search for: year2

...","T","T","T","T","T","T"),nrow=18,ncol=1) #CHANGE THIS!!! nrow=# of T and Cs or A-Ks (same #) hab <- data.frame(hab) #Year effects year <- matrix(c("Year1","Year1","Year1","Year1","Year2","Year2","Year2","Year2","Year2","Year1","Year1","Year1","Year1","Year2","Year2","Year2","Year2","Year2"),nrow=18,ncol=1) #CHANGE THIS!!! to match number of A-Ks A...

...stations lat <- dat1[,2] #latitude of station lon <- dat1[,3] #longitude of station year <- dat1[,4] # data year month <- dat1[,5] #data month ppt <- dat2 #precipiation data #### station's missing zero years ### ppt2 <- matrix(ppt[,which(stnid == '140365')],ncol=1) year2 <- year[which(stnid == '140365')] year2 <- year2[which(duplicated(year2) == FALSE)] month2 <- month[which(stnid == '140365')] ppt3 <- matrix(ppt[,which(stnid == '140405')],ncol=1) year3 <- year[which(stnid == '140405')] year3 <- year3[which(duplic...

...rying 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 doing: damages<-matrix(c(3,4, 2,2), ncol=2, dimnames=list(answer=c("yes", "no"), year=c("year1", year2))) chisq.test(damages) Does that make sense? Should I maybe be doing a different...

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 [[alternative HTML v...

...(nrow(country1)), replace = T)) unemployment2 <- as.matrix(sample(country2$lagunemployment, size = (nrow(country2)), replace = T)) unemployment3 <- as.matrix(sample(country3$lagunemployment, size = (nrow(country3)), replace = T)) country.year1 <- as.matrix(cbind(country1$Country, country1$Year2)) country.year2 <- as.matrix(cbind(country2$Country, country2$Year2)) country.year3 <- as.matrix(cbind(country3$Country, country3$Year2)) country1.2 <- as.data.frame(cbind(country.year1, exp1, tax1, gdp1, unemployment1)) country2.2 <- as.data.frame(cbind(country.year2, exp2, tax2, gdp2...

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", xlim=c(1,3)) > lines(second.vector, type="b", lty=2) It's clunky, though, and I have a bunch of these to do. Can I streamline it? TIA. Jamie [[a...

Given the data structure below, how can I create a bar plot for the values of disc for each area grouped by year? bar <-structure(list(year = c(2003, 2003, 2003, 2003, 2003, 2003, 2003, 2007, 2007, 2007, 2007, 2007, 2007, 2007), area = structure(c(6L, 4L, 1L, 2L, 3L, 5L, 7L, 6L, 4L, 1L, 2L, 3L, 5L, 7L), .Label = c("AK", "ID", "MT", "NW",

...n't been able to figure it out for two different types of plots. Below is my test case: library(ggplot2) library(gridExtra) #generate test precipitation data year<-c(2000,2001,2002,2003,2004) precip<-c(46,100,80,74,20) yp<-data.frame(year, precip) #generate test fecal coliform data year2<-c(2000,2000,2000,2000,2000,2000,2000,2000,2000,2000, 2001,2001,2001,2001,2001,2001,2001,2001,2001,2001, 2002,2002,2002,2002,2002,2002,2002,2002,2002,2002, 2003,2003,2003,2003,2003,2003,2003,2003,2003,2003, 2004,2004,2004,2004,2004,2004,2004,2004,2004,2004) fc<-sample(1:1000, 50) yfc<-data...

...series: year = 2001 g = 1.9 plot(1) legend('top', legend=substitute(paste(year, ': ', gamma, '=', g), list(year=year, g=g)) ) My problem starts, when I want to put more than one series of data in the plot and accordingly need one legend row per data series: year1 = 2001 year2 = 2005 g1 = 1.9 g2 = 1.7 plot(1) legend('top', legend=c( substitute(paste(year, ': ', gamma, '=', g), list(year=year1, g=g1)), substitute(paste(year, ': ', gamma, '=', g), list(year=year2, g=g2)) ) ) This obviously does not produce the desired result....

...untry.name", timevar="year", options=list(width=700, height=600)) plot(M) This works fine and I can smoothly move back and forth between the scatter plots and the line plots. ## here I express the date as a Date class object - arbibrarily assigning each year to June 1st. subData$year2 <- as.Date(ISOdate(subData$year, 6, 1 )) M2 <- gvisMotionChart(subData, idvar="country.name", timevar="year2", options=list(width=700, height=600)) plot(M2) Using Chrome, this plot is very slow to load and it appears when pressing play that the date field fills in each...

