R for psychologists

Professor Thom Baguley, Psychology, Nottingham Trent University

Thomas.Baguley@ntu.ac.uk

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0. Overview

1. What is R?

2. Why use R?

3. R basics

4. R graphics

5. Linear models in R

6. ANOVA and ANCOVA

6. Writing your own R functions

8. Simulation and bootstrapping

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1. What is R?

R is a software environment for statistical analysis

“the lingua franca of computational statistics”

De Leeuw & Mair (2007)

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2. Why use R? It is:

- free- open source- cross-platform (Mac, Linux, PC)

It has:

- excellent basic statistics functions- powerful and versatile graphics

- hundreds of user-contributed packages

- a large community of users

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3. R basics

Some common R objects:

- characters (text strings e.g., 'a' or ’1’)

- numbers (numbers like 2 or 1e+3)

- vectors (1D set of numbers or other elements)

- data frames (like vectors organized in columns)

- matrices (r by c array of numbers)

- lists (objects that contain other objects)

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Assignment

Input:

> vector1 <- 6

> vector1

Output:

[1] 6

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Arithmetic

> 6 + 6 * 2

[1] 18

> vector1^2

[1] 36

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Calling functions

> c(1, 9, 25) > log(vector1)

[1] 1 9 25 [1] 1.791759

> rnorm(1, 100, 15) ?rnorm

[1] 123.5336 help(rnorm)

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Loading data

> dat1 <- read.csv('pride.csv')

> library(foreign)

> h05 <- read.spss('Hayden_2005.sav', to.data.frame = TRUE)

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Addressing the contents of an object

> vector1[1]

[1] 6

> days <- h05$days

> days[1:9]

[1] 2864 1460 2921 2921 2921 1460 2921 1460 31

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Addressing the contents of an object

> vector1[1]

[1] 6

> days <- h05$days

> days[1:9]

[1] 2864 1460 2921 2921 2921 1460 2921 1460 31

Or combine the steps with h05$days[1:9]

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Data frames

> is.data.frame(h05)

[1] TRUE

> dim(h05) > names(h05)

[1] 43 2 [1] "names" "days”

- this data frame has 43 rows and 2 named columns

- it can be helpful to think of a data frame as a set of named variables (vectors) bound into columns

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Matrix objects (matrices)

> cells <- c(3677,56,3924,31)

> cat.names <- list(c("Before", "After"), c("Alive", "Dead"))

> checklist <- matrix(cells, 2, 2, byrow=TRUE, dimnames= cat.names)

> checklist

Alive Dead

Before 3677 56

After 3924 31

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The power of objects

> chisq.test(checklist)

Pearson's Chi-squared test with Yates' continuity correction

data: checklist

X-squared = 8.1786, df = 1, p-value = 0.004239

> chisq.test(c(2, 18))

Chi-squared test for given probabilities

data: c(2, 18)

X-squared = 12.8, df = 1, p-value = 0.0003466

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Calling functions: defaults and named arguments

> chisq.test(checklist, correct=FALSE)

Pearson's Chi-squared test

data: checklist

X-squared = 8.8072, df = 1, p-value = 0.003000

- the default argument (for matrix input is) correct=TRUE

- setting correct=FALSE over-rides the default

- because unnamed arguments to functions are matched in order, this argument must be named (otherwise naming the arguments is optional)

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Exercise 1 (R Basics)

- entering data

- working with objects

- simple statistical functions

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4. R graphics

n = 43

> boxplot(h05$days)

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> hist(h05$days) > plot(density(h05$days))

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Distribution functions

e.g., Normal distribution

dnorm(x, mean = 0, sd = 1) pnorm(q, mean = 0, sd = 1, lower.tail = TRUE) qnorm(p, mean = 0, sd = 1, lower.tail = TRUE) rnorm(n, mean = 0, sd = 1)

> rdat <- rnorm(30, 100, 15)

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> curve(dnorm(x), xlim=c(-4,4), col='purple')> curve(dt(x, 1), col='red', add=TRUE, lty=3)

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Exercise 2 (R Graphics)

- exploratory plots

- plot parameters

- plotting distribution functions

- plotting a serial position curve (optional)

