similar to: 2.0.0 binaries for SCO available

----------------------------------- Has anyone been able to compile a working version of Samba using the native compiler on SCO Openserver 5.0? The compile completes successfully. But I get the following on smbclient: #------------------------------------- $ smbclient \\\\larry\\c number of interfaces returned is: 2 Impossible netmask 255.255.255.0 - using defaults error connecting to

I have been building 2.0beta5 on a SCO OS5 box using ggg 2.7.1 and get the following error (every time rpc_msg.h is included). Compiling smbd/connection.c In file included from /usr/include/rpc/rpc.h:42, from include/includes.h:317, from smbd/connection.c:22: /usr/include/rpc/rpc_msg.h:54: parse error before `1' *** Error code 1 (bu21) The problem is with

Does anyone have an XPM version of the new Samba logo? (Suitable for an AfterStep/fvwm etc wharf). Cheers and thanks, Stephen Davies ======================================================================== Stephen Davies Consulting scldad@sdc.com.au Adelaide, South Australia. Voice: 61-8-82728863 Computing & Network

Here (thanks to "MSgt Donovan Chin" <chind@ogn.af.mil>) is what you have to do to get the dialog box back: Start RegEdit Double click on the following in order: HKEY_LOCAL_MACHINE Software Microsoft Windows CurrentVersion Network Real Mode Net Over on the right hand side, you will see one line with AutoLogon Click once on AutoLogon and press the Delete key. NOTE: the only line

Hello all. I am sure that this is an FAQ but I cannot find a succinct answer anywhere. My question is: What is the correct way to access a Samba share when the Samba server has a private IP adress (eg 192.168.x.y), the (Windows via PPP) client is assigned a relevant private address (eg 192.168.x.z) and the username/password used to login to the client is not known to the server and/or is

Dear all, Could you please help me how to get the output as I described in the following example? x<-c(543, 543, 543, 543, 551 , 551 ,1128 ,1197, 1197) diff<-x-lag(x) diff [1] NA 0 0 0 8 0 577 69 0 How to index the occasions in x repeatedly if the diff<15? if diff>=15, it will give a new index. I want the output be like y. y<-c(1,1,1,1,1,1,2,3,3) Thank you so

Dear all, Could you please help me how to get the output as I described in the following example? x<-c(543, 543, 543, 543, 551 , 551 ,1128 ,1197, 1197) diff<-x-lag(x) diff [1] NA 0 0 0 8 0 577 69 0 How to index the occasions in x repeatedly if the diff<15? if diff>=15, it will give a new index. I want the output be like y. y<-c(1,1,1,1,1,1,2,3,3) Thank you so

Hi All, I have a vector of data on which I am operating. The equation with which I want to operate on the vector has a value k. I want to run the equation and output a new vector, each time replacing k with each value from the sequence I defined. I have thought about using for loops and such, but this seems like overkill. I am wondering if there is a simple solution that would allow me to

My company is based in Australia and we have a need to be able to present CLI (ANI) and DNIS to an analog extension. Currently our PABX vendor is saying they can only deliver CLI between the first and second cadence. The system will collect calls via an aggregate of ISDN PRI services from the PSTN and then direct them to the analog extensions in a hunt group configuration. It is important that

I thnk what you want is ?outer. e.g.: outer(Data -min(Data),value,FUN = "+") Whether this works for your real task, however, may depend on details and complexities that you have omitted. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom

Hello Everyone, I have three variables: Waist circumference (WC), serum triglyceride (TG) level and gender. Waist circumference and serum triglyceride is numeric and gender (male and female) is categorical. From these three variables, I want to calculate the "Lipid Accumulation Product (LAP) Index". The equation to calculate LAP is different for male and females. I am giving both

Hi, there, Suppose I have two vector say x=c(1 2 3 4 5) and y=(2 3 6 7). Then I want to combine these two vector together and get z=c(1 2 3 4 5 6 7) with 2 and 3 only appear once. I want to extend this one to a general case(say more than 100 elements in x and y and each time I don't know which elements are the same). Do you happen to know how to do this and which command should use? Thank

Well, something like: LAP <- ifelse(gender =='male', (WC-65)*TG, (WC-58)*TG) The exact code depends on whether your variables are in a data frame or list or whatever, which you failed to specify. If so, ?with may be useful. Cheers, Bert On Fri, Nov 3, 2023 at 3:43?AM Md. Kamruzzaman <mkzaman.m at gmail.com> wrote: > Hello Everyone, > I have three variables: Waist

Hello, for reasons I want to know the return value of TYPEOF(R_NilValue), I expect it to be NILSXP, but I can't find this documented anywhere. Ideally, I'd like to see the source definition of R_NilValue, but after fair bit of searching I cannot find an obviously location for this. Would someone please point me in the right direction? thanks -- CRICOS provider code 00123M

Hello Everyone, I was analysing big survey data using survey packages on RStudio. Survey package allows survey data analysis with the design effect.The survey package included functions for all other statistical analysis except two-proportion z tests. I was trying to calculate the difference in prevalence of Diabetes and Prediabetes between the year 2011 and 2017 (with 95%CI). I was able to

df$LAP <- with(df, ifelse(G=='male', (WC-65)*TG, (WC-58)*TG)) That will do both calculations and merge the two vectors appropriately. It will use extra memory, but it should be much faster than a 'for' loop. Regards, Jorgen Harmse. ------------------------------ Message: 8 Date: Fri, 3 Nov 2023 11:10:49 +1030 From: "Md. Kamruzzaman" <mkzaman.m at gmail.com>

Good Day list, Need assistance determining the best place to read up on whether Asterisk can help me out. I have a situation where I need to do the following <PRI from Telco> ------- <Analog Channel Bank>------------<Proprietary Box> | | | | | | <PRI Port 1 of Digium Quad T1> <PRI Port 2 of Digium Quad T1> | | | | | |

I want to use question numbers from my survey data (fake data below) as markers in a scatterplot. I'm using "as.character" to convert question numbers to characters. However, plot truncates the 0 (zero) off of question 10. How can I get the ending zero so I can add questions 11 - 20? # Sample code below Question <- c(1,2,3,4,5,6,7,8,9,10) Performance <-

Hello, I run R on Windows 2000. When I load lattice library and plot some lme class objects, the graphic window displays a dark-green color. Graphics look ok on the screen, but lines and points become invisible whem I send the plot to a printer because of the background color. Are there ways to reset the background color? Both par(bg=) and windows() do not work. Any ideas? Richard

Dear Md Kamruzzaman, To answer your second question first, you could just use the svychisq() function. The difference-of-proportion test is equivalent to a chisquare test for the 2-by-2 table. You don't say how you computed the confidence intervals for the two separate proportions, but if you have their standard errors (and if not, you should be able to infer them from the confidence