Fitting linear model / ANOVA by group [duplicate] - r

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Linear Regression and group by in R
(10 answers)
Closed 6 years ago.
I'm trying to run anova() in R and running into some difficulty. This is what I've done up to now to help shed some light on my question.
Here is the str() of my data to this point.
str(mhw)
'data.frame': 500 obs. of 5 variables:
$ r : int 1 2 3 4 5 6 7 8 9 10 ...
$ c : int 1 1 1 1 1 1 1 1 1 1 ...
$ grain: num 3.63 4.07 4.51 3.9 3.63 3.16 3.18 3.42 3.97 3.4 ...
$ straw: num 6.37 6.24 7.05 6.91 5.93 5.59 5.32 5.52 6.03 5.66 ...
$ Quad : Factor w/ 4 levels "NE","NW","SE",..: 2 2 2 2 2 2 2 2 2 2 ...
Column r is a numerical value indicating which row in the field an individual plot resides
Column c is a numerical value indicating which column an individual plot resides
Column Quad corresponds to the geographical location in the field to which each plot resides
Quad <- ifelse(mhw$c > 13 & mhw$r < 11, "NE",ifelse(mhw$c < 13 & mhw$r < 11,"NW", ifelse(mhw$c < 13 & mhw$r >= 11, "SW","SE")))
mhw <- cbind(mhw, Quad)
I have fit a lm() as follows
nov.model <-lm(mhw$grain ~ mhw$straw)
anova(nov.model)
This is an anova() for the entire field, which is testing grain yield against straw yield for each plot in the dataset.
My trouble is that I want to run an individual anova() for the Quad column of my data to test grain yield and straw yield in each quadrant.
perhaps a with() might fix that. I have never used it before and I am in the process of learning R currently. Any help would be greatly appreciated.

I think you are looking for by facility in R.
fit <- with(mhw, by(mhw, Quad, function (dat) lm(grain ~ straw, data = dat)))
Since you have 4 levels in Quad, you end up with 4 linear models in fit, i.e., fit is a "by" class object (a type of "list") of length 4.
To get coefficient for each model, you can use
sapply(fit, coef)
To produce model summary, use
lapply(fit, summary)
To export ANOVA table, use
lapply(fit, anova)
As a reproducible example, I am taking the example from ?by:
tmp <- with(warpbreaks,
by(warpbreaks, tension,
function(x) lm(breaks ~ wool, data = x)))
class(tmp)
# [1] "by"
mode(tmp)
# [1] "list"
sapply(tmp, coef)
# L M H
#(Intercept) 44.55556 24.000000 24.555556
#woolB -16.33333 4.777778 -5.777778
lapply(tmp, anova)
#$L
#Analysis of Variance Table
#
#Response: breaks
# Df Sum Sq Mean Sq F value Pr(>F)
#wool 1 1200.5 1200.50 5.6531 0.03023 *
#Residuals 16 3397.8 212.36
#---
#Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
#$M
#Analysis of Variance Table
#
#Response: breaks
# Df Sum Sq Mean Sq F value Pr(>F)
#wool 1 102.72 102.722 1.2531 0.2795
#Residuals 16 1311.56 81.972
#
#$H
#Analysis of Variance Table
#
#Response: breaks
# Df Sum Sq Mean Sq F value Pr(>F)
#wool 1 150.22 150.222 2.3205 0.1472
#Residuals 16 1035.78 64.736
I was aware of this option, but not familiar with it. Thanks to #Roland for providing code for the above reproducible example:
library(nlme)
lapply(lmList(breaks ~ wool | tension, data = warpbreaks), anova)
For your data I think it would be
fit <- lmList(grain ~ straw | Quad, data = mhw)
lapply(fit, anova)
You don't need to install nlme; it comes with R as one of recommended packages.

Related

What is wrong with my syntax in lme4::lmer() for a split-split plot design with unbalanced repeated measures?

