Displaying 20 results from an estimated 3000 matches similar to: "Dummy (factor) based on a pair of variables"
2009 Jun 26
3
Compute correlation matrix for panel data with specific ordering
Hello All,
I have a panel date - here a small-scale example:
df <-
data.frame(cbind(rep(c("AUT","BEL","DEN","GER"),4),cbind(rep(c(1999,2000,2001,2002),4)),sample(10,16,replace=T)))
names(df) <- c("country","year","x")
SORT <- c("GER","BEL","DEN","AUT")
I need to compute the
2007 Nov 28
1
Order observations in a dataframe
Dear All,
Suppose I have the following dataframe:
country;weight;group
bul;10;1
cze;12;1
grc;12;1
hun;12;1
prt;12;1
rom14;1
fra;29;2
ita;29;2
gbr;29;2
aut;10;3
bel;12;3
The "group" variable denotes the id-number of a group of countries. How can
I re-label the groups in the descending order of their cumulative "weight",
which wound be:
country;weight;group
fra;29;1
ita;29;1
2008 Jan 14
2
Permutations of variables in a dataframe
Hallo All,
I would like to apply a function to all permutations of variables in a
dataframe (except the first). What is the best way to achieve this?
I produce the permutations using:
nvar <- ncol(dat) - 1
perms <- as.matrix( expand.grid(rep( list(1:0) , nvar ))[ , nvar:1] )
Thanks in advance
Serguei
Test-dataframe, comma-delimited:
code,wav,w,area,gdp,def,pop,coast,milspend,agr
2006 Dec 14
3
Delete all dimnames
Hello, how can I get rid of all dimnames so that:
$amat
Var3 Var2 Var1
8 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0
7 1 1 1 0 1 0 0 0 1 0 0 0 0 0 0
6 1 1 0 1 0 1 0 0 0 1 0 0 0 0 0
5 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0
4 1 0 1 1 0 0 1 0 0 0 0 1 0 0 0
3 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0
2 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0
1 1 0 0 0 0 0 0 0 0 0 0 0 0
2007 Nov 28
4
Replacing values job
Hallo,
I have two vectors of different lengths which contain the same set of
values:
X < -c(2,6,1,7,4,3,5)
Y <- c(1,1,6,4,6,1,4,1,2,3,6,6,1,2,4,4,5,4,1,7,6,6,4,4,7,1,2)
How can I replace the values in Y with the index (!) of the corresponding
values in X. So 2 appears in X in the first coordinate, so all 2’s in Y
should be replaced by 1, etc.
Thank you for your help,
Serguei
2005 Nov 23
2
vector of permutated products
Given an x-vector with, say, 3 elements, I would like to compute the
following vector of permutated products
(1-x1)*(1-x2)*(1-x3)
(1-x1)*(1-x2)*x3
(1-x1)*x2*(1-x3)
x1*(1-x2)*(1-x3)
(1-x1)*x2*x3
x1*(1-x2)*x3
x1*x2*(1-x3)
x1*x2*x3
Now, I already have the correctly sorted matrix of permutations! So, the
input looks something like:
#input
x<-c(0.3,0.1,0.2)
Nx<-length(x)
Ncomb<-2^Nx
2005 Dec 04
4
Construct a data.frame in a FOR-loop
Say I have a FOR-loop for computing powers (just a trivial example)
for(i in 1:5)
{
x<-i^2
y<-i^3
}
How can I create a data.frame and a 3D plot of (i,x(i),y(i)), i.e. for
each iteration
Thanks,
Serguei Kaniovski
--
___________________________________________________________________
??sterreichisches Institut f??r Wirtschaftsforschung (WIFO)
Name: Serguei Kaniovski
2006 Sep 22
3
Compiling a contingency table of counts by case
I have asked a similar question before but this time
the problem is somewhat more involved. I have the
following data:
case;name;x
1;Joe;1
1;Mike;1
1;Zoe;1
2;Joe;1
2;Mike;0
2;Zoe;1
2;John;1
3;Mike;1
3;Zoe;0
3;Karl;0
I would like to count the number of "case"
in which any two "name"
a. both have "x=1",
b. the first has "x=0" - the second has
2007 Feb 06
1
Questions on counts by case
Hi all,
for the data below I would like to
1. generate a dummy variable for each group "gr" of the same composition by
people, then save each portion in a separate file,
2. compute the frequency of "1"'s in "x" for each person by group
"gr". So, "mike" will have freq=2/3, as he has two "1" and one "0" in 3
groups.
