Displaying 20 results from an estimated 1000 matches similar to: "Selecting cases from matrices stored in lists"
2011 Jun 02
2
Counting occurrences in a moving window
Hi list, based on the following data.frame I would like to create a variable
that indicates the number of occurrences of A in the 3 years prior to the
current year:
DF = data.frame(read.table(textConnection(" A B
8025 1995
8026 1995
8029 1995
8026 1996
8025 1997
8026 1997
8025 1997
8027 1997
8026 1999
8027 1999
8028 1995
8029 1998
8025 1997
8027 1997
8026 1999
8027 1999
2011 Aug 25
3
Selections in lists
Hi,
I have produced a list g and I would like to reduce the amount of
information contained in each object in g.
For each matrix I would like to keep the values where the column name equals
g[year][[1]][[x]] and the row names equals g[year][[1]][[-x]]. So in
g$`1999`$`8029`, year = 1999 and x = 8029. I have been experimenting with
the subset function, but have been unsuccesful. Thanks for your
2011 Aug 15
1
Selecting section of matrix
Hi,
I have a question concerning the selection of data. Let's say that given
list h created below, I would like to select a section of the 1999 matrix.
For a case (rownames and colnames) I would like to select the cells that
have a value > 0. So for case 8025
8025 8026 8027
8025 1 1 1
8026 1 1 1
8027 1 1 1
And for case 8028
8028 8029
8028 1
2011 Apr 29
4
For loop and sqldf
Hi list,
Can anyone tell my why the following does not work? Thanks a lot! Your help
is very much appreciated.
DF = data.frame(read.table(textConnection(" B C D E F G
8025 1995 0 4 1 2
8025 1997 1 1 3 4
8026 1995 0 7 0 0
8026 1996 1 2 3 0
8026 1997 1 2 3 1
8026 1998 6 0 0 4
8026 1999 3 7 0 3
8027 1997 1 2 3 9
8027 1998 1 2 3 1
8027 1999
2011 Apr 14
1
Create matrices for time series
Hi list, I would like to use the following data.frame to generate matrices
over a 3 year moving window:
DF = data.frame(read.table(textConnection(" A B C
80 8025 1995
80 8026 1995
80 8029 1995
81 8026 1996
82 8025 1997
82 8026 1997
83 8025 1997
83 8027 1997
84 8026 1999
84 8027 1999
85 8028 1995
85 8029 1998"),head=TRUE,stringsAsFactors=FALSE))
Function to be
2011 May 26
1
Divide matrix into multiple smaller matrices
Hi list,
Using the script below, I have generated two lists (c and h) containing
yearly matrices. Now I would like to divide the matrices in c into multiple
matrices based on h. The number of matrices should be equal to:
length(unique(DF1$B))*length(h). So each unique value in DF1$B get's a
yearly matrix. Each matrix should contain all values from c where element
cij is 1. An example for
2011 Jun 14
1
Multiply list objects
Hi,
I am trying to use the objects from the list below to create more objects.
For each year in h I am trying to create as many objects as there are B's
keeping only the values of B. Example for 1999:
$`1999`$`8025`
B
B 8025 8026 8027 8028 8029
8025 1 1 1 0 0
8026 1 0 0 0 0
8027 1 0 0 0 0
8028 0 0 0 0 0
8029
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?
#
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 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
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
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
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 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),
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
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
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
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