I am using the dplyr and broom combination and try to fitting regression models depending on the condition inside of the data groups. Finally I want to extract the regression coefficients by each group.
So far I'm getting the same fitting results for all groups (Each group is separated with letters a:f) . It's the main problem.
library(dplyr)
library(minpack.lm)
library(broom)
direc <- rep(rep(c("North","South"),each=20),times=6)
V <- rep(c(seq(2,40,length.out=20),seq(-2,-40,length.out=20)),times=1)
DQ0 = c(replicate(2, sort(runif(20,0.001,1))))
DQ1 = c(replicate(2, sort(runif(20,0.001,1))))
DQ2 = c(replicate(2, sort(runif(20,0.001,1))))
DQ3 = c(replicate(2, sort(runif(20,0.001,1))))
No = c(replicate(1,rep(letters[1:6],each=40)))
df <- data.frame(direc,V,DQ0,DQ1,DQ2,DQ3,No)
fit conditions can be described as follows;
direc=North and if V<J1 do fitting with the equation exp((-t_pw)/f0*exp(-del1*(1-V/J1)^2)) else if direc=Southand V>J2
do fitting with the same equation. In both case, if V<J1& V>J2 are not satisfied return 1 for each case.
UPDATE
I found that conditional nls can be possible conditional-formula-for-nls with the suggestion in this link.
nls_fit=nlsLM(DQ0~ifelse(df$direc=="North"&V<J1, exp((-t_pw)/f0*exp(-del1*(1-V/J1)^2)),1)*ifelse(df$direc=="South"&V>J2, exp((-t_pw)/f0*exp(-del2*(1-V/J2)^2)),1)
,data=df,start=c(del1=1,J1=15,del2=1,J2=-15),trace=T)
nls_fit
Nonlinear regression model
model: DQ0 ~ ifelse(df$direc == "North" & V < J1, exp((-t_pw)/f0 * exp(-del1 * (1 - V/J1)^2)), 1) * ifelse(df$direc == "South" & V > J2, exp((-t_pw)/f0 * exp(-del2 * (1 - V/J2)^2)), 1)
data: df
del1 J1 del2 J2
1.133 23.541 1.079 -20.528
residual sum-of-squares: 16.93
Number of iterations to convergence: 4
Achieved convergence tolerance: 1.49e-08
On the other hand when I try to fit other columns such as DQ1,DQ2 and DQ3;
I tried
nls_fit=nlsLM(df[,3:6]~ifelse(.....
Error in nls.lm(par = start, fn = FCT, jac = jac, control = control, lower = lower, :
evaluation of fn function returns non-sensible value!
now the problem came down to multiple column fitting. How can I fit multiple columns DQ0:DQ3 ? I checked how to succinctly write a formula with many variables from a data frame? but couldn't find the solution to use in my data frame.
In addition when I do fitting for DQ0 column inside of its groups
as you can see from the output same Del and J parameters are produced for all groups a:f
df_new<- df%>%
group_by(No)%>%
do(data.frame(model=tidy()))>%
ungroup
df_new
Source: local data frame [24 x 6]
No model.term model.estimate model.std.error model.statistic model.p.value
1 a del1 1.132546 9024.255 1.255002e-04 0.9999000
2 a J1 23.540764 984311.373 2.391597e-05 0.9999809
3 a del2 1.079182 27177.895 3.970809e-05 0.9999684
4 a J2 -20.527520 2362268.839 -8.689748e-06 0.9999931
5 b del1 1.132546 9024.255 1.255002e-04 0.9999000
6 b J1 23.540764 984311.373 2.391597e-05 0.9999809
7 b del2 1.079182 27177.895 3.970809e-05 0.9999684
8 b J2 -20.527520 2362268.839 -8.689748e-06 0.9999931
9 c del1 1.132546 9024.255 1.255002e-04 0.9999000
10 c J1 23.540764 984311.373 2.391597e-05 0.9999809
.. .. ... ... ... ... ...
