Lab 2
Nasif Hossain
2022-07-07
Description
Lab 2 is based on the creation of a vector, matrix, list of variables, and dataframe. Finally, we will investigate the association of two variables by a simple linear regression model as well as a scatter plot for visual inspection.
Data Structures
Data structures are used to store data. R supports multiple data structures but there are 4 fundumetnal data structures that an R practitioner needs to know. They are:
1. Vectors: The fundamental (atomic)
data structure. R vectorizes elements and stores them in
vectors. Unlike MATLAB, which stores data in row-vectors, R
stores data in column vectors. Vectors can only store one particular
data type, i.e. a vector can only hold character data type but not a
mixture of character and numeric data types.
x <- c(55, 65, 90, 86) # create a vector through concatenation by using c()
x## [1] 55 65 90 86y <- c(75, 60, 92, 80)
y## [1] 75 60 92 80z <- x + y # vector addition
z## [1] 130 125 182 166u <- c(200, 300, 400, 450)
u## [1] 200 300 400 450z-u## [1] -70 -175 -218 -284Names<- c("Rob", "James", "Alberto", "Maya") # Character vectors
Names## [1] "Rob" "James" "Alberto" "Maya"- Matrices: Matrices are built by either combining vectors row wise
(
rbind()) or by using columns (cbind()). Matrices can also be created by using thematrix()function (use ?matrix to look up the arguments). Similar to vectors, matrices also only support one single data type in a single matrix.
m <- matrix(1:12, nrow = 3, ncol = 4, byrow = FALSE)
m## [,1] [,2] [,3] [,4]
## [1,] 1 4 7 10
## [2,] 2 5 8 11
## [3,] 3 6 9 12n <- cbind(x, y)
n## x y
## [1,] 55 75
## [2,] 65 60
## [3,] 90 92
## [4,] 86 80v<- rbind(x, y)
v## [,1] [,2] [,3] [,4]
## x 55 65 90 86
## y 75 60 92 80Some basic matrix operations:
t(n) # create a transpose## [,1] [,2] [,3] [,4]
## x 55 65 90 86
## y 75 60 92 80n %*% t(n) # create a square matrix## [,1] [,2] [,3] [,4]
## [1,] 8650 8075 11850 10730
## [2,] 8075 7825 11370 10390
## [3,] 11850 11370 16564 15100
## [4,] 10730 10390 15100 13796solve((t(n) %*% n)) # inverse## x y
## x 0.002319781 -0.002232726
## y -0.002232726 0.0021904503. Lists: Lists are the building blocks of Data
Frames. They can hold multiple different data types in a single list.
Use as.list() to convert to a list or list() to transform
vectors/matrices of different data types into a single list.
# Unlike vectors and matrices, lists can hold multiple datatypes
list.1 <- list(names, x, y)
list.1; str(list.1)## [[1]]
## function (x) .Primitive("names")
##
## [[2]]
## [1] 55 65 90 86
##
## [[3]]
## [1] 75 60 92 80## List of 3
## $ :function (x)
## $ : num [1:4] 55 65 90 86
## $ : num [1:4] 75 60 92 80list.2 <- list(names, n, m)
list.2; str(list.2)## [[1]]
## function (x) .Primitive("names")
##
## [[2]]
## x y
## [1,] 55 75
## [2,] 65 60
## [3,] 90 92
## [4,] 86 80
##
## [[3]]
## [,1] [,2] [,3] [,4]
## [1,] 1 4 7 10
## [2,] 2 5 8 11
## [3,] 3 6 9 12## List of 3
## $ :function (x)
## $ : num [1:4, 1:2] 55 65 90 86 75 60 92 80
## ..- attr(*, "dimnames")=List of 2
## .. ..$ : NULL
## .. ..$ : chr [1:2] "x" "y"
## $ : int [1:3, 1:4] 1 2 3 4 5 6 7 8 9 10 ...4. Data Frames: Data frames are the fundamental data
structure to hold numerous data types. Base R performs very
well with data frame, however the modern “tidyverse” universe uses
something called a “tibble”.