I am trying to use the lme4 package in R and function lmer() to fit a model for my split-split plot design. I would have used a repeated measures ANOVA if I did not have a small number of observations missing, but the missing data should be no problem with a linear mixed effects model.
My data frame (data) has a simple structure with four factors and a numeric outcome variable called all_vai. Note that in this example data frame, not all levels of all factors are crossed even though they would be in my real data (except for the missing observations). It shouldn't matter for my question, which is an attempt to fix problematic syntax.
collected_vai <- rnorm(125, mean = 6, sd = 1)
missing <- rep(NA, times = 3)
all_vai <- c(collected_vai, missing)
year1 <- rep(2018, times = 32)
year2 <- rep(2019, times = 32)
year3 <- rep(2020, times = 32)
year4 <- rep(2021, times = 32)
year <- c(year1, year2, year3, year4)
disturbance_severity <- rep(c(0,45,65,85), each = 32)
treatment <- rep(c("B" , "T"), each = 64)
replicate <- rep(c("A", "B", "C", "D"), each = 32)
data = data.frame(all_vai, year, disturbance_severity, treatment, replicate)
data$year <- as.factor(data$year)
data$disturbance_severity <- as.factor(data$disturbance_severity)
data$treatment <- as.factor(data$treatment)
data$replicate <- as.factor(data$replicate)
Here is the model I ran for an identical data set with a different (normally distributed) numeric outcome and no missing observations -- i.e., this is the model I would be running if I didn't have unbalanced repeated measures now due to missing data:
VAImodel1 <- aov(all_vai ~ disturbance_severity*treatment*year + Error(replicate/disturbance_severity/treatment/year), data = data)
summary(VAImodel1)
When I run this, I get the error message: "Warning message:
In aov(mean_vai ~ disturbance_severity * treatment * Year + Error(Replicate/disturbance_severity/treatment/Year), :
Error() model is singular"
I have observations from different years nested within different treatments, which are nested within different disturbance severities, and all of this nested within replicates (which are experimental blocks). So I tried using this structure in lme4:
library(lme4)
library(lmerTest)
VAImodel2 <- lmer(all_vai ~ (year|replicate:disturbance_severity:treatment) + disturbance_severity*treatment*year, data = data)
summary(VAImodel2)
And this is the error message I get: "Error: number of observations (=125) <= number of random effects (=128) for term (Year | Replicate:disturbance_severity:treatment); the random-effects parameters and the residual variance (or scale parameter) are probably unidentifiable"
Next I tried simplifying my model so that I was not running out of degrees of freedom, by removing the treatment variable and interaction term, like so:
VAImodel3 <- lmer(all_vai ~ (year|replicate:disturbance_severity) + disturbance_severity*year, data = data)
summary(VAImodel3)
This time I get a different error: "boundary (singular) fit: see ?isSingular
Warning message:
Model failed to converge with 1 negative eigenvalue: -1.2e-01 "
Thank you in advance for any help.
Your problem is wrong data preparation!!
Let's start by defining values for your variables year, disturbance_severity, treatment, replicate.
library(tidyverse)
set.seed(123)
yars = 2018:2021
disturbances = c(0,45,65,85)
treatments = c("B" , "T")
replicates = c("A", "B", "C", "D")
n = length(yars)*length(disturbances)*length(treatments)*length(replicates)*1
nNA=3
Please note that I first created the variables yars, disturbances, treatments and replicates with all the allowed values.
Then I calculated the amount of data in n (you can increase the last value in the multiplication from 1 e.g. to 10) and determined how many values will be missing in the variable nNA.
The key aspect is the use of the function expand.grid(yars, disturbances, treatments, replicates) which will return the appropriate table with the correct distribution of values.
Look at the first few lines of what expand.grid returns.
Var1 Var2 Var3 Var4
1 2018 0 B A
2 2019 0 B A
3 2020 0 B A
4 2021 0 B A
5 2018 45 B A
6 2019 45 B A
7 2020 45 B A
8 2021 45 B A
9 2018 65 B A
10 2019 65 B A
11 2020 65 B A
12 2021 65 B A
13 2018 85 B A
14 2019 85 B A
15 2020 85 B A
16 2021 85 B A
17 2018 0 T A
18 2019 0 T A
This is crucial here.
The next step is straight ahead. We create a tibble sequence and put it in the aov function.
data = tibble(sample(c(rnorm(n-nNA, mean = 6, sd = 1), rep(NA, nNA)), n)) %>%
mutate(expand.grid(yars, disturbances, treatments, replicates)) %>%
rename_with(~c("all_vai", "year", "disturbance_severity", "treatment", "replicate"))
VAImodel1 <- aov(all_vai ~ disturbance_severity*treatment*year +
Error(replicate/disturbance_severity/treatment/year), data = data)
summary(VAImodel1)
output
Error: replicate
Df Sum Sq Mean Sq F value Pr(>F)
disturbance_severity 1 0.1341 0.1341 0.093 0.811
treatment 1 0.0384 0.0384 0.027 0.897
Residuals 1 1.4410 1.4410
Error: replicate:disturbance_severity
Df Sum Sq Mean Sq F value Pr(>F)
disturbance_severity 1 0.1391 0.1391 0.152 0.763
treatment 1 0.1819 0.1819 0.199 0.733
year 1 1.4106 1.4106 1.545 0.431
Residuals 1 0.9129 0.9129
Error: replicate:disturbance_severity:treatment
Df Sum Sq Mean Sq F value Pr(>F)
treatment 1 0.4647 0.4647 0.698 0.491
year 1 0.8127 0.8127 1.221 0.384
Residuals 2 1.3311 0.6655
Error: replicate:disturbance_severity:treatment:year
Df Sum Sq Mean Sq F value Pr(>F)
treatment 1 2.885 2.8846 3.001 0.144
year 1 0.373 0.3734 0.388 0.560
treatment:year 1 0.002 0.0015 0.002 0.970
Residuals 5 4.806 0.9612
Error: Within
Df Sum Sq Mean Sq F value Pr(>F)
treatment 1 0.03 0.031 0.039 0.8430
year 1 1.29 1.292 1.662 0.2002
treatment:year 1 4.30 4.299 5.532 0.0206 *
Residuals 102 79.26 0.777
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Now there are no model is singular errors!!

