similar to: for-loop with multiple variables changing

Displaying 20 results from an estimated 4000 matches similar to: "for-loop with multiple variables changing"

2006 Nov 09
4
data storage/cubes and pointers in R
Hi all, I am faced with the situation where I want to store/analyze relatively large, organized sets of numerical data, which depend on a number of conditions (biological properties, exposure times, concentrations etc etc). Imagine about a hundred dataframes of a few thousand numerical values, with some annotation in text for some entries. Intuitively, I would like to be able to slice
2005 Jan 25
2
"disregarded projections" warning when fitting lm model
Hi all, I'm fitting a linear model (using lm) to some 2500 data points. The model consists of 4 single terms and two combined terms. I get the following warning message: "Extra arguments projections are just disregarded. in: lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) " Can anybody clarify this ? I don't seem to find any pointer to what this might
2005 Jan 24
1
proj() function for lm objects
Dear all, I'm trying to find a clear explanation of what the 'proj(lm)' function produces after having fit a linear model using 'lm'. I find the help page on the proj() function highly unclear (surely part to my limited knowledge of statistics). Can anybody provide a pointer to a clearer explanation, preferable containing some examples of the calculations involved ?
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
2006 Jan 20
3
Selecting data frame components by name - do you know a shorter way?
Hi! I suspect there must be an easy way to access components of a data frame by name, i.e. the input should look like "name1 name2 name3 ..." and the output be a data frame of those components with the corresponding names. I ´ve been trying for hours, but only found the long way to do it (which is not feasible, since I have lots of components to select):
2020 Jul 23
5
Off Topic bash question
I have a simple script: #!/bin/bash # index=0 total=0 names=() ip=() while read -r LINE do NODENAME=` echo $LINE | cut -f 1 -d ','` IP=` echo $LINE | cut -f 2 -d ','` names[index]="$NODENAME" ip[index]="$IP" index=`expr index+1` total=`expr total+1` done <<< $(cat list.txt) simple file: more list.txt name1,ip1 name2,ip2 name3,ip3 output when
2001 May 23
1
Passing a string variable to Surv
Hi, I am trying to write a function to automate multiple graph generation. My data looks like: Table of numeric values with the following headers: timeM1 statusM1 xM1 timeM2 statusM2 xM2 timeM3 statusM3 xM3 1 2 3 4 5 6 Where M1,M2, M3 hve no similarity except they have a max string length of 7. Examples are mcw0045, adl0003, lei0101. Now, what I want to do is Function(M1, M2,
2008 May 25
3
naming components of a list
Hi I have a character vector with thousands of names which looks like this: > V=c("Fred", "Mary", "SAM") > V [1] "Fred" "Mary" "SAM" > class(V) [1] "character" I would like to change it to a list: > L=as.list(V) > L [[1]] [1] "Fred" [[2]] [1] "Mary" [[3]] [1] "SAM" but I need to
2017 Jul 16
0
Arranging column data to create plots
On Sat, 15 Jul 2017, Michael Reed via R-help wrote: > Dear All, > > I need some help arranging data that was imported. It would be helpful if you were to use dput to give us the sample data since you say you have already imported it. > The imported data frame looks something like this (the actual file is > huge, so this is example data) > > DF: > IDKey X1 Y1 X2 Y2
2017 Jun 04
2
Warning from reshape2 when melting a data frame with uneven number of columns.