Related
I have data set from a incomplete lattice design study that I have imported into R from excel and would like to conduct a PBIB.test. However, after running the function as shown below, the output shows object Area not found, even after repeated times.
library("agricolae", lib.loc = "~/R/win-library/3.3")
Rdata2 <- PBIB.test("BlockNo", "AccNo", "Rep", Area, k = 9, c("REML"), console = TRUE)
Error in data.frame(v1 = 1, y) : object 'Area' not found
What is the problem?
See below for a sample application of PBIB.test, based on the agricolae tutorial.
First, create some sample data.
# Construct the alpha design with 30 treatments, 2 repetitions, and block size = 3
Genotype <- c(paste("gen0", 1:9, sep= ""), paste("gen", 10:30, sep= ""));
r <- 2;
k <- 3;
s <- 10;
b <- s * r;
book <- design.alpha(Genotype, k, r,seed = 5);
# Source dataframe
df <- book$book;
Create a vector of response values.
# Response variable
response <- c(
5,2,7,6,4,9,7,6,7,9,6,2,1,1,3,2,4,6,7,9,8,7,6,4,3,2,2,1,1,2,
1,1,2,4,5,6,7,8,6,5,4,3,1,1,2,5,4,2,7,6,6,5,6,4,5,7,6,5,5,4);
Run PBIB.test
model <- with(df, PBIB.test(block, Genotype, replication, response, k = 3, method="REML"))
head(model);
#$ANOVA
#Analysis of Variance Table
#
#Response: yield
# Df Sum Sq Mean Sq F value Pr(>F)
#Genotype 29 72.006 2.4830 1.2396 0.3668
#Residuals 11 22.034 2.0031
#
#$method
#[1] "Residual (restricted) maximum likelihood"
#
#$parameters
# test name.t treatments blockSize blocks r alpha
# PBIB-lsd Genotype 30 3 10 2 0.05
#
#$statistics
# Efficiency Mean CV
# 0.6170213 4.533333 31.22004
#
#$model
#Linear mixed-effects model fit by REML
# Data: NULL
# Log-restricted-likelihood: -73.82968
# Fixed: y ~ trt.adj
# (Intercept) trt.adjgen02 trt.adjgen03 trt.adjgen04 trt.adjgen05 trt.adjgen06
# 6.5047533 -3.6252940 -0.7701618 -2.5264354 -3.1633495 -1.9413054
#trt.adjgen07 trt.adjgen08 trt.adjgen09 trt.adjgen10 trt.adjgen11 trt.adjgen12
# -3.0096514 -4.0648738 -3.5051139 -2.8765561 -1.7111335 -1.6308755
#trt.adjgen13 trt.adjgen14 trt.adjgen15 trt.adjgen16 trt.adjgen17 trt.adjgen18
# -2.2187974 -2.3393290 -2.0807215 -0.3122845 -3.4526453 -1.0320169
#trt.adjgen19 trt.adjgen20 trt.adjgen21 trt.adjgen22 trt.adjgen23 trt.adjgen24
# -3.1257616 0.2101325 -1.7632411 -1.9177848 -1.0500345 -2.5612960
#trt.adjgen25 trt.adjgen26 trt.adjgen27 trt.adjgen28 trt.adjgen29 trt.adjgen30
# -4.3184716 -2.3071359 1.2239927 -1.3643068 -1.4354599 -0.4726870
#
#Random effects:
# Formula: ~1 | replication
# (Intercept)
#StdDev: 8.969587e-05
#
# Formula: ~1 | block.adj %in% replication
# (Intercept) Residual
#StdDev: 1.683459 1.415308
#
#Number of Observations: 60
#Number of Groups:
# replication block.adj %in% replication
# 2 20
#
#$Fstat
# Fit Statistics
#AIC 213.65937
#BIC 259.89888
#-2 Res Log Likelihood -73.82968
I have the following data frame:
lm mean resids sd resids resid 1 resid 2 resid 3 intercept beta
1 0.000000e+00 6.2806844 -3.6261548 7.2523096 -3.6261548 103.62615 24.989340
2 -2.960595e-16 8.7515899 -5.0527328 10.1054656 -5.0527328 141.96786 -1.047323
3 -2.960595e-16 5.9138984 -3.4143908 6.8287817 -3.4143908 206.29046 -26.448694
4 3.700743e-17 0.5110845 0.2950748 -0.5901495 0.2950748 240.89801 -35.806642
5 7.401487e-16 6.6260504 3.8255520 -7.6511040 3.8255520 187.03479 -23.444762
6 5.921189e-16 8.7217431 5.0355007 -10.0710014 5.0355007 41.43239 3.138396
7 0.000000e+00 5.5269434 3.1909823 -6.3819645 3.1909823 -119.90628 27.817845
8 -1.480297e-16 1.0204260 -0.5891432 1.1782864 -0.5891432 -180.33773 35.623363
9 -5.921189e-16 6.9488186 -4.0119023 8.0238046 -4.0119023 -64.72245 21.820226
10 -8.881784e-16 8.6621512 -5.0010953 10.0021906 -5.0010953 191.65339 -5.218767
Each row represents an estimated linear model with window length 3. I used rollapply on a separate dataframe with the function lm(y~t) to extract the coefficients and intercepts into a new dataframe, which I have combined with the residuals from the same model and their corresponding means and residuals.