2007 Feb 26
1
Adding duplicates by rows
Hi,
I am trying to add duplicates of matrix "mat" by row. Commands
subset(mat,duplicated(rownames(mat)))
or
mat[which(duplicated(rownames(mat))),]
return only half of the required indices. How can I find the remaining
ones, ie the matches, so that I can add them up?
Thanks,
Serguei
___________________________________________________________________
Austrian Institute of Economic
2008 Dec 24
1
Implementing a linear restriction in lm()
Dear All!
I want to test a coeffcient restriction beta=1 in a univariate model lm
(y~x). Entering
lm((y-x)~1) does not help since anova test requires the same dependent
variable. What is the right way to proceed?
Thank you for your help and marry xmas,
Serguei Kaniovski
________________________________________
Austrian Institute of Economic Research (WIFO)
2006 Oct 03
1
Reshape into a contingency table/Fisher's test
Dear all,
how can I "reshape"/"cast" the following matrix
00;01;10;11
John.Mike;123;313;12;31
John.Jim;54;57;39;36
John.Steve;135;47;47;74
Mike.Jim;63;37;27;16
Mike.Steve;15;15;5;61
Jim.Steve;6;10;34;35
into a set of stacked 2x2 contingency tables
0;1
John;123;12
Mike;313;31
John;54;39
Jim;57;36
John;135;47
Steve;47;16
...
so that the "fisher.test" and
2006 Jul 18
2
A contingency table of counts by case
Here is an example of the data.frame that I have,
df<-data.frame("case"=rep(1:5,each=9),"id"=rep(1:9,times=5),"x"=round(runif(length(rep(1:5,each=9)))))
"case" represents the cases,
"id" the persons, and
"x" is the binary state.
I would like to know in how many cases any two persons
a. both have "1",
b. the first has
2006 Dec 04
1
Count cases by indicator
Hi,
In the data below, "case" represents cases, "x" binary states. Each
"case" has exactly 9 "x", ie is a binary vector of length 9.
There are 2^9=512 possible combinations of binary states in a given
"case", ie 512 possible vectors. I generate these in the order of the
decimals the vectors represent, as:
2007 Dec 05
1
Information criteria for kmeans
Hello,
how is, for example, the Schwarz criterion is defined for kmeans? It should
be something like:
k <- 2
vars <- 4
nobs <- 100
dat <- rbind(matrix(rnorm(nobs, sd = 0.3), ncol = vars),
matrix(rnorm(nobs, mean = 1, sd = 0.3), ncol = vars))
colnames(dat) <- paste("var",1:4)
(cl <- kmeans(dat, k))
schwarz <- sum(cl$withinss)+ vars*k*log(nobs)
Thanks
2006 Oct 04
1
Optim: Function definition
Hi all,
I apply "optim" to the function "obj", which minimizes the goodness of
fit statistic and obtains Pearson minimum chi-squared estimate for x[1],
x[2] and x[3]. The vector "fr" contains the four observed frequencies.
Since "fr[i]" appears in the denominator, I would like to substitute "0"
in the sum if fr[i]=0.
I tried an
2005 Dec 03
1
Correlation matrix from a vector of pairwise correlations
I've a vector of pairwise correlations in the order low-index element
precedes the high-index element, say:
corr(1,2)=0.1, corr(1,3)=0.2, corr(2,3)=0.3, corr(3,4)=0.4
How can I construct the corresponding correlation matrix?
I tried using the "combn"-function in "combinat" package:
library(combinat)
combn(c(0.1,0.2,0.3,0.4),2)
, but to no avail...
Thank you for your
2011 Oct 19
1
Estimating bivariate normal density with constrains
Dear R-Users
I would like to estimate a constrained bivariate normal density, the
constraint being that the means are of equal magnitude but of opposite
signs. So I need to estimate four parameters:
mu (meanvector (mu,-mu))
sigma_1 and sigma_2 (two sd deviations)
rho (correlation coefficient)
I have looked at several packages, including Gaussian mixture models in
Mclust, but I am not sure
2006 Mar 03
3
Two quick questions
Hi all,
1. How to construct a date from three variables year, month, and day,
where all three are integers?
2. I have a dataframe by date and sector. I would like to add-up all
entries for all variable with identical date and sector, replacing the
original entries, i.e. emulate the STATA command "collapse (sum) var1
var2 var3, by(date sector)".
Thank you,
Serguei Kaniovski
2007 Feb 22
4
Sorting rows of a binary matrix
Hallo,
The command:
x <- 3
mat <- as.matrix(expand.grid(rep(list(0:1), x)))
generates a matrix with 2^x columns containing the binary representations
of the decimals from 0 to (2^x-1), here from 0 to 7. But the rows are not
sorted in this order.
How can sort the rows the ascending order of the decimals they represent,
preferably without a function which converts binaries to decimals