Year <- seq(1981, 1997, by = 1)
Observations <- sample(1000, 17, replace = TRUE) # Generate 17 random numbers between 0-1000
df.1 <- data.frame(Year, Observations) # Create a data frame
Year## [1] 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995
## [16] 1996 1997Observations## [1] 864 70 888 741 785 570 788 740 697 182 297 425 905 264 294 873 703Observations## [1] 864 70 888 741 785 570 788 740 697 182 297 425 905 264 294 873 703df.1## Year Observations
## 1 1981 864
## 2 1982 70
## 3 1983 888
## 4 1984 741
## 5 1985 785
## 6 1986 570
## 7 1987 788
## 8 1988 740
## 9 1989 697
## 10 1990 182
## 11 1991 297
## 12 1992 425
## 13 1993 905
## 14 1994 264
## 15 1995 294
## 16 1996 873
## 17 1997 703class(df.1) # check the class of object## [1] "data.frame"str(df.1) # check the structure of the object and its respective columns## 'data.frame': 17 obs. of 2 variables:
## $ Year : num 1981 1982 1983 1984 1985 ...
## $ Observations: int 864 70 888 741 785 570 788 740 697 182 ...Preliminary Statistical Analysis
Now, let’s look at doing some preliminary statistical analysis.
linear.model.1 <- lm(Observations ~ Year, data = df.1) # perform linear regression
summary(linear.model.1) # check the summary and the coefficients##
## Call:
## lm(formula = Observations ~ Year, data = df.1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -576.7 -253.5 109.6 179.5 342.2
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 15759.419 27970.539 0.563 0.581
## Year -7.625 14.063 -0.542 0.596
##
## Residual standard error: 284.1 on 15 degrees of freedom
## Multiple R-squared: 0.01922, Adjusted R-squared: -0.04616
## F-statistic: 0.294 on 1 and 15 DF, p-value: 0.5956plot(df.1$Year, df.1$Observations, col = "red", xlab = "Year", ylab = "Observations")
abline(linear.model.1) # Create a Linear Regression line using abline()
par(mfrow = c(2,2)) # create a 2*2 matrix for the plots
plot(linear.model.1) # create a plot of the residuals
Excercises
Excercises-1:
Create a data-frame called “Results” by combining the vectors “Names”, “x”, and “y”. Rename “x” as “Result_1” and “y” as “Result_2”.
Answer-1:
Results<-data.frame(Names,x, y)
Results## Names x y
## 1 Rob 55 75
## 2 James 65 60
## 3 Alberto 90 92
## 4 Maya 86 80Rename column of dataframe
# way-1
colnames(Results)<-c("Names", "Result_1", "Result_2")
Results## Names Result_1 Result_2
## 1 Rob 55 75
## 2 James 65 60
## 3 Alberto 90 92
## 4 Maya 86 80# way-2
Results<-data.frame(Names,x, y)
names(Results)[2]<-paste("Result_1")
names(Results)[3]<-paste("Result_2")
Results## Names Result_1 Result_2
## 1 Rob 55 75
## 2 James 65 60
## 3 Alberto 90 92
## 4 Maya 86 80# way-3: by using dplyr
Results<-data.frame(Names,x, y)
library(dplyr)
Results<-Results %>%
rename(
Result_1 = x,
Result_2 = y
)
Results## Names Result_1 Result_2
## 1 Rob 55 75
## 2 James 65 60
## 3 Alberto 90 92
## 4 Maya 86 80Excercises-2:
Create a Scatter-Plot of “Result_1” on the x-axis “Result_2” on the y-axis.
Answer-2:
plot(Results$Result_1, Results$Result_2,
main="Scatter Plot of Result_1 and Result_2",
xlab = "Result_1",
ylab="Result_1")
Excercises-3:
Perform a Linear Regression by regressing “Result_2” on “Result_1”. (Hint: use the lm() function)
Answer-3:
model<-lm(Result_1~Result_2, data =Results )Excercises-4:
Plot the residuals and summarize the results.
Answer-4:
par(mfrow = c(2,2)) # create a 2*2 matrix for the plots
plot(model) # create a plot of the residuals
summary(model)##
## Call:
## lm(formula = Result_1 ~ Result_2, data = Results)
##
## Residuals:
## 1 2 3 4
## -17.449 5.850 2.480 9.119
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.9559 49.4618 0.120 0.915
## Result_2 0.8866 0.6374 1.391 0.299
##
## Residual standard error: 14.63 on 2 degrees of freedom
## Multiple R-squared: 0.4917, Adjusted R-squared: 0.2376
## F-statistic: 1.935 on 1 and 2 DF, p-value: 0.2988Excercises-5:
Interpret your results.
Answer-5:
From the summary statistics of the linear model, we found that
Result_2 is positively associated with
Result_1 because the estimated coefficient is positive
(0.8866). However, the result is not significate (p-value: 0.299).
According to the fitted model, less than half of the variability can be
explained (R-squared: 0.4917, Adjusted R-squared: 0.2376)