Since the window length is 3, it implies that there are 3 residuals as shown, per model, in resid 1, resid 2 and resid 3. The mean and sd of these are included accordingly.
I am seeking to predict the next observation, in essence, k+1, where k is the window length, using the intercept and beta.
Recall that lm1 takes observations 1,2,3 to estimate the intercept and the beta, and lm2 takes 2,3,4, lm3 takes 3,4,5, etc. The function for the prediction should be:
predict_lm1 = intercept_lm1 + beta_lm1*(k+1)
Where k+1 = 4. For lm2:
predict_lm2 = intercept_lm2 + beta_lm2*(k+1)
Where k+1 = 5.
Clearly, k increases by 1 every time I move down one row in the dataset. This is because the explanatory variable is time, t, which is a sequence increasing by one per observation.
Should I use a for loop, or an apply function here?
How can I make a function that iterates down the rows and calculates the predictions accordingly with the information found in that row?
Thanks.
EDIT:
I managed to find a possible solution by writing the following:
n=nrow(dataset)
for(i in n){
predictions = dataset$Intercept + dataset$beta*(k+1)
}
However, k does not increase by 1 per iteration. Thus, k+1 is always = 4.
How can I make sure k increases by 1 accordingly?
EDIT 2
I managed to add 1 to k by writing the following:
n=nrow(dataset)
for(i in n){
x = 0
x[i] = k + 1
preds = dataset$`(Intercept)` + dataset$t*(x[i])
}
However, the first prediction is overestimated. It should be 203, whereas it is estimated as 228, implying that it sets the explanatory variable as 1 too high.
Yet, the second prediction is correct. I am not sure what I am doing wrong. Any advice?
EDIT 3
I managed to find a solution as follows:
n=nrow(dataset)
for(i in n){
x = k + 1
preds = dataset$`(Intercept)` + dataset$t*(x)
x = x + 1
}
Your loop is not iterating:
dataset <- read.table(text="lm meanresids sdresids resid1 resid2 resid3 intercept beta
1 0.000000e+00 6.2806844 -3.6261548 7.2523096 -3.6261548 103.62615 24.989340
2 -2.960595e-16 8.7515899 -5.0527328 10.1054656 -5.0527328 141.96786 -1.047323
3 -2.960595e-16 5.9138984 -3.4143908 6.8287817 -3.4143908 206.29046 -26.448694
4 3.700743e-17 0.5110845 0.2950748 -0.5901495 0.2950748 240.89801 -35.806642
5 7.401487e-16 6.6260504 3.8255520 -7.6511040 3.8255520 187.03479 -23.444762
6 5.921189e-16 8.7217431 5.0355007 -10.0710014 5.0355007 41.43239 3.138396
7 0.000000e+00 5.5269434 3.1909823 -6.3819645 3.1909823 -119.90628 27.817845
8 -1.480297e-16 1.0204260 -0.5891432 1.1782864 -0.5891432 -180.33773 35.623363
9 -5.921189e-16 6.9488186 -4.0119023 8.0238046 -4.0119023 -64.72245 21.820226
10 -8.881784e-16 8.6621512 -5.0010953 10.0021906 -5.0010953 191.65339 -5.218767", header=T)
n <- nrow(dataset)
predictions <- data.frame()
for(i in 1:n){
k <- i ##not sure where k is coming from but put it here
predictions <- rbind(predictions, dataset$intercept[i] + dataset$beta[i]*(k+1))
}
predictions
This question already has answers here:
Linear Regression and group by in R
(10 answers)
Closed 6 years ago.