Wrong degrees of freedom in lsmeans and SE calculation in R

I have this sample data:
Sample Replication Days
1 1 10
1 1 14
1 1 13
1 1 14
2 1 NA
2 1 5
2 1 18
2 1 20
1 2 16
1 2 NA
1 2 18
1 2 21
2 2 15
2 2 7
2 2 12
2 2 14
I have four observations for each sample with a total of 64 samples in each of the two replications. In total, I have 512 values for both the replications. I also have some missing values designated as 'NA'. I prformed ANOVA for Mean values for each Sample for each Rep that I generated using
library(tidyverse)
df <- Data %>% group_by(Sample, Rep) %>% summarise(Mean = mean(Days, na.rm = TRUE))
curve.anova <- aov(Mean~Rep+Sample, data=df)
Result of anova is:
> summary(curve.anova)
Df Sum Sq Mean Sq F value Pr(>F)
Rep 1 6.1 6.071 2.951 0.0915 .
Sample 63 1760.5 27.945 13.585 <2e-16 ***
Residuals 54 111.1 2.057
I created a table for mean and SE values,
ANOVA<-lsmeans(curve.anova, ~Sample)
ANOVA<-summary(ANOVA)
write.csv(ANOVA, file="Desktop/ANOVA.csv")
A few lines from file are:
Sample lsmean SE df lower.CL upper.CL
1 24.875 1.014145417 54 22.84176086 26.90823914
2 25.5 1.014145417 54 23.46676086 27.53323914
3 31.32575758 1.440722628 54 28.43728262 34.21423253
4 26.375 1.014145417 54 24.34176086 28.40823914
5 26.42424242 1.440722628 54 23.53576747 29.31271738
6 25.5 1.014145417 54 23.46676086 27.53323914
7 28.375 1.014145417 54 26.34176086 30.40823914
8 24.875 1.014145417 54 22.84176086 26.90823914
9 21.16666667 1.014145417 54 19.13342752 23.19990581
10 23.875 1.014145417 54 21.84176086 25.90823914
df for all 64 samples is 54 and the error bars in the ggplot are mostly equal for all the Samples. SE values are larger than the manually calculated values. Based on anova results, df=54 is for residuals.
I want to double check the ANOVA results so that they are correct and I am correctly generating lsmeans and SE to plot a bargraph using ggplot with confirdence interval error bars.
I will appreciate any help. Thank you!
After reading your comments, I think your workflow as an issue. Basically, when you are applying your anova test, you are doing it on means of the different samples.
So, in your example, when you are doing :
curve.anova <- aov(Mean~Rep+Sample, data=df)
You are comparing these values:
> df
# A tibble: 4 x 3
# Groups: Sample [2]
Sample Replication Mean
<dbl> <dbl> <dbl>
1 1 1 12.8
2 1 2 18.3
3 2 1 14.3
4 2 2 12
So, basically, you are comparing two groups with two values per group.
So, when you tried to remove the Replication group, you get an error because the output of:
df = Data %>% group_by(Sample %>% summarise(Mean = mean(Days, na.rm = TRUE))
is now:
# A tibble: 2 x 2
Sample Mean
<dbl> <dbl>
1 1 15.1
2 2 13
So, applying anova test on that dataset means that you are comparing two groups with one value each. So, you can't compute residuals and SE.
Instead, you should do it on the full dataset without trying to calculate the mean first:
anova_data <- aov(Days~Sample+Replication, data=Data)
anova_data2 <- aov(Days~Sample, data=Data)
And their output are:
> summary(anova_data)
Df Sum Sq Mean Sq F value Pr(>F)
Sample 1 16.07 16.071 0.713 0.416
Replication 1 9.05 9.054 0.402 0.539
Residuals 11 247.80 22.528
2 observations deleted due to missingness
> summary(anova_data2)
Df Sum Sq Mean Sq F value Pr(>F)
Sample 1 16.07 16.07 0.751 0.403
Residuals 12 256.86 21.41
2 observations deleted due to missingness
Now, you can apply lsmeans:
A_d = summary(lsmeans(anova_data, ~Sample))
A_d2 = summary(lsmeans(anova_data2, ~Sample))
> A_d
Sample lsmean SE df lower.CL upper.CL
1 15.3 1.8 11 11.29 19.2
2 12.9 1.8 11 8.91 16.9
Results are averaged over the levels of: Replication
Confidence level used: 0.95
> A_d2
Sample lsmean SE df lower.CL upper.CL
1 15.1 1.75 12 11.33 19.0
2 13.0 1.75 12 9.19 16.8
Confidence level used: 0.95
It does not change a lot the mean and the SE (which is good because it means that your replicate are consistent and you don't have too much variabilities between those) but it reduces the confidence interval.
So, to plot it, you can:
library(ggplot2)
ggplot(A_d, aes(x=as.factor(Sample), y=lsmean)) +
geom_bar(stat="identity", colour="black") +
geom_errorbar(aes(ymin = lsmean - SE, ymax = lsmean + SE), width = .5)
Based on your initial question, if you want to check that the output of ANOVA is correct, you can mimick fake data like this:
d2 <- data.frame(Sample = c(rep(1,10), rep(2,10)),
Days = c(rnorm(10, mean =3), rnorm(10, mean = 8)))
Then,
curve.d2 <- aov(Days ~ Sample, data = d2)
ANOVA2 <- lsmeans(curve.d2, ~Sample)
ANOVA2 <- summary(ANOVA2)
And you get the following output:
> summary(curve.d2)
Df Sum Sq Mean Sq F value Pr(>F)
Sample 1 139.32 139.32 167.7 1.47e-10 ***
Residuals 18 14.96 0.83
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> ANOVA2
Sample lsmean SE df lower.CL upper.CL
1 2.62 0.288 18 2.02 3.23
2 7.90 0.288 18 7.29 8.51
Confidence level used: 0.95
And for the plot
ggplot(ANOVA2, aes(x=as.factor(Sample), y=lsmean)) +
geom_bar(stat="identity", colour="black") +
geom_errorbar(aes(ymin = lsmean - SE, ymax = lsmean + SE), width = .5)
As you can see, we get lsmeans for d2 close to 3 and 8 what we set at the first place. So, I think your output are correct. Maybe your data do not present any significant differences and the computation of SE are the same because the distribution of your data are the same. It is what it is.
I hope this answer helps you.
Data
df = data.frame(Sample = c(rep(1,4), rep(2,4),rep(1,4), rep(2,4)),
Replication = c(rep(1,8), rep(2,8)),
Days = c(10,14,13,14,NA,5,18,20,16,NA,18,21,15,7,12,14))

How to set up a 2 way repeated measures ANCOVA using R?

