similar to: an issue about removing "NA"s from an array

Displaying 20 results from an estimated 10000 matches similar to: "an issue about removing "NA"s from an array"

2017 Oct 12
0
comparing two strings from data
It's generally a very good idea to examine the structure of data after you have read it in. str(data2) would have shown you that read.csv() turned your strings into factors, and that's why the == operator no longer does what you think it does. use ... data_2 <- read.csv("excel_data.csv", stringsAsFactors = FALSE) ... to turn this off. Also, the %in% operator will achieve
2017 Oct 13
1
comparing two strings from data
Combining and completing the advice from Greg and Boris the complete solution is two lines: data_2 <- read.csv("excel_data.csv", stringsAsFactors = FALSE) match_list <- match( data_2$data1, data_2$data2 ) The vector match_list will have the matching position when it exists and NA's otherwise. Its length will be the same as the length of data_2$data1. You should get
2017 Oct 12
4
comparing two strings from data
Hi, I have two columns that contain numbers along with letters (as shown below) and have different lengths. Each entry in the first column is likely to be found in the second column at most once. For each entry of the first column, if that entry is found in the second column, I would like to get the corresponding index. For instance, if the first entry of the first column is 5th entry in the
2010 Dec 29
5
linear regression for grouped data
Hi, I have been examining large data and need to do simple linear regression with the data which is grouped based on the values of a particular attribute. For instance, consider three columns : ID, x, y, and I need to regress x on y for each distinct value of ID. Specifically, for the set of data corresponding to each of the 4 values of ID (76,111,121,168) in the below data, I should invoke
2002 Dec 16
1
help
Hi I download the R, but I dont know how to get the script (syntax) file and run it. I would be very pleased for any help. Regards. Yasin -- Yasin Al-tawarah Tel: (01782) 583652 E-mail: mad26 at keele.ac.uk
2011 Nov 30
1
s/n ratio detection etc...
Hi everybody, I' ve been following this list for a while now. Is there a way to detect the individual and cumulative s/n ratio values for the incoming calls in Asterisk or any other Call Center solution?... -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.digium.com/pipermail/asterisk-users/attachments/20111130/d0d53c1f/attachment.htm>
2008 Jun 17
1
invalid arguments RUNIF
Hi all I would be grateful you can help me with my problem. I try to run an optimization code . in one line I have runif in order to sample the PDF. I get this error while i run it. Error in runif(1, f$d[[n.of.u.vars + n.of.o.vars + j]][[2]][1], f$d[[n.of.u.vars + : invalid arguments Here is a part of that code: # initialize random numeber generator if (seed>0)
2012 Aug 30
2
Identifying and Removing NA Columns and factor Columns with more than x Levels
Hi, How do you subset a dataframe so that you only have columns: 1. that contain one or more NAs? 2. that contain factors with greater than or equal to 32 levels? How do you remove from a dataframe columns** 3. with one or more NA's? 4. that contain factors with greater than or equal to 32 levels? ** I know how to remove columns at a basic level but I am trying
2012 Feb 25
5
which is the fastest way to make data.frame out of a three-dimensional array?
foo <- rnorm(30*34*12) dim(foo) <- c(30, 34, 12) I want to make a data.frame out of this three-dimensional array. Each dimension will be a variabel (column) in the data.frame. I know how this can be done in a very slow way using for loops, like this: x <- rep(seq(from = 1, to = 30), 34) y <- as.vector(sapply(1:34, function(x) {rep(x, 30)})) month <- as.vector(sapply(1:12,
2013 Apr 27
2
Polynomial Regression and NA coefficients in R
Hey all, I'm performing polynomial regression. I'm simulating x values using runif() and y values using a deterministic function of x and rnorm(). When I perform polynomial regression like this: fit_poly <- lm(y ~ poly(x,11,raw = TRUE)) I get some NA coefficients. I think this is due to the high correlation between say x and x^2 if x is distributed uniformly on the unit interval
2010 Oct 06
1
replaces a matrix of "NA"s in an array with the previous matrix with numbers
Dear list, Does anyone know if there is a function that replaces a matrix of "NA"s in an array with the previous matrix with numbers? For example, I have an array ab <- array(dim=c(3,3,15),dimnames=list(rows=1:3,cols=1:3,dim=times)) . Select out put from the array is: , , dim = 0.478356969557745 cols rows 1 2 3 1 0.4921053 0 0.5078947 2 0.0000000 0
2011 Dec 06
4
how to view/edit large matrix/array in R?