I want to apply lm() to observations grouped by subject, but cannot work out the sapply syntax. At the end, I want a dataframe with 1 row for each subject, and the intercept and slope (ie, rows of: subj, lm$coefficients[1] lm$coefficients[2])
set.seed(1)
subj <- rep(c("a","b","c"), 4) # 4 observations each on 3 experimental subjects
ind <- rnorm(12) #12 random numbers, the independent variable, the x axis
dep <- rnorm(12) + .5 #12 random numbers, the dependent variable, the y axis
df <- data.frame(subj=subj, ind=ind, dep=dep)
s <- (split(df,subj)) # create a list of observations by subject
I can pull a single set of observations from s, make a dataframe, and get what I want:
df2 <- as.data.frame(s[1])
df2
lm1 <- lm(df2$a.dep ~ df2$a.ind)
lm1$coefficients[1]
lm1$coefficients[2]
I am having trouble looping over all the elements of s and getting the data into the final form I want:
lm.list <- sapply(s, FUN= function(x)
(lm(x[ ,"dep"] ~ x[,"ind"])))
a <-as.data.frame(lm.list)
I feel like I need some kind of transpose of the structure below; the columns (a,b,c) are what I want my rows to be, but t(a) does not work.
head(a)
a
coefficients 0.1233519, 0.4610505
residuals 0.4471916, -0.3060402, 0.4460895, -0.5872409
effects -0.6325478, 0.6332422, 0.5343949, -0.7429069
rank 2
fitted.values 0.74977179, 0.09854505, -0.05843569, 0.47521446
assign 0, 1
b
coefficients 1.1220840, 0.2024222
residuals -0.04461432, 0.02124541, 0.27103003, -0.24766112
effects -2.0717363, 0.2228309, 0.2902311, -0.2302195
rank 2
fitted.values 1.1012775, 0.8433366, 1.1100777, 1.0887808
assign 0, 1
c
coefficients 0.2982019, 0.1900459
residuals -0.5606330, 1.0491990, 0.3908486, -0.8794147
effects -0.6742600, 0.2271767, 1.1273566, -1.0345665
rank 2
fitted.values 0.3718773, 0.2193339, 0.5072572, 0.2500516
assign 0, 1
By the sounds of it, this might be what you're trying to do:
sapply(s, FUN= function(x)
lm(x[ ,"dep"] ~ x[,"ind"])$coefficients[c(1, 2)])
# a b c
# (Intercept) 0.71379430 -0.6817331 0.5717372
# x[, "ind"] 0.07125591 1.1452096 -1.0303726
Other alternatives, if this is what you're looking for
I've seen it noted that in general, if you're splitting and then using s/lapply, you can usually just jump straight to by and skip the split step:
do.call(rbind,
by(data = df, INDICES=df$subj, FUN=function(x)
lm(x[, "dep"] ~ x[, "ind"])$coefficients[c(1, 2)]))
# (Intercept) x[, "ind"]
# a 0.7137943 0.07125591
# b -0.6817331 1.14520962
# c 0.5717372 -1.03037257
Or, you can use one of the packages that lets you do such sorts of calculations more conveniently, like "data.table":
library(data.table)
DT <- data.table(df)
DT[, list(Int = lm(dep ~ ind)$coefficients[1],
Slo = lm(dep ~ ind)$coefficients[2]), by = subj]
# subj Int Slo
# 1: a 0.7137943 0.07125591
# 2: b -0.6817331 1.14520962
# 3: c 0.5717372 -1.03037257
How about nlme::lmList?
library(nlme)
coef(lmList(dep~ind|subj,df))
## (Intercept) ind
## a 0.7137943 0.07125591
## b -0.6817331 1.14520962
## c 0.5717372 -1.03037257
You can transpose this if you want.
