similar to: Laptops unable to access Windows SMB share

On Wed, 22 Jan 2025 22:04:31 +0000 Stephen Brandli via samba <samba at lists.samba.org> wrote: > Long-time Samba user. Love it! > > I have a dozen or so computers accessing several file (SMB) shares. > All but two are in fixed locations. Some of those have fixed IPs, > the rest DHCP IPs but they don't move between subnets. The last two, > both laptops, are the

> > When at Unsecure 2, the laptops will not access the Windows server's > file share. Remember they will access the Samba file server on the same > subnet, Secure 1. The error says it can't find the share and asks to check > the name. To me, this is DNS and routing. Have laptops use a VPN to Sec2 whenever in Unsecure 2, and make sure that Sec1 is also routable through

Thanks for the reply. I agree it's reasonable to expect me to update to see if that fixes the problem. I have started that very long process. I would not have bothered you all except, in the meantime, we have a real problem with the laptops. Also, since the fixed machines do not have the problem, I can see only two reasons for it: (1) The laptops are relying on broadcast, which isn't

I am experiencing some problems at working with lists at high level. In the following "coef" contains the original DWT coefficients organized in a list. Thorugh applying the following two commands: coef.abs <- lapply(unlist(coef,recursive=FALSE,use.names =TRUE),abs) coef.abs.sorted <- sort(unlist(coef.abs),decreasing=TRUE) I get vector "coef.abs.sorted" containing

Dear R-list members, I've had a hard time trying to solve a non-linear system (nls) of equations which structure for the equation i, i=1,...,4, is as follows: f_i(d_1,d_2,d_3,d_4)-k_i(l,m,s) = 0 (1) In the expression above, both f_i and k_i are known functions and l, m and s are known constants. I would like to estimate the vector d=(d_1,d_2,d_3,d_4) which is solution

System: Cenots Linux 6.9 Application: nut-2.7.5-0.20170613gitb1314c6 [with usb 0.1 from distro] Device: Mustek PowerMust 1060 LCD Comunication log file: dump.txt We are looking at the possibility of successful communicating with this device UPS Mustek PowerMust 1060 LCD. PS: wolfy on the list gives me assistance and i can install any new compiled nut version from sources. Thanks, Catalin.

Hello R-Community, I have the following matching problem: I have two data.frames, one with an observation every month (per company ID), and one with an observation every quarter (per company ID; note that quarter means fiscal quarter; therefore 1Q = Jan, Feb, Mar is not necessarily correct and also, a fiscal quarter is not necessarily 3 month long). For every month and company, I want to get the

For learning purposes mainly I attempted to implement hashes/maps/dictionaries (Python lingua) as S4 classes, see the coding below. I came across some rough S4 edges, but in the end it worked (for one dictionary). When testing ones sees that the dictionaries D1 and D2 share their environments D1 at hash and D2 at hash, though I thought a new and empty environment would be generated each time

I wonder if there is a more efficient way to do this task. Suppose I have two data frames, such as d1 <- data.frame(x = c(1,2,3), y = c(4,5,6), z = c(7,8,9)) d2 <- d1[, c('y', 'x')] The first dataframe d1 has more variables than d2 and the variable columns are in a different order. So, what I want to do is compare the two frames on the variables that are common between

I'm trying to do the following: For each ordered pair of a data frame (D1) containing longitudes and latitudes and unique point IDs, calculate the distance to every point in another data frame (D2) also containing longitudes, latitudes and point IDs, and return to a new variable in D1 the point ID of the nearest element of D2. Dramatis personae (mostly self-explanatory): D1$long

Hi, I have 2 date vectors d1 and d2. d1 <- structure(c(14526, 14495, 14464, 14433, 14402, 14371, 14340, 14309, 14278, 14247, 14216, 14185), class = "Date") d2 <- structure(c(14526, 14509, 14488, 14466, 14453, 14441, 14396, 14388, 14343, 14333, 14310, 14281), class = "Date") I would like to create another dataframe with columns d1 and d2, where d1 is the original d1

Hello, Another way would be n <- nrow(expand.grid(1:nrow(D1), 1:nrow(D2))) D5 <- data.frame(distance=integer(n),difference=integer(n)) D5[] <- do.call(rbind, lapply(seq_len(nrow(D1)), function(i) t(sapply(seq_len(nrow(D2)), function(j){ c(distance=sqrt(sum((D1[i,1:2]-D2[j,1:2])^2)),difference=(D1[i,3]-D2[j,3])^2) } )))) identical(D3, D5) In my first answer I forgot to say that

I'm running samba 2.0.5a on a Sun Sparc 5 with Solaris 2.6 and trying to use smbclient to copy an entire directory tree to a Windows NT 4.0 box. I'm using the recurse command and can create first level directories but I am unable to create new subdirectories in any of them. For example I created the following directory structure on the Sun: 1 % ls -R .: d1/ d2/ ./d1: f11 f12

Hello, The obvious way would be to preallocate the resulting data.frame, to expand an empty one on each iteration being a time expensive operation. n <- nrow(expand.grid(1:nrow(D1), 1:nrow(D2))) D4 <- data.frame(distance=integer(n),difference=integer(n)) k <- 0 for (i in 1:nrow(D1)){ for (j in 1:nrow(D2)) { k <- k + 1 D4[k, ] <-

Otras variantes con y sin paquetes adicionales... > sapply(split(datIn$Gain, as.factor(datIn$Diet)), mean) d1 d2 d3 280 278 312 > by(datIn$Gain, datIn$Diet, mean) datIn$Diet: d1 [1] 280 -------------------------------------------------------------- datIn$Diet: d2 [1] 278 -------------------------------------------------------------- datIn$Diet: d3 [1] 312 > > library(dplyr) >

Hi Berend, I see you are one of the contributors to the rbecnhmark package. I am sorry that I am bothering you again. I have tried to run your code (slightly tweaked) involving the benchmark function, and I am getting the following error message. What am I doing wrong? Error in benchmark(d1 <- s1(df), d2 <- s2(df), d3 <- s3(df), d4 <- s4(df), : could not find function

Dear All, on rbind:ing together a number of data.frames, I found that character variables are converted into factors. Since this occurred for a data identifier, it was a little inconvenient and, to me, unexpected. (The help page explains the general procedure used. I also found that on forming a data frame, character variables are converted to factors. The help page on read.table has the

Hi I need to use rep() to get a vector out, but I have spotted something very strange. See the reproducible example below. N <- 79 seg <- 5 segN <- N / seg # = 15.8 d1 <- seg - ( segN - floor(segN) ) * seg d1 # = 1 rep(2, d1) # = numeric(0), strange - why doesn't it print one "2"? rep(2, 1) # 2, ok rep(2, d1 / 1,1) # 2, this

Estimados Cuando existia epicalc, había una manera muy fácil de determinar la media de una variable (en esta caso Gain) por grupos, en este caso (Diet). ?Como se puede hacer ahora? Diet Gain 1 d1 270 2 d1 300 3 d1 280 4 d1 280 5 d1 270 6 d2 290 7 d2 250 8 d2 280 9 d2 290 10 d2 280 11 d3 290 12 d3 340 13 d3 330 14 d3 300 15 d3 300

Hello Initial Disclaimer: I used this tutorial. I know its in german but the entries to be made in the configuration files are marked and should be able to identify if there are any errors. https://techbotch.org/blog/ups-setup/index.html After several power failures and damaged fuses I decided to give Eaton & Nut another try. Eaton because the model I bought seems to be the one on the