head, tail and fix commands don't really work well if I have large matrix/array for which I would like to be able to scroll up and dow, left and right ... Could anybody please help me? Thanks [[alternative HTML version deleted]]
2012 Feb 14
3
execute array of functions
Hi all, I'm trying to get the min and max of a sequence of number using a loop like the folllowing. Can anyone point me to why it doesn't work. Thanks. type <- c("min","max") n <- 1:10 for (a in 1:2) { print(type[a](n)) } -- Muhammad
2012 Jun 07
3
conditional statement to replace values in dataframe with NA
Hello and thanks for helping. #some data L3 <- LETTERS[1:3] dat1 <- data.frame(cbind(x=1, y=rep(1:3,2), fac=sample(L3, 6, replace=TRUE))) #When x==1 and y==1 I want to replace the 1 values with NA #I can select the rows I want: dat2<-subset(dat1,x==1 & y==1) #replace the 1 with NA dat2$x<-rep(NA,nrow(dat2) dat2$y<-rep(NA,nrow(dat2) #select the other rows and rbind
2010 Dec 27
3
linear regression with dates
Hi, I am trying to do simple linear regression using dates in R but receiving error messages. With the data shown below, I would like to regress x on y. x y 11/12/1999 56.8 11/29/1999 17.9 01/04/2000 27.4 1/14/2000 96.8 1/31/2000 49.5 R gives the following error messages after reading the linear regression command: Error in storage.mode(y) <-
2011 Dec 09
3
gam, what is the function(s)
Hello, I'd like to understand 'what' is predicting the response for library(mgcv) gam? For example: library(mgcv) fit <- gam(y~s(x),data=as.data.frame(l_yx),family=binomial) xx <- seq(min(l_yx[,2]),max(l_yx[,2]),len=101) plot(xx,predict(fit,data.frame(x=xx),type="response"),type="l") I want to see the generalized function(s) used to predict the response
2012 May 25
2
Collecting results of a test with array
Dear contributors I have tried this experiment: x<-c() for (i in 1:12){ x[i]<-list(cbind(x1[i],x2[i])) #this is a list of 12 couples of time series I am using to perform a test } # that compares them 2 by 2 # ################# #trace statistic test<-data.frame() cval<-array( , dim=c(2,3,12)) for (i in 2:12){ for (k in 1:2){ for (j in 1:3){ result[k,j,i]<-
2013 Feb 18
2
error: Error in if (is.na(f0$objective)) { : argument is of length zero
Dear all, I tried running the following syntax but it keeps running for about 4 hours and then i got the following errors: Error in if (is.na(f0$objective)) { : argument is of length zero In addition: Warning message: In is.na(f0$objective) : is.na() applied to non-(list or vector) of type 'NULL' Here is the syntax itself: library('nloptr') library('pracma') #
2012 May 20
5
removeing only rows/columns with "na" value from square ( symmetrical ) matrix.
I have some square matrices with na values in corresponding rows and columns. M<-matrix(1:2,10,10) M[6,1:2]<-NA M[10,9]<-NA M<-as.matrix(as.dist(M)) print (M) 1 2 3 4 5 6 7 8 9 10 1 0 2 1 2 1 NA 1 2 1 2 2 2 0 1 2 1 NA 1 2 1 2 3 1 1 0 2 1 2 1 2 1 2 4 2 2 2 0 1 2 1 2 1 2 5 1 1 1 1 0 2 1 2 1 2 6 NA NA 2 2 2 0 1 2 1 2 7 1 1 1 1 1 1 0 2 1 2 8
2013 May 13
2
reduce three columns to one with the colnames
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