Displaying 20 results from an estimated 200 matches similar to: "Assign name to a name"
2006 Aug 24
1
Using a 'for' loop : there should be a better way in R
I need to apply a yearly inflation factor to some
wages and supply some simple sums by work category. I
have gone at it with a brute force "for" loop approach
which seems okay as it is a small dataset. It looks
a bit inelegant and given all the warnings in the
Intro to R, etc, about using loops I wondered if
anyone could suggest something a bit simpler or more
efficent?
Example:
2010 Jan 30
2
drawing a line that shifts from solid to broken
I am graphing longitudinal data from three time points. I'd like to draw a
solid line from point 1 to point 2, and then a dashed line from point 2 to
point 3. It works if I do it in two steps:
> first.vector <- c(mean(year1$variable1), mean(year2$variable1))
> second.vector <- c(NA, mean(year2$variable1), mean(year3$variable1))
> plot(first.vector, type="b",
2013 Apr 03
1
linear model coefficients by year and industry, fitted values, residuals, panel data
Hi R-helpers,
My real data is a panel (unbalanced and with gaps in years) of thousands of firms, by year and industry, and with financial information (variables X, Y, Z, for example), the number of firms by year and industry is not always equal, the number of years by industry is not always equal.
#reproducible example
firm1<-sort(rep(1:10,5),decreasing=F)
year1<-rep(2000:2004,10)
2012 Jun 24
2
Defining multiple variables in a loop
Good day,
For lack of a better solution (or perhaps I am ignorant to something
more elegant), I have been bootstrapping panel data by hand so to
speak and I would like to know if there is a way to define multiple
variables in a loop using the loop variable. I found a post (here:
https://stat.ethz.ch/pipermail/r-help/2002-October/026305.html ) that
discussed naming multiple variables but it
2010 Aug 20
1
Shifting of Principal amount while calculating Present Value
Dear R Helpers
I have following data -
cash_flow = c(7, 7, 107) # 107 = Principle 100 + interest of 7%
t = c(1,2,3)
and zero rate table as
rating year1 year2 year3
AAA 3.60 4.17 4.73
AA 3.65 4.22
4.78
A 3.72 4.32 4.93
BBB 4.10
2013 Jun 07
4
matched samples, dataframe, panel data
I R-helpers
#I have a data panel of thousands of firms, by year and industry and
#one dummy variable that separates the firms in two categories: 1 if the firm have an auditor; 0 if not
#and another variable the represents the firm dimension (total assets in thousand of euros)
#I need to create two separated samples with the same number os firms where
#one firm in the first have a corresponding
2011 Aug 03
1
Need help with xyplot
Consider I have the following data:
AgeRange AgeOfPerson PersonNo FriendsAtYear0 FriendsAtYear1 FriendsAtYear2 FriendsAtYear3 FriendsAtYear4 FriendsAtYear5
10 - 12 11 1 0 1 2 2 3 3
10 - 12 12 2 0 1 2 2 3 3
15 - 18 13 3 1 2 3 4 6 7
15 - 18 14 4 1 3 4 5 7 7
30 - 40 33 5 3 5 5 6 8 9
30 - 40 36 6 4 4 4 4 4 4
I want to plot the number of friends against number of years, as to show how friendships
2008 Dec 10
2
how to merge panel data stored by variable?
Hi,
I have two datasets stored in tab-separated format in the following way
file1:
country year1 year2
Germany var1 var1
Hungary var1 var1
file2:
country year1 year2
Germany var2 var2
Hungary var2 var2
I can easily read in these files, but how can I merge them as a panel
dataset?
Thanks,
Viktor
2010 Aug 24
2
chisq.test on samples of different lengths
Hello,
I am trying to see whether there has been a significant difference in whether people experienced damages from wildlife in two different years. I therefore have two columns:
year 1:
yes
no
no
no
yes
yes
no
year 2:
no
yes
no
yes
I wanted to do a chisq.test, but if I enter it this way:
chisq.test(year1, year2)
I get the error saying the columns are two different lengths. So then I tried
2006 Dec 14
1
legend/plotmath/substitute problem
Dear R Experts,
I am trying to produce a legend for a series of plots which are
generated in a loop. The legend is supposed to look like this:
2000: gamma=1.8
where gamma is replaced by the greek letter and both the year and the
value of gamma are stored in variables.