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5. Linear models in R

> expenses <- read.csv('expenses.csv')

> t.test(majority ~ problem, data = expenses)

Welch Two Sample t-test

data: majority by problem t = -3.7477, df = 505.923, p-value = 0.000199alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval: -2044.457 -638.160 sample estimates:mean in group 0 mean in group 1 7092.712 8434.021

Data courtesy of Mark Thompson http://markreckons.blogspot.com/

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> expenses <- read.csv('expenses.csv')

> lm(majority ~ problem, data = expenses)

Call:

lm(formula = majority ~ problem, data = expenses)

Coefficients:

(Intercept) problem 7093 1341

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> lm.out <- lm(majority ~ problem, data = expenses)

> max(rstandard(lm.out))

[1] 2.629733

> AIC(lm.out)

[1] 12674.85

> lm.out$coefficients

(Intercept) problem

7092.712 1341.308 > ?lm

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> summary(lm.out)

Call:lm(formula = majority ~ problem, data = expenses)

Residuals: Min 1Q Median 3Q Max -8397.0 -3403.5 -260.9 3118.8 11543.3

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 7092.7 218.9 32.397 < 2e-16 ***problem 1341.3 357.0 3.758 0.000187 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 4395 on 644 degrees of freedomMultiple R-squared: 0.02145, Adjusted R-squared: 0.01993 F-statistic: 14.12 on 1 and 644 DF, p-value: 0.0001871

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> glm(problem~I(majority/10000), family='binomial', data = expenses)

Call: glm(formula = problem ~ I(majority/10000), family = "binomial", data = expenses)

Coefficients:

(Intercept) I(majority/10000)

-1.0394 0.6878

Degrees of Freedom: 645 Total (i.e. Null); 644 Residual

Null Deviance: 855.5

Residual Deviance: 841.6 AIC: 845.6

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> lrm.out <- glm(problem ~ I(majority/10000), family='binomial', data = expenses)

> plot(election$majority, lrm.out$fitted, col='dark green')

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Exercise 3 (Linear models)

- using a formula in a call to a model function

- linear models

- plotting a regression line

- generalized linear models (optional)

- plotting confidence bands (optional)

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6. ANOVA

> factor1 <- gl(10,4)> factor 1

[1] 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4Levels: 1 2 3 4

> library(foreign)> diag.data <- read.spss("diagram.sav", to.data.frame = T)

> diag.fit <- lm(descript ~ group, data=diag.data)

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Regression model (dummy coding)

> diag.fit

Call:

lm(formula = descript ~ group, data = diag.data)

Coefficients:

(Intercept) groupPicture groupFull Diagram groupSegmented

17.8 0.5 4.6 9.5

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aov()

> aov(diag.fit)

Call:

aov(formula = diag.fit)

Terms:

group Residuals

Sum of Squares 583.7 2440.2

Deg. of Freedom 3 36

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anova()

> summary(aov(diag.fit))

> anova(diag.fit)

Analysis of Variance Table

Response: descript

Df Sum Sq Mean Sq F value Pr(>F)

group 3 583.7 194.567 2.8704 0.04977 *

Residuals 36 2440.2 67.783

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

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Factorial ANOVA and ANCOVA

> lm(DV ~ factor1 + factor2 + factor1:factor2, data=data2)

> lm(DV ~ factor1 * factor2, data=data2)

> lm(descript ~ group + time, diag.data)

> lm(descript ~ 0 + group + scale(time, scale=F), diag.data)

> diag.ancov <- lm(descript ~ group + scale(time, scale=F), diag.data)

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Type I and Type II Sums of squares

> drop1(diag.ancov, test = 'F')

Single term deletions

Model:

descript ~ group + scale(time, scale = F)

Df Sum of Sq RSS AIC F value Pr(F)

<none> 2154.4 169.46

group 3 821.45 2975.9 176.38 4.4484 0.009478 **

scale(time, scale = F) 1 285.78 2440.2 172.44 4.6427 0.038149 *

- sequential sums of squares (Type I) tests are given by default

- hierarchical sums of squares (Type II) tests are given by the drop1() function

[Software such as SAS and SPSS defaults to unique (Type III) SS]

