similar to: How can we compare corresponding values of x and y (first value of x exacly matches with the first value of y)?

In this following example Id 4 is duplicated with Id 1. Like this I want both Ids (Duplicated and Duplicated with). Can anyone help? df <- data.frame( "Publication" = c(1, 2, 3, 1, 4, 5, 2, 3), "Reference" = c("a", "b", "c", "a", "d", "e", "b", "c"), "Id"= c(1, 2, 3, 4,

x=c(5, 8, 28, 29, 30) y=c(5, 8, 28, 29, 31) How can we compare these two vectors, whether each element is exactly matched with the elements in the other vector ? How can we get the non matched elements from both the vectors? Can anyone help? ________________________________ Notice: The information contained in this electronic mail message is intended only for the use of the designated recipient.

Hi everyone Can anyone help? x = c("Treatment", "Planned", "Actual", "Demographics", "Outcomes") for (i in x) { df <- paste(i)[, c("Reference_ID", "Arm_ID")] unique(df) -> uniq as.data.frame(table(uniq$Reference_ID)) -> y i.Arms<- paste(y$Var1, y$Freq, sep="_") Print_Message("Check for the Total No of

Suppose I want the output as "Trial and a sheet" without quotes x=c("a", "b", "c") print("Trial and x[1] sheet") Getting "Trial and x[1] sheet" Can anyone help? Notice: The information contained in this electronic mail message is intended only for the use of the designated recipient. This message is privileged and confidential. and the

Why the result is coming as NULL. Can anyone help. I want to find the outliers for a reference setwd("D:/AZ") library("RODBC") cdb_cnct <- odbcConnectExcel("Book1.xls") cdb_frame <- sqlFetch(cdb_cnct, "Sheet1") odbcClose(cdb_cnct) rm(cdb_cnct) x<- cdb_frame$Publication >=1990 & cdb_frame$Publication <=2012 invalid <- cdb_frame[!x,

Hello, Say all the below five vectors should have same elements in any situation. How to find the non matching vectors among these five, if so how we can find the non matching elements of those vectors? Can anyone help? a=c(1,2,3) b=c(1,2,3,4) c=c(1,2,3) d=c(1,2,3) e=c(1,4,5) identical(a,b,c,d,e) Visit us at Booth No. 5 at 2012 ChemOutsourcing Conference, 10-13 Sept 2012, Ocean Place Resort -

a=c(1,2,3) b=c(23, 24, 25) x=c("a", "b") #if (length(x[1]) == 0) {cat("x[1] is having 3 elements")} Suppose I want to send the vector a into the Object A, um getting only "a" as the ouput for Object A but not getting required output as the vector with the elements 1, 2, 3 with the following code A<- x[1] How to code it? Can anyone help? Notice: The

Hi Everyone, Can anyone help why the errors are coming and rectify it? invalid.ids <- c(1,3,5) if (length(invalid.ids)==0) { cat("No Errors found") } else {

Why the error is coming? even though the length of outcome.new$compkey and outcome.new$armkey were exactly same. Can anyone help? setwd("D:/AZ") library("RODBC") cdb_cnct <- odbcConnectExcel("AZIF_DC_GVK_NSCLC_MSALL_287papers_02072012_141450_v1_4.xls") outcomes <- sqlFetch(cdb_cnct, "Outcomes_info") odbcClose(cdb_cnct) rm(cdb_cnct)

Even though the sheet name Publication exists. The error is coming. Can anyone help? library("RODBC") cdb_cnct <- odbcConnectExcel("Copy of AZIF_DC_GVK_NSCLC_MSALL_287papers_02072012_141450_v1_4.xls") cdb_frame <- sqlFetch(cdb_cnct, "Publication") odbcClose(cdb_cnct) Error in odbcTableExists(channel, sqtable) : 'Publication': table not found on

key1.=c(1, 2, 3) key2.=c(2) if (identical(key1.,key2.) == "TRUE") { cat("No Errors found") } if (length(setdiff(key1., key2.)) !=0) {