I have a 2-way repeated measures design (3 x 2), and I would like to get figures out how to calculate effect sizes (partial eta squared).
I have a matrix with data in it (called a) like so (repeated measures)
A.a A.b B.a B.b C.a C.b
1 514.0479 483.4246 541.1342 516.4149 595.5404 588.8000
2 569.0741 550.0809 569.7574 599.1509 621.4725 656.8136
3 738.2037 660.3058 812.2970 735.8543 767.0683 738.7920
4 627.1101 638.1338 641.2478 682.7028 694.3569 761.6241
5 599.3417 637.2846 599.4951 632.5684 626.4102 677.2634
6 655.1394 600.9598 729.3096 669.4189 728.8995 716.4605
idata =
Caps Lower
A a
A b
B a
B b
C a
C b
I know how to do a repeated measures ANOVA with the car package (type 3 SS is standard in my field although I know that it results in a logical error.. if somebody wants to explain that to me like I'm 5 I would love to understand it):
summary(Anova(lm(a ~ 1),
idata=idata,type=3,
idesign=~Caps*Lower)),
multivariate=FALSE)
I think what I want to do is take this part of the summary print out:
Univariate Type III Repeated-Measures ANOVA Assuming Sphericity
SS num Df Error SS den Df F Pr(>F)
(Intercept) 14920141 1 153687 5 485.4072 3.577e-06 ***
Caps 33782 2 8770 10 19.2589 0.000372 ***
Lower 195 1 13887 5 0.0703 0.801451
Caps:Lower 2481 2 907 10 13.6740 0.001376 **
And use it to calculate partial ETA squared. So, if I'm not mistaken, I need to take the SS from the first column and divide it by (itself + SS Error for that row) for each effect. Is this the correct way to go about it? If so, how do I do it? I can't figure out how to reference values from the summary print out.
The partial eta-squared can be calculated with the etasq function in heplots package
library(car)
mod <- Anova(lm(a ~ 1),
idata = idata,
type = 3,
idesign = ~Caps*Lower)
mod
library(heplots)
etasq(mod, anova = TRUE)
Since you are asking about the calculations:
From ?etasq: 'For univariate linear models, classical η^2 = SSH / SST and partial η^2 = SSH / (SSH + SSE). These are identical in one-way designs.'.
If you wish to inspect the code for the calculations of η^2 for a model with a class as in the example, you may use getS3method(f = "etasq", class = "Anova.mlm").
I am currently learning R. I have no previous knowledge of STATA.
I want to reanalyze a study which was done in Stata (xtpcse linear regression with panel-corrected standard errors). I could not find the model or more detailed code in Stata or any other hint how to rewrite this in R. I have the plm package for econometrics installed for R. That's as far as I got.
The first lines of the .do file from STATA are copied below (I just saw that it's pretty unreadable. Here is a link to the txt file in which I copied the .do content: http://dl.dropbox.com/u/4004629/This%20was%20in%20the%20.do%20file.txt).
I have no idea of how to go about this in a better way. I tried google-ing STATA and R comparison and the like but it did not work.
