similar to: winbind and getent group

Hai, > -----Oorspronkelijk bericht----- > Van: samba [mailto:samba-bounces at lists.samba.org] Namens > Andreas Hauffe via samba > Verzonden: dinsdag 22 augustus 2017 11:26 > Aan: samba at lists.samba.org > Onderwerp: Re: [Samba] Winbind with krb5auth for trust users > > Hi, > > thanks for the fast answer. > > All DCs (local and trusted domain) running on

See inline comments: On Tue, 22 Aug 2017 12:20:04 +0200 Andreas Hauffe via samba <samba at lists.samba.org> wrote: > Hi, > > hier are the file. I replaced the real domain/realm name by > "search&replace", so there should not be a typping error in my file > concernig the realm or domain names. > > Regards, > Andreas > > client:~ # more

getent group is not working in an opensuse 13.1 member server for an active directory samba 4 domain wbinfo -u, wbinfo -g, wbinfo -t and getent passwd | grep SIENIC are working, these are my configuration files and the output of the commands. Note: the domain controller has samba installed from source (4.1.11), the member server has the distro packages installed (4.1.0)

Hi, hier are the file. I replaced the real domain/realm name by "search&replace", so there should not be a typping error in my file concernig the realm or domain names. Regards, Andreas client:~ # more /etc/hostname client.loc.example.de client:~ # more /etc/hosts # # hosts         This file describes a number of hostname-to-address #               mappings for the TCP/IP

Hi, I'm trying a basic setup : samba 4.2 on vm as ad dc, linux server as a dc member with samba shares and win 7 as a ad member and samba client. Unix attrs are assigned, windows auth and linux kinit work ok. But when I try to access samba share from windows a get an error above in my log.smb: check_ntlm_password: Checking password for unmapped user [KURSK]\[video]@[EVENT] with the new

Hi, I'm trying to set up a small network over IPv6. It will have IPv4 too but the dhcp server may not work and Ubuntu (9.04) automatically configure a .local domain IPv6 addresses, so I must run the samba server and clients over IPv6. Only one machine will act as a server, but samba is up and running in all of them. The problem is that smbclient can't resolve the server's name, my

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

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

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

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

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

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.

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

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) >

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

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

Hi: I am a new R user. I have the following question and would appreciate your input Data1 (data frame 1) p1,d1,d2 (p1 is text and d1 and d2 are numeric) xyz,10,25 Data2 (data frame 2) p1,d1,d2 xyz,11,15 Now I want to create a new data frame that looks like so below. The fields d1 and s2 are summed by the product key. Data3 p1,d1,d2 xyz,21 (sum of 10 from Data1 and 11 from Data2),40 (sum of 25