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Repeated measures (within-subjects) factors

- a weakness in the base R installation, but easily done using packages such as ez, nlme or lme4 (the latter two able to fit a wider range of repeated measures models)

> pride.long <- read.csv("pride_long.csv")

> pride.mixed <- aov(accuracy ~ emotion*condition + Error(factor(participant)/emotion), pride.long)

> summary(pride.mixed)

[Software such as SAS and SPSS defaults to unique (Type III) SS]

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Multiple comparisons and contrasts

- you can run contrasts by changing default contrasts in R

- Fisher LSD, Tukey’s HSD and p adjustment (e.g., Hochberg, Holm, Bonferroni, FDR) in R base installation

- powerful modified Bonferroni (e.g., Shaffer, Westfall) and general linear contrasts available in multcomp package

- flexible contrast and estimable functions in gmodels package

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Exercise 4 (ANOVA)

- factors

- aov() and anova()

- ANCOVA

- repeated measures (within-subjects)

- contrasts and multiple comparisons (optional)

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7. Writing your own functions

sd.pool <- function(sd.1, n.1, sd.2, n.2 = n.1){num <- (n.1-1)*sd.1^2+(n.2-1)*sd.2^2denom <- n.1+n.2-2output <- sqrt(num/denom)output}

> sd.pool(6.1, 20, 5.3, 25)[1] 5.66743

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Exercise 5 (Write a function)

- pick a simple statistical procedure

- write a function to automate it

e.g., pooled standard deviation

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8. Simulation and bootstrapping

- distribution functions

> rnorm(100, 10, 1)

- R boot package

- sample() and replicate()

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Bootstrap CIs for a median or timmed mean

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The bootstrap (and other Monte Carlo methods)

Bootstrapping involves repeatedly resampling with replacement from a data set

e.g., Data set = {0,1}

7 simulated samples:

{1,0}, {0,1}, {0,0}, {0,0}, {0,1}, {0,0}, {1,1}

Bootstrapping effectively treats the sample distribution as the population distribution

> replicate(7, sample(x,2,replace=TRUE))

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Bootstrapping using R

> library(boot)> x1.boot <- boot(samp,function(x,i) median(x[i]),R=10000)> boot.ci(x1.boot)

BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONSBased on 10000 bootstrap replicates

CALL : boot.ci(boot.out = x1.boot)

Intervals : Level Normal Basic 95% (-0.1747, 0.6948 ) (-0.1648, 0.6614 )

Level Percentile BCa 95% (-0.1538, 0.6724 ) (-0.2464, 0.6620 ) Calculations and Intervals on Original Scale

Normal 95% CI for mean (using t) [-2.00, 1.33]

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Exercise 6 (Bootstrapping)

- resample data

- construct a simple percentile bootstrap

- run a BCa bootstrap using the boot package (optional)

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9. Advanced statistical modelling in R

Multilevel modeling nlme package

lme4 package

MCMCglmm package

Meta-analysis meta package

metafor package

… and many, many more other packages

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metafor package

> m.e <- c(10.7, 16.24, 10.03, 3.65, 5.73)

> n.e <- c(31, 57, 9, 17, 10)

> m.c <- c(2.87, 6.83, 7.18, 4.65, 7.47)

> n.c <- c(14, 52, 9, 18, 10)

> sd.pooled <- c(7.88, 9.84, 8.67, 6.34, 5.74)

> diff <- m.c - m.e

> se.diffs <- sd.pooled * sqrt(1/n.e + 1/n.c)

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> install.packages('metafor')> library(metafor)> meta.out <- rma(yi=diff, sei=se.diffs, method = 'FE')> meta.out

Fixed-Effects Model (k = 5)

Test for Heterogeneity: Q(df = 4) = 21.0062, p-val = 0.0003

Model Results:

estimate se zval pval ci.lb ci.ub -4.1003 1.0750 -3.8141 0.0001 -6.2073 -1.9933 ***

---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

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> funnel(trimfill(meta.out))

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> forest(rma(yi=diff, sei=se.diffs))

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