All data for the study I want to replicate are here:
https://umdrive.memphis.edu/rblanton/public/ISQ_data
---STATA---
Group variable: c_code Number of obs = 265
Time variable: year Number of groups = 27
Panels: correlated (unbalanced) Obs per group: min = 3
Autocorrelation: common AR(1) avg = 9.814815
Sigma computed by pairwise selection max = 14
Estimated covariances = 378 R-squared = 0.8604
Estimated autocorrelations = 1 Wald chi2(11) = 8321.15
Estimated coefficients = 15 Prob > chi2 = 0.0000
------------------------------------------------------------------------------
| Panel-corrected
food | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
lag_food | .8449038 .062589 13.50 0.000 .7222316 .967576
ciri | -.010843 .0222419 -0.49 0.626 -.0544364 .0327504
human_cap | .0398406 .0142954 2.79 0.005 .0118222 .0678591
worker_rts | -.1132705 .0917999 -1.23 0.217 -.2931951 .066654
polity_4 | .0113995 .014002 0.81 0.416 -.0160439 .0388429
market_size | .0322474 .0696538 0.46 0.643 -.1042716 .1687665
income | .0382918 .0979499 0.39 0.696 -.1536865 .2302701
econ_growth | .0145589 .0105009 1.39 0.166 -.0060224 .0351402
log_trade | -.3062828 .1039597 -2.95 0.003 -.5100401 -.1025256
fix_dollar | -.0351874 .1129316 -0.31 0.755 -.2565293 .1861545
fixed_xr | -.4941214 .2059608 -2.40 0.016 -.897797 -.0904457
xr_fluct | .0019044 .0106668 0.18 0.858 -.0190021 .0228109
lab_growth | .0396278 .0277936 1.43 0.154 -.0148466 .0941022
english | -.1594438 .1963916 -0.81 0.417 -.5443641 .2254766
_cons | .4179213 1.656229 0.25 0.801 -2.828227 3.66407
-------------+----------------------------------------------------------------
rho | .0819359
------------------------------------------------------------------------------
. xtpcse fab_metal lag_fab_metal ciri human_cap worker_rts polity_4 market
> income econ_growth log_trade fix_dollar fixed_xr xr_fluct lab_growth
> english, pairwise corr(ar1)
Update:
I just tried Vincent's code. I tried the pcse2 and vcovBK code, and they both worked (even though I'm not sure what to do with the correlation matrix that comes out of vcocBK).
However, I still have troubles reproducing the estimates of the regression coefficients in the paper I'm reanalyzing. I'm following their recipe as good as I can, the only step I'm missing is, I think, the part where in Stata "Autocorrelation: common AR(1)" is done. The paper I'm analyzing says: "OLS regression using panel corrected standard errors (Beck/Katz '95), control for first order correlation within each panel (corr AR1 option in Stata)."
How do I control for first order correlation within each panel in R?
Here is what I did so far on my data:
## run lm
res.lm <- lm(total_FDI ~ ciri + human_cap + worker_rts + polity_4 + lag_total + market_size + income + econ_growth + log_trade + fixed_xr + fix_dollar + xr_fluct + english + lab_growth, data=D)
## run pcse
res.pcse <- pcse2(res.lm,groupN="c_code",groupT="year",pairwise=TRUE)
As Ramnath mentioned, the pcse package will do what Stata's xtpcse does. Alternatively, you could use the vcovBK() function from the plm package. If you opt for the latter option, make sure you use the cluster='time' option, which is what the Beck & Katz (1995) article suggests and what the Stata command implements.
The pcse package works well, but there are some issues that makes a lot of intuitive user inputs unacceptable, especially if your dataset is unbalanced. You might want to try this re-write of the function that I coded a while ago. Just load the pcse package, load the pcse2 function, and use it by following the instructions in the pcse documentation. IMHO, the function pasted below is cleaner, more flexible and more robust than the one provided by the pcse folks. Simple benchmarks also suggest that my version may be 5 to 10 times faster than theirs, which may matter for big datasets.
Good luck!
library(Matrix)
pcse2 <- function(object, groupN, groupT, pairwise=TRUE){
## Extract basic model info
groupT <- tail(as.character((match.call()$groupT)), 1)
groupN <- tail(as.character((match.call()$groupN)), 1)
dat <- eval(parse(text=object$call$data))
## Sanity checks
if(!"lm" %in% class(object)){stop("Formula object must be of class 'lm'.")}
if(!groupT %in% colnames(dat)){stop(paste(groupT, 'was not found in data', object$call$data))}
if(!groupN %in% colnames(dat)){stop(paste(groupN, 'was not found in data', object$call$data))}
if(anyDuplicated(paste(dat[,groupN], dat[,groupT]))>0){stop(paste('There are duplicate groupN-groupT observations in', object$call$data))}
if(length(dat[is.na(dat[,groupT]),groupT])>0){stop('There are missing unit indices in the data.')}
if(length(dat[is.na(dat[,groupN]),groupN])>0){stop('There are missing time indices in the data.')}