...4,10) industry1<-rep(20,50) X<-rnorm(50) Y<-rnorm(50) Z<-rnorm(50) data1<-data.frame(firm1,year1,industry1,X,Y,Z) data1 colnames(data1)<-c("firm","year","industry","X","Y","Z") firm2<-sort(rep(11:15,3),decreasing=F) year2<-rep(2001:2003,5) industry2<-rep(30,15) X<-rnorm(15) Y<-rnorm(15) Z<-rnorm(15) data2<-data.frame(firm2,year2,industry2,X,Y,Z) data2 colnames(data2)<-c("firm","year","industry","X","Y","Z") firm3<-sort(rep(16:20,4),...

I would like to know how I can assign a name to a name. I have a dataset that has different years in it. I am writing scripts using R and I would like to give a month a generic name and then use the generic name to do different analysis. The reason for the generic name would be so that I only have to change one thing if I wanted to change the year. For example. Year1 = 1999 datayear <-

...is currently two data frames. I haven?t actually tried to run a svychisq with the two frames yet as I am waiting on the data for the second frame. I?m trying to plan out the logic ahead of time. So, let?s say I have survey design frame Year1 with variables Wt1 and MyVar, and survey design frame Year2 with variables Wt2 and MyVar ? remember the variable name is the same in both frames. My first inclination is that I would use a statement like: svychisq(~MyVar+MyVar, Year1+Year2, statistic=?Chisq?) However, it seems from reading the help files that I can only use the ?svychisq? function if I de...

...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: cat1 <- c( 1,1,6,1,1,5) cat2 <- c( 1,2,3,4,5,6) cat3 <- c( 5,4,6,7,8,8) cat4 <- c( 1,2,1,2,1,2) years <- c( 'year1', 'year2', 'year3', 'year3', 'year1', 'year1') id <- c('a','a','b','c','c','a') df1 <- data.frame(id,years,cat1,cat2, cat3, cat4) nn <- levels(df1$id) # levels for outer loop hh <- levels(df1$years) # level...

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            4.67             5.25 For each of these ratings I need to calculate the Present Val...

...<- data.frame(j1999=c(0,0,0,0,1,0), j2000=c(NA, 1, 1, 1, 0, 0), j2001=c(1, 0, 1, 0, 0, 0), year=c(2001, 2000, 2000, 2000, 1999, NA)) library(gsubfn) x <- apply(df==1, 1, which) giveYear <- function(df) { return( as.numeric(gsubfn("^[^0-9]+", "", names(df)[1])) ) } df$year2 <- sapply(x, giveYear) Thanks in advance! Cheers!! Albert-Jan ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ All right, but apart from the sanitation, the medicine, education, wine, public order, irrigation, roads, a fresh water system, and public health, what have...

...Estimate Std. Error z value Pr(>|z|) (Intercept) -1.513e-01 1.356e+00 -0.1116 0.91112 VAR2 -1.440e-02 6.429e-02 -0.2241 0.82271 VAR3 3.130e-04 1.413e-03 0.2215 0.82469 year1 -2.671e-01 5.719e-01 -0.4670 0.64050 year2 4.888e-01 4.792e-01 1.0199 0.30780 year3 -2.720e+00 1.169e+00 -2.3260 0.02002 * year4 1.287e+00 4.211e-01 3.0569 0.00224 ** year5 -1.180e-01 4.195e-01 -0.2812 0.77855 year6 7.367e-01 4.249e-01 1.7340 0.08292 ....

...: lmgroup.data <- lmList (variable1 ~ treatmentB | year/treatmentA, data = data, na.action = na.omit) When I call the object, I see : Call: Model: variable1 ~ treatmentB | year/treatmentA Data: data Coefficients: (Intercept) treatment B year2/treatmentA0 44.08387 81.11284 year2/treatmentA1 66.61333 155.62163 year3/treatmentA0 60.55125 72.83121 year3/treatmentA1 63.62340 161.92080 Degrees of freedom: 188 total; 176 residual Residual standard error:...

...389,23456,2367,3892,5438,37824, 23,2897,3456,7690,6022,3678,9431,2890) data1<-data.frame(firm1,year1,industry1,dummy1,dimension1) data1 colnames(data1)<-c("firm","year","industry","dummy","dimension") firm2<-sort(rep(11:15,3),decreasing=F) year2<-rep(2001:2003,5) industry2<-rep(30,15) dummy2<-c(0,0,0,0,0,0,1,1,1,1,1,1,1,0,1) dimension2<-c(12456,781,32489,2345,5754,8976,3245,2120,345,2341,5678,10900,12900,123,2345) data2<-data.frame(firm2,year2,industry2,dummy2,dimension2) data2 colnames(data2)<-c("firm","ye...

...nnaire 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 completed both questionnaires. Thus, some subjects have the outcome only for year1 and others only for year2. If I try wilcox.test(Outcome~Year, paired=T), I receive an error "not the same length". Is there a fast method to remove those subjects with missing outcomes? Thanks for any help. Regards, Denis Aydin _____________________________________________ Denis Aydin, MSc Swiss Tropical a...