similar to: Using a function inside a function

Displaying 20 results from an estimated 1000 matches similar to: "Using a function inside a function"

2011 Jun 09
1
Error: missing values where TRUE/FALSE needed
I'm writing a function and keep getting the following error message. myfunc <- function(lst) { lst <- list(roots = c("car insurance", "auto insurance"), roots2 = c("insurance"), prefix = c("cheap", "budget"), prefix2 = c("low cost"), suffix = c("quote", "quotes"), suffix2 = c("rate",
2011 Jun 09
1
Trying to make code more efficient
I have a repetative task in R and i'm trying to find a more efficient way to perform the following task. lst <- list(roots = c("car insurance", "auto insurance"), roots2 = c("insurance"), prefix = c("cheap", "budget"), prefix2 = c("low cost"), suffix = c("quote", "quotes"),
2011 Jun 09
2
Problem with a if statement inside a function
I have a really long functions, and at the end of the function, I am using a if statement to tag certain keywords based on whether they have certain values contained in them. However, the if statement doesn't seem to work. When I had split up the commands into various functions, it worked fine, but I'm not sure what going on now that it's combined into a single function. myfunc
2011 Jun 06
1
Merge two columns of a data frame
I have the following data: prefix <- c("cheap", "budget") roots <- c("car insurance", "auto insurance") suffix <- c("quote", "quotes") prefix2 <- c("cheap", "budget") roots2 <- c("car insurance", "auto insurance") roots3 <- c("car insurance", "auto
2011 Jun 07
1
Adding values to the end of a data frame
Let's say that I'm trying to write a functions that will allow me to automate a process where I examine all possible combinations of various string groupings. Each time I run the one function, I want to include the new values to the end of a data frame. The data frame will basically be one column with a lot of rows. roots <- c("car insurance", "auto insurance")
2005 Mar 31
1
IDMAP LDAP problems
Hi, I running samba-3.0.13-1 on RH9 (openldap-2.0.27-8,krb5-1.2.7-10,nss_ldap-202-5) and configured as show below, my intention is only to make IDMAP storage in LDAP using winbind. I've looked on SAMBA3 by example book and relatives official guide on the site. First I have try to run samba and winbind retriving users and groups from ADS and storing them in winbindd_idmap.tdb and
2005 Mar 30
0
IDMAP storage in LDAP using winbind
Hi, I running samba-3.0.13-1 on RH9 (openldap-2.0.27-8,krb5-1.2.7-10,nss_ldap-202-5) and configured as show below, my intention is only to make IDMAP storage in LDAP using winbind. I've looked on SAMBA3 by example book and relatives official guide on the site. First I have try to run samba and winbind retriving users and groups from ADS and storing them in winbindd_idmap.tdb and
2003 Apr 07
4
subsetting a dataframe
How does one remove a column from a data frame when the name of the column to remove is stored in a variable? For Example: colname <- "LOT" newdf <- subset(olddf,select = - colname) The above statement will give an error, but thats what I'm trying to accomplish. If I had used: newdf <- subset(olddf,select = - LOT) then it would have worked, but as I said the column
2004 Jul 16
3
still problems with predict!
Hi all, I still have problems with the predict function by setting up the values on which I want to predict ie: original df: p1 (193 obs) variates y x1 x2 rm(list=ls()) x1<-rnorm(193) x2<-runif(193,-5,5) y<-rnorm(193)+x1+x2 p1<-as.data.frame(cbind(y,x1,x2)) p1 y x1 x2 1 -0.6056448 -0.1113607 -0.5859728 2 -4.2841793 -1.0432688 -3.3116807 ...... 192
2011 Nov 11
2
One step way to create data frame with variable "variable names"?