I have not seen an example of this particular test being done in many forums I have been scouring using R. I have seen an example of a 1 way ANCOVA here or 2 way ANOVA (repeated measures) here. I have presented here my version of it after jumping through several hoops and I am not sure if it is correct.
I am trying to setup a statistical test for results from a clinical trial. A simplified version of the study is as follows. In this study I have divided my subjects into two groups (Condition A & B) with n=3 subjects per group. Dependent variable (DV) was measured in these subjects over 4 time points. I aim to conduct a two way ANCOVA (repeated measures). To ask the questions:- is the Response variable (DV) affected by the Condition-timepoint interaction with bodywt as a covariate.
Note: I have to stick to "traditional" aov function and not lme or lmer since my reports will be compared to SPSS generated output-- see discussion here that aov and lme are not the same.
My example code:
SubjectID <- rep(1:6, each= 4)
Condition <- c(rep('A',12),rep('B',12))
bodywt <- rep(16:20, each= 4)
timepoint <- rep(1:4, 6)
DV <- c(42,51,63,57,46,60,63,61,41,55,62,57,73,56,53,58,50,60,56,54,52,57,54,54)
data <- data.frame(SubjectID = SubjectID,
Condition = Condition,
bodywt = bodywt,
timepoint = timepoint,
DV = DV)
#Setting up my 2 way ANCOVA repeated measures
stat_result <- aov(DV~Condition*timepoint + bodywt:Condition + Error(SubjectID), data)
summary(stat_result)
#Error: SubjectID
# Df Sum Sq Mean Sq
#Condition 1 0.8036 0.8036
#
#Error: Within
# Df Sum Sq Mean Sq F value Pr( #F)
#Condition 1 41.8 41.8 1.341 0.26282
#timepoint 1 126.1 126.1 4.045 0.06042 .
#Condition:timepoint 1 323.4 323.4 10.377 0.00501 **
#Condition:bodywt 2 81.7 40.9 1.311 0.29540
#Residuals 17 529.8 31.2
#---
#Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
I know that I am on the right track because when I remove the covariate bodywt, I Notice that the F value and Pr(F) for Condition:timepoint are predictably affected.
stat_result <- aov(DV~Condition*timepoint + bodywt + Error(SubjectID), data)
summary(stat_result)
#Error: SubjectID
# Df Sum Sq Mean Sq
#Condition 1 0.8036 0.8036
#
#Error: Within
# Df Sum Sq Mean Sq F value Pr( #F)
#Condition 1 41.8 41.8 1.231 0.28175
#timepoint 1 126.1 126.1 3.714 0.06989 .
#bodywt 1 0.5 0.5 0.015 0.90435
#Condition:timepoint 1 323.4 323.4 9.527 0.00636 **
#Residuals 18 611.0 33.9
#---
#Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
My question is that:
Is my code the correct way to set up a 2 way repeated measures ANCOVA?
(bonus question) What would be a best post hoc test to compare means between the two groups. Again I would prefer staying away from lme or lmer.
Thanks in advance

How can I compare regression coefficients across three (or more) groups using R?

Sometimes your research may predict that the size of a regression coefficient may vary across groups. For example, you might believe that the regression coefficient of height predicting weight would differ across three age groups (young, middle age, senior citizen). Below, we have a data file with 3 fictional young people, 3 fictional middle age people, and 3 fictional senior citizens, along with their height and their weight. The variable age indicates the age group and is coded 1 for young people, 2 for middle aged, and 3 for senior citizens.
