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...m x86_64 Ubuntu 22.04.1 LTS # Hostname: roundcube-internal-server-1 auth_mechanisms = plain login disable_plaintext_auth = no listen = * lock_method = dotlock mail_always_cache_fields = date.received mail_cache_min_mail_count = 5000 mail_location = mbox:~/mail:INBOX=/var/mail/%u:INDEX=/var/indexes/isical.ac.in/%n mail_privileged_group = mail mail_sort_max_read_count = 100 mailbox_list_index_include_inbox = yes mailbox_list_index_very_dirty_syncs = yes maildir_very_dirty_syncs = yes mbox_dotlock_change_timeout = 1 mins mbox_lazy_writes = no mbox_write_locks = dotlock fcntl namespace inbox { inb...

Dear R users, I want to plot a one variable continuous function f(x) vs x, x=[0,1]. Say for example: f(x)= x^2. Now, using the command plot(f~x) I will get a curve where the range of x-axis is [0,1] with all equispaced label. But, I need something else, and that is: my curve will be such that 80% on x-axis the range would be [0,0.5] and the rest 20% would contain [0.5,1]. Let me draw informally

Dear Sir/madam, I'm getting a problem with a R-code which calculate Fisher Information Matrix for Hybrid Censored Weibull Distribution. My problem is that: when I take weibull(scale=1,shape=2) { i.e shape>1} I got my desired result but when I take weibull(scale=1,shape=0.5) { i.e shape<1} it gives error : Error in integrate(int2, lower = 0, upper = t) : the integral is probably

Dear Sir/Madam, I'm getting a problem with a R-code which converts a data frame to a matrix. It first generate a (m^(n-m) * m) matrix A and then regenerate another matrix B having less dimension than A which satisfy some condition. Now I wish to assign each row of B to a vector as individual. My problem is when I set any choice of (n,m) except m=1 it works fine but setting m=1 I got the

...canner # Scanners # USB Ethernet, requires mii #device aue # ADMtek USB ethernet #device cue # CATC USB ethernet #device kue # Kawasaki LSI USB ethernet options IPFIREWALL options IPFIREWALL_VERBOSE options IPDIVERT options TCP_DROP_SYNFIN options NETATALK options ELSA_QS1PCI device isic pseudo-device "i4bq921" pseudo-device "i4bq931" pseudo-device "i4b" pseudo-device "i4btrc" 4 pseudo-device "i4bctl" pseudo-device "i4brbch" 4 #pseudo-device "i4btel" 2 #pseudo-device "i4bisppp" 4 #pseude-device "i4...

Dear R-listers, I am giving part of my R code : ########################################################### n=15 m=1 library("partitions") library("gregmisc") library("combinat") x = t(restrictedparts(n-m,m)) l = length(x[,1]) for(u in 1:l){ A= unique(matrix( unlist(permn(x[u,])), ncol=m, byrow=TRUE )) }

Hello, I understand that the/ demean/ argument in the *ar()* function to fit an autoregressive model selects the best AR model fitted to the mean deleted observations. What is the purpose of using this demean procedure at all? Its seems silly as the post doesn't deal with R problems.... Thanks -- View this message in context:

Hello I am facing a curious problem.I have a time series data with which i want to fit auto-regressive model of order p, where p runs from 1:9.I am using a for loop which will fit an AR(p) model for each value of p using the *ar.ols* function. I am using the following code for ( p in 1:9){ a=ar.ols (x=data.ts, order.max=p, demean=T, intercept=T) } Specifying the *order.max* to be p, it gives me a