Suppose plotx <- "someName" modx <- "otherName" plotxRange <- c(10,20) modxVals <- c(1,2,3) It often happens I want to create a dataframe or object with plotx or modx as the variable names. But can't understand syntax to do that. I can get this done in 2 steps, creating the data frame and then assigning names, as in newdf <- data.frame( c(1, 2, 3, 4),
2012 May 14
2
Error in names(x) <- value: 'names' attribute must be the same length as the vector
Dear R-helpers, I am stuck on an error in R: When I run my code (below), I get this error back: Error in names(x) <- value : 'names' attribute must be the same length as the vector Then when I use traceback(), R gives me back this in return: `colnames<-`(`*tmp*`, value = c(""Item", "Color" ,"Number", "Size")) I'm not exactly
2003 Sep 05
2
eliminating a large subset of data from a frame
I have a data frame with 155,000 rows. One of the columns represents the user id (of which about 10,000 are unique). I am able to isolate 1000 of these user ids (stored in a list) that I want to eliminate from the data set, but I don't know of an efficient way to do this. Certainly this would be slow: newdf<-df for(i in listofbadusers) { newdf<-subset(tmp,uid!=i) } is there a better
2018 Apr 27
5
predict.glm returns different results for the same model
Hi all, Very surprising (to me!) and mystifying result from predict.glm(): the predictions vary depending on whether or not I use ns() or splines::ns(). Reprex follows: library(splines) set.seed(12345) dat <- data.frame(claim = rbinom(1000, 1, 0.5)) mns <- c(3.4, 3.6) sds <- c(0.24, 0.35) dat$wind <- exp(rnorm(nrow(dat), mean = mns[dat$claim + 1], sd = sds[dat$claim + 1])) dat <-
2012 Feb 25
1
Unexpected behavior in factor level ordering
Hello, Everybody: This may not be a "bug", but for me it is an unexpected outcome. A factor variable's levels do not retain their ordering after the levels function is used. I supply an example in which a factor with values "BC" "AD" (in that order) is unintentionally re-alphabetized by the levels function. To me, this is very bad behavior. Would you agree? #
2005 Dec 07
4
Maintaining factors when copying from one data frame to another
Greetings all: OK, this is bugging the @#@%* out of me. I know the answer is simple and straightforward but for the life of me I cannot find it in the documentation, in the archives, or in my notes (because I know I've encountered this in the past). My problem is: I have a data frame with columns A, B, C, D, and E. A, B, and E are factors and C and D are numeric. I need a new data frame with
2017 Jul 16
3
Arranging column data to create plots
Dear All, I need some help arranging data that was imported. The imported data frame looks something like this (the actual file is huge, so this is example data) DF: IDKey X1 Y1 X2 Y2 X3 Y3 X4 Y4 Name1 21 15 25 10 Name2 15 18 35 24 27 45 Name3 17 21 30 22 15 40 32 55 I would like to create a new data frame with the following NewDF: IDKey X Y Name1 21 15 Name1
2012 Jul 14
3
Can't understand syntax
OK, I need help!! I've been searching, but I don't understand the logic of some this dataframe addressing syntax. What is this type of code called? test [["v3"]] [is.na(test[["v2"]])] <-10 #choose column v3 where column v2 is == 4 and replace with 10 and where is it documented? The code below works for what I want to do (find the non-missing value in a row),
2011 Sep 21
1
Problem with predict and lines in plotting binomial glm
Problems with predict and lines in plotting binomial glm Dear R-helpers I have found quite a lot of tips on how to work with glm through this mailing list, but still have a problem that I can't solve. I have got a data set of which the x-variable is count data and the y-variable is proportional data, and I want to know what the relationship between the variables are. The data was
2012 Apr 20
1
predictOMatic for regression. Please try and advise me
I'm pasting below a working R file featuring a function I'd like to polish up. I'm teaching regression this semester and every time I come to something that is very difficult to explain in class, I try to simplify it by writing an R function (eventually into my package "rockchalk"). Students have a difficult time with predict and newdata objects, so right now I'm
2007 Oct 29
3
how to split data.frame by row?
hi, if I have 20 x 3 data.frame, how to split it into 10 x 6 (moving the lower part of 10x3 to column) or 5 x 12 thanks -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III