So, how can I compare regression coefficients (slope mainly) across three (or more) groups using R?
Sample data:
age height weight
1 56 140
1 60 155
1 64 143
2 56 117
2 60 125
2 64 133
3 74 245
3 75 241
3 82 269
There is an elegant answer to this in CrossValidated.
But briefly,
require(emmeans)
data <- data.frame(age = factor(c(1,1,1,2,2,2,3,3,3)),
height = c(56,60,64,56,60,64,74,75,82),
weight = c(140,155,142,117,125,133,245,241,269))
model <- lm(weight ~ height*age, data)
anova(model) #check the results
Analysis of Variance Table
Response: weight
Df Sum Sq Mean Sq F value Pr(>F)
height 1 25392.3 25392.3 481.5984 0.0002071 ***
age 2 2707.4 1353.7 25.6743 0.0129688 *
height:age 2 169.0 84.5 1.6027 0.3361518
Residuals 3 158.2 52.7
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
slopes <- emtrends(model, 'age', var = 'height') #gets each slope
slopes
age height.trend SE df lower.CL upper.CL
1 0.25 1.28 3 -3.84 4.34
2 2.00 1.28 3 -2.09 6.09
3 3.37 1.18 3 -0.38 7.12
Confidence level used: 0.95
pairs(slopes) #gets their comparisons two by two
contrast estimate SE df t.ratio p.value
1 - 2 -1.75 1.82 3 -0.964 0.6441
1 - 3 -3.12 1.74 3 -1.790 0.3125
2 - 3 -1.37 1.74 3 -0.785 0.7363
P value adjustment: tukey method for comparing a family of 3 estimates
To determine whether the regression coefficients "differ across three age groups" we can use anova function in R. For example, using the data in the question and shown reproducibly in the note at the end:
fm1 <- lm(weight ~ height, DF)
fm3 <- lm(weight ~ age/(height - 1), DF)
giving the following which is significant at the 2.7% level so we would conclude that there are differences in the regression coefficients of the groups if we were using a 5% cutoff but not if we were using a 1% cutoff. The age/(height - 1) in the formula for fm3 means that height is nested in age and the overall intercept is omitted. Thus the model estimates separate intercepts and slopes for each age group. This is equivalent to age + age:height - 1.
> anova(fm1, fm3)
Analysis of Variance Table
Model 1: weight ~ height
Model 2: weight ~ age/(height - 1)
Res.Df RSS Df Sum of Sq F Pr(>F)
1 7 2991.57
2 3 149.01 4 2842.6 14.307 0.02696 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Note 1: Above fm3 has 6 coefficients, an intercept and slope for each group. If you want 4 coefficients, a common intercept and separate slopes, then use
lm(weight ~ age:height, DF)
Note 2: We can also compare a model in which subsets of levels are the same. For example, we can compare a model in which ages 1 and 2 are the same to models in which they are all the same (fm1) and all different (fm3):
fm2 <- lm(weight ~ age/(height - 1), transform(DF, age = factor(c(1, 1, 3)[age])))
anova(fm1, fm2, fm3)
If you do a large number of tests you can get significance on some just by chance so you will want to lower the cutoff for p values.
Note 3: There are some notes on lm formulas here: https://sites.google.com/site/r4naturalresources/r-topics/fitting-models/formulas
Note 4: We used this as the input:
Lines <- "age height weight
1 56 140
1 60 155
1 64 143
2 56 117
2 60 125
2 64 133
3 74 245
3 75 241
3 82 269"
DF <- read.table(text = Lines, header = TRUE)
DF$age <- factor(DF$age)