Hi All, Can you please help me, why this code in not generating line chart? library(ggplot2) par(mfrow=c(1,3)) #qplot(TIME1, BASCHGA, data=Orange1, geom= c("point", "line"), colour= ACTTRT) unique(Orange1$REFID) -> refid for (i in refid) { Orange2 <- Orange1[i == Orange1$REFID, ] pdf('PGA.pdf') qplot(TIME1, BASCHGA, data=Orange2, geom= c("line"),

Hi all, A small problem raising. Suppose the follow raw data is stored in a csv file. And the error is coming as NULL with the following code. Can you please help? Thank you in advance. SUBJCET 9999 9999 9999 9999 9999 1Ito 1Ito 1Ito 1Ito 1Ito 1Ito 9999 9999 9999 9999 read.csv("Noname3.csv") -> FD FD$SUBJECT NULL Regards Devarayalu [[alternative HTML version

Dear list, I have the following data set id 1 2 3 4 5 6 7 8 9 10 disease a b c d e f g h i j age 23 40 32 34 25 32 22 35 29 21 city NY LD NY SG NY LD VG SA LD SG sex 1 1 2 2 2 2 1 1 1 2 treat_a y y y y treat_b n n n n n n ques1_1 2 4 5 6 8 3 1 2 4 5 ques1_2 6 4 5 12 10 9 8 4 5 7

After importing data from Excel through ODBC. In the inclusion dataset, class(inclusion$Value) is coming as factor. After filtering the data, length(inclusion$Value == 0),the answer is coming as 4879, but actually Value contains only 225 rows. So how can I get the length as 225. Can anyone help? setwd("D:/AZ") library("RODBC") cdb_cnct <-

Hi all, Can anybody debug the following programme, as I am getting some Junk graphs in the pdf. Please find the attached raw data file. Thank you Regards Rayalu library(ggplot2) setwd("D:\\General Check list") library(RODBC) conn <- odbcConnectExcel ("Book1.xls") Orange1 <- sqlFetch (conn, "Sheet3") odbcClose(conn) rm(conn) #CDAI Change

Hi all, I have a simple question: I have a character vector like this [1] "10_1" "1_1" "11_1" "2_1" "3_1" "4_1" "5_1" "6_1" "7_1" "8_1" [11] "9_1" "1_3" "2_3" "3_3" "4_3" "1_4" "2_4" "3_4"

Dear list, I have the following data set id 1 2 3 4 5 6 7 8 9 10 disease a b c d e f g h i j age 23 40 32 34 25 32 22 35 29 21 city NY LD NY SG NY LD VG SA LD SG sex 1 1 2 2 2 2 1 1 1 2 treat_a y y y y treat_b n n n n n n ques1_1 2 4 5 6 8 3 1 2 4 5 ques1_2 6 4 5 12 10 9 8 4 5 7

Dear all I would like to ask you if an assignment can be done inside a lapply statement. For example I would like to covert a double nested for loop for (i in c(1:dimx)){ for (j in c(1:dimy)){ Powermap[i,j] <- Pr(c(i,j),c(PRX,PRY),f) } } to something like that: ij<-expand.grid(i=seq(1:dimx),j=(1:dimy)) unlist(lapply(1:nrow(ij),function(rowId) { return

Suppose I have "Orange1" data like this. TIME1 RESPONSE RESPSTAT RESPUT REFID ARM SUBARM 0 299.5 Mean 8-item scale 68 0 0 0 287 Median 8-item scale 68 0 0 0 303.9 Mean 8-item scale 68 1 0 0 286 Median 8-item scale 68 1 0 0 295 Mean 8-item scale 85 0 0 12 254 Mean 8-item scale 85 0 0 0 287 Mean 8-item scale 85 1 0 12 216 Mean