv <- I_ext - ( + g_Na * (m_Na(v))^3 * h_Na * (E_Na - v) + ((g_CaT * m_CaT * h_CaT) + (g_CaS * m_CaS)) * (E_Ca - v) + (g_K * m_K^4) * (E_K - v) + (g_KCa * m_KCa * h_KCa * (E_K - v)) + (g_Af * h_Af + g_As * h_As) * m_A^3 * (E_K - v) + (g_L * (E_L - v))) dh_Na <- k_h * (0.08 * (1 - h_Na) * exp(-0.125 * (v + 39)) - h_Na / (1 + exp(-(0.2 * (2 + 8))))) dm_Ca <- -C_Ca * (g_CaT * m_CaT * h_CaT + g_...

...; # ? Use `print(n = ...)` to see more rows >> ? > >> >> Here's my attempts. You can see some of the things I've tried in the >> commented out sections: >> weights %>% >> ????? group_by(year(Date)) %>% >> ????? summarize( >> ????????? m_K = mean(K, na.rm = TRUE), >> ????????? m_J = mean(J, na.rm = TRUE), >> ????????? ) %>% >> ????? ggplot(aes(x = `year(Date)`)) + >> ????? geom_point(aes(y = m_K, color = "red")) + >> ????? geom_smooth(aes(y = m_K, color = "red")) + >> ????? g...

See below. On 5/16/23 10:52, Christopher Ryan wrote: > I"m more of a lattice guy than a ggplot guy, but perhaps this is part of > the problem: > > ..... > ? ? ?geom_point(aes(y = m_K, color = "red")) +? ##### >> you've > associated "K" with the color red > ? ? ?geom_smooth(aes(y = m_K, color = "red")) + > ? ? ?geom_point(aes(y = m_J, color = "blue")) +? ?###### >> and "J" > with the...

Yes, you're right. In fact, it's just an adaptation of a matlab command and the author advises using N^4 replications that's why it's the default in the function. The bistochastic matrix is not my subject of interest, but I need it to perform some random tranformation of a vector of incomes. Florent Bresson ----- Message d'origine ---- De : Richard M. Heiberger <rmh at

Thanks, I tried someting like this, but computation takes times for large matrices btransf <- function(y,X=length(y)^4) { N <- length(y) bm <- matrix(rep(1/N,N^2),N,N) for(j in 1:X){ coord <- sample(1:N,4,replace=T) d <- runif(1,0,min(bm[coord[1],coord[2]],bm[coord[3],coord[4]]))

Hello, i use i.e. the cluster-package and the fanny-object to construct some cluster's and get the membership'S for every "Person" & Cluster ! My finally problem is to match efficient the data and the results's togehter in one table! Until now i use the ODBC-Package and save both in a access-file and copy them together, but i can't be sure that the sorting is the

Hi, I would like to know whether there exist algorithms to compute the coefficients or, at least, the degree of the minimal polynomial of a square matrix A (over the field of complex numbers)? I don't know whether this would require symbolic computation. If not, has any of the algorithms been implemented in R? Thanks very much, Ravi. P.S. Just for the sake of completeness, a

Hello list. I'm trying to rebuild the 2.6.9.67.0.20.EL kernel, but it fails even without modifications. How did I try it? Created a (non-root) build environment (not a mock ) Installed the kernel.scr.rpm and did a rpmbuild -ba --target=`uname -m` kernel-2.6.spec 2> prep-err.log | tee prep-out.log The build failed at the end: Processing files: kernel-xenU-devel-2.6.9-67.0.20.EL Checking