R - plm regression with time in posix-format

I have little experience with panel data in R, and am trying to run a simple panel regression with the plm-package. When converting my dataframe to a pdata.frame, however, my time index-variable is transformed to a factor variable. This means that if I want to regress a dependent variable as a function of time, the regression generates a long list of dummy-variables for time and calculates individual coefficients for each. I just want the average effect per time unit (ie. average monthly increase/decrease in points).
Example dataframe:
ID Date Points
1 1/11/2014 2
1 1/12/2014 4
1 1/1/2015 6
1 1/2/2015 8
2 1/11/2014 1
2 1/12/2014 2
2 1/1/2015 3
2 1/2/2015 4
Say the example dataframe structure is ID = int, Date = POSIXct, Points = int.
I then convert it to a pdata.frame with index ID and Date:
panel <- pdata.frame(dataframe, c("ID", "Date"))
And run a plm fixed effects regression:
fixed <- plm(Points ~ Date, data=panel, model="within")
summary(fixed)
The resulting coefficients are then broken down by each month as dummies.
I want to treat my time-variable as a continuous variable, so I get only one coefficient for Date. How can I do this? Is there a way to avoid formatting the time index-variable as a factor in panel dataframes?
I think you need to create a separate clock or time counter from panel$Date to use in your model. For example:
library(dplyr)
dataframe <- dataframe %>%
group_by(ID) %>%
mutate(clock = seq_along(ID))
panel <- pdata.frame(dataframe, c("ID", "Date"))
That produces these data:
ID Date Points clock
1-2014-11-01 1 2014-11-01 2 1
1-2014-12-01 1 2014-12-01 4 2
1-2015-01-01 1 2015-01-01 6 3
1-2015-02-01 1 2015-02-01 8 4
2-2014-11-01 2 2014-11-01 1 1
2-2014-12-01 2 2014-12-01 2 2
2-2015-01-01 2 2015-01-01 3 3
2-2015-02-01 2 2015-02-01 4 4
That produces this output:
> fixed <- plm(Points ~ clock, data=panel, model="within")
> summary(fixed)
Oneway (individual) effect Within Model
Call:
plm(formula = points ~ clock, data = panel, model = "within")
Balanced Panel: n=2, T=4, N=8
Residuals :
Min. 1st Qu. Median 3rd Qu. Max.
-0.750 -0.375 0.000 0.375 0.750
Coefficients :
Estimate Std. Error t-value Pr(>|t|)
clock 1.50000 0.22361 6.7082 0.001114 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Total Sum of Squares: 25
Residual Sum of Squares: 2.5
R-Squared : 0.9
Adj. R-Squared : 0.5625
F-statistic: 45 on 1 and 5 DF, p-value: 0.0011144

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