Here is a small reproducible example: data <- structure(list(V1 = structure(1:3, .Label = c("Name1", "Name2", "Name3"), class = "factor"), V2 = structure(c(1L, 3L, 2L), .Label = c("nam1", "name-1", "name_12"), class = "factor"), V3 = structure(1:3, .Label = c("nam2", "nam_34",
2012 Mar 17
2
Reading then transposing from file
Hi, I'm an R beginner and I'm struggling with what should be a rudimentary task. My data is along these lines: ID name1 name2 name3 name4 Class 0 1 0 2 Var1 A B C A Var2 B C C A Var3 C A B A etc. I'm using the following: foo <- data.frame(t(read.table("file", header=FALSE))) but of course now it's not using ID, Class, etc. as column names. As you can imagine,
2009 Sep 04
2
transforming a badly organized data base into a list of data frames
Dear R-ers! I have a badly organized data base in Excel. Once I read it into R it looks like this (all variables become factors because of many spaces and other characters in Excel):
2006 Nov 29
1
Extract some character from a character vector of length 1
the content of th character vector (of length 1) is as follows: a <- "something2 ....pat1 name1 pat2 something2....pat1 name2 pat2....pat1 name3 pat2 " I would like to extract the character bewteen pat1 and pat2. That's to say, I would like to get a vecter of c("name1", "name2","name3"). What I did is use strsplit() twise. But I wonder if there
2000 Oct 16
2
renaming an object
Say I have a file called exp.batch which contains 2 cols The first col contains names of R objects the user would like to use. The second col contains the file names which will be read in using read.table i.e. exp.batch may look like this..... name1 complex/filename/path1.txt name2 complex/filename/path2.txt name3 complex/filename/path3.txt name4 complex/filename/path4.txt I want to have a
2002 Jul 18
3
Oddity with names
Hi all, I'm using R 1.5.1 on Windows 2000. The following snippet of code doesn't seem to do anything - no error is reported, and there is no name change. names(myFrame[,c(1:3)]) <- c("name1", "name2", "name3") This code however works nicely: names(myFrame)[c(1:3)] <- c("name1", "name2", "name3") Can anyone suggest why
2010 May 26
1
Xen guest does not autostart
I have a virtual machine stack which was purely Centos 5.4 the last time I rebooted and experienced this problem: one of the guests does not start automatically after reboot. [root at farm1 xen]# pwd /etc/xen [root at farm1 xen]# ls -l auto total 0 lrwxrwxrwx 1 root root 8 Dec 11 17:25 name1 -> ../name1 lrwxrwxrwx 1 root root 8 May 5 21:10 name2 -> ../name2 lrwxrwxrwx 1 root root 8 Nov
2009 Jul 23
2
Constructing lists (yet, again)
This is an attempt to rescue an old R-help question that apparently received no response from the oblivion of collective silence, and besides I'm also curious about the answer > From: Griffith Feeney (gfeeney at hawaii.edu) > Date: Fri 28 Jan 2000 - 07:48:45 EST wrote (to R-help) > Constructing lists with > > list(name1=name1, name2=name2, ...) > > is tedious when
2010 Feb 10
4
Readjusting the OUTPUT csv file
Dear R helpers   I have some variables say ABC, DEF, PQR, LMN and XYZ. I am choosing any three varaibles at random at a time for my analysis and name these files as input1.csv, input2.csv and input3.csv. So if I choose variables say ABC, DEF and PQR, I am passing the specifications of these variables to input1.csv, input2.csv and input3.csv respectively.   This means in another case even if I
2009 Jan 19
2
How to assign names in a list
Hi All How can you associate names with a list when names have not been assigned? For example if you have a list like this: list2<-list(1,2,3) list2 [[1]] [1] 1 [[2]] [1] 2 [[3]] [1] 3 How do you make it look like this with names? : f1<-1 f2<-2 f3<-3 list1<-list(name1=f1, name2=f2, name3=f3) list1 $name1 [1] 1 $name2 [1] 2 $name3 [1] 3 Thanks for any help. I expect
2010 May 13
2
need help in igraph package of R
hi I am struck with a problem in igraph package of R. My problem is as follows I want to plot a power law fit for my data (in .net format --- pajek format) syntax for that in R is g <- read.graph("filename.net", "pajek") d <- degree (g, mode="in") power.law.fit (d+1, 2) it gives me desired out put if my if input a single file but I want to use a variable