similar to: Error: level sets of factors are different?

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 <-

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

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,

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

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.

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,

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 {

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 -

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

df <- data.frame ( "RowId" = 4:7, "x"=c("1_1", "2_2", "3_3", "3_3"), "y"=c("1_1", "3_3", "2_2", "3_3") ) How can we compare corresponding values of x and y (first value of x exacly matches with the first value of y)? If they were not matced

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