## Expand model frame to include groupT, groupN, resid columns.
f <- as.formula(object$call$formula)
f.expanded <- update.formula(f, paste(". ~ .", groupN, groupT, sep=" + "))
dat.pcse <- model.frame(f.expanded, dat)
dat.pcse$e <- resid(object)
## Extract basic model info (part II)
N <- length(unique(dat.pcse[,groupN]))
T <- length(unique(dat.pcse[,groupT]))
nobs <- nrow(dat.pcse)
is.balanced <- length(resid(object)) == N * T
## If balanced dataset, calculate as in Beck & Katz (1995)
if(is.balanced){
dat.pcse <- dat.pcse[order(dat.pcse[,groupN], dat.pcse[,groupT]),]
X <- model.matrix(f, dat.pcse)
E <- t(matrix(dat.pcse$e, N, T, byrow=TRUE))
Omega <- kronecker((crossprod(E) / T), Matrix(diag(1, T)) )
## If unbalanced and pairwise, calculate as in Franzese (1996)
}else if(pairwise==TRUE){
## Rectangularize
rectangle <- expand.grid(unique(dat.pcse[,groupN]), unique(dat.pcse[,groupT]))
names(rectangle) <- c(groupN, groupT)
rectangle <- merge(rectangle, dat.pcse, all.x=TRUE)
rectangle <- rectangle[order(rectangle[,groupN], rectangle[,groupT]),]
valid <- ifelse(is.na(rectangle$e),0,1)
rectangle[is.na(rectangle)] <- 0
X <- model.matrix(f, rectangle)
X[valid==0,1] <- 0
## Calculate pcse
E <- crossprod(t(matrix(rectangle$e, N, T, byrow=TRUE)))
V <- crossprod(t(matrix(valid, N, T, byrow=TRUE)))
if (length(V[V==0]) > 0){stop("Error! A CS-unit exists without any obs or without any obs in a common period with another CS-unit. You must remove that unit from the data passed to pcse().")}
Omega <- kronecker(E/V, Matrix(diag(1, T)))
## If unbalanced and casewise, caluate based on largest rectangular subset of data
}else{
## Rectangularize
rectangle <- expand.grid(unique(dat.pcse[,groupN]), unique(dat.pcse[,groupT]))
names(rectangle) <- c(groupN, groupT)
rectangle <- merge(rectangle, dat.pcse, all.x=TRUE)
rectangle <- rectangle[order(rectangle[,groupN], rectangle[,groupT]),]
valid <- ifelse(is.na(rectangle$e),0,1)
rectangle[is.na(rectangle)] <- 0
X <- model.matrix(f, rectangle)
X[valid==0,1] <- 0
## Keep only years for which we have the max number of observations
large.panels <- by(dat.pcse, dat.pcse[,groupT], nrow) # How many valid observations per year?
if(max(large.panels) < N){warning('There is no time period during which all units are observed. Consider using pairwise estimation.')}
T.balanced <- names(large.panels[large.panels==max(large.panels)]) # Which years have max(valid observations)?
T.casewise <- length(T.balanced)
dat.balanced <- dat.pcse[dat.pcse[,groupT] %in% T.balanced,] # Extract biggest rectangular subset
dat.balanced <- dat.balanced[order(dat.balanced[,groupN], dat.balanced[,groupT]),]
e <- dat.balanced$e
## Calculate pcse as in Beck & Katz (1995)
E <- t(matrix(dat.balanced$e, N, T.casewise, byrow=TRUE))
Omega <- kronecker((crossprod(E) / T.casewise), Matrix(diag(1, T)))
}
## Finish evaluation, clean and output
salami <- t(X) %*% Omega %*% X
bread <- solve(crossprod(X))
sandwich <- bread %*% salami %*% bread
colnames(sandwich) <- names(coef(object))
row.names(sandwich) <- names(coef(object))
pcse <- sqrt(diag(sandwich))
b <- coef(object)
tstats <- b/pcse
df <- nobs - ncol(X)
pval <- 2*pt(abs(tstats), df, lower.tail=FALSE)
res <- list(vcov=sandwich, pcse=pcse, b=b, tstats=tstats, df=df, pval=pval, pairwise=pairwise,
nobs=nobs, nmiss=(N*T)-nobs, call=match.call())
class(res) <- "pcse"
return(res)
}
Look at the pcse package, which considers panel corrected standard errors. You certainly have to look at the documentation in STATA to figure out the assumptions made and cross check that with pcse.