Everything works fine as long as I have only one data series:
year = 2001
g = 1.9
plot(1)
legend('top',
2011 Jul 27
2
for loop help
I am having a hard time putting the below into a loop, where it pulls out ppt from all he stations I have versus having to go through and hard code the data to the specific stations. I tried
stnID <- stnid[which(duplicated(stnid)==FALSE)]
for(i in 1:length(stnID))
{
ppt[i] <- ppt[which(stnid==[i])]
}
but it doesn't like to use the which function inside a for loop? Any idea's here is
2003 Nov 17
1
Generalized linear model
Hi all!
I am fitting a Poisson model, using the following command:
> fit2<-glm(canc~id1+year1+time+lnpa,family=poisson)
where 'id1', 'year1' and 'time' are factors. I defined them with:
> id1<-C(factor(id1), treatment)
and 'lnpa' is a continuous variable.
The 'summary' function gives me all the effects estimates, that is, for id1,
I
2010 Feb 24
1
Remove missing observations
Hi everyone
I have the following problem: My dataframe has 3 variables: ID, Year and
and an outcome variable. The dataframe contains repeated measurements
because the subjects filled out a questionnaire every year. The time span
covers 2 years.
Now I want to check if there is a significant change in the outcome over
the 2 years with a paired wilcox.test. The problem: Not every subject
2011 Apr 18
1
Comparing two lines - Ancova: lm or aov?
Hello!
I have measurements (length and volume) of fish collected in two years. I
want to know if the the relationship between length and volume is the same
for both years. The number of fish measured is different for each year. I
don't know whether lm or aov is more appropriate to use.
Here are the two output options:
Call:
lm(formula = Volume ~ Length * Year)
Residuals:
Min 1Q
2012 Feb 14
1
cumsum function to determine plankton phenology
Apologies for the empty email earlier!
I have species abundance data sampled at a weekly frequency or
sometimes monthly depending on the year.
The goal is to identify the dates in an annual cycle in which the
cumulative abundance of a species reaches some threshold.
Here's an example of the data for 1 species over an annual period:
"mc_pheno" is the object created from this data:
2008 Oct 12
0
false convergence (8) after removal of the two-way interaction
Dear All,
I am working with a generalized linear mixed-effects model with poisson error using the lme4 package in R. I created a model with the lmer function including some main effects, three two way interactions and two random effects.
The model works well, but I have troubles when removing on of the two-way interactions. The Warning message: "In mer_finalize(ans) : false convergence
2012 Feb 19
2
barplot with more than 1 variable
Dear R listers,
I am trying to produce a simple (for a stata user) barplot with 4
countries on the x axis, each country observed in 2 subsequent years
and 3 variables.
Basically, I should have three bars for each year for each country. I
am attaching the chart I made in Stata, but I am not sure you'll
manage to see it!
I did the following:
#here I create the data-set TUSE2. The vectors mw, st
2011 Jul 04
4
How to build a matrix of number of appearance?
I have a matrix of claims at year1 that I get simply by
claims<-read.csv(file="Claims.csv")
qq1<-claims[claims$Year=="Y1",]
I have MemberID and ProviderID for every claim in qq1 both are integers
An example for the type of questions that I want to answer is
how many times ProviderID number 345 appears together with MemberID 23 in
the table qq1
In order to answer
2011 Dec 01
0
Error message: object 'A' not found
I ran the following code:
And I run into problems with the last line of code (when it says
hn<-......). I keep getting an error code: Error in distsamp(~hab ~ 1,
peldist, keyfun = "halfnorm", output = "density", :
object 'A' not found
I would appreciate any and all help.
rm(list=ls(all=TRUE)) #clear the computer's
2004 Jul 24
0
rbind()
hi. I'm merging two datasets.
one of them is 51 rows, and a typical row looks like this:
midwar[midwar$dispnum==89,]
dispnum synch sanum saname sbnum sbname year1 yearn ainit binit fatala
158 89 0 220 FRN 230 SPN 1822 1823 1 1 4
fatalb key1 keyn warnum year1.war yearn.war awon
158 5 2202301822 2202301823 1 1823 1823