# James Evans - 1910316 # Statistical Programming # Assignment 1 #TASK 1 #1. Create a vector v1 that contains the values 1 to 16. # Set up and display vector v1, using concatenate, which contains the values 1 to 16. v1<-c(1:16) v1 #2. Add the values 17 to 20 to the end of v1. # Change and display vector v1, using concatenate, which will now contain the values 1 to 20. v1<-c(v1,17:20) v1 #3. Delete elements 5, 6, 7 and 13 from v1. You should use a one method to remove elements 5, 6, 7 and another method to remove element 13. # To delete elements 5,6 and 7; first concatenate elements 1 to 4, and attach elements 8 to 20 to the right hand side. v1<-c(v1[1:4],v1[8:20]) v1 # To then delete element 13 (which now has the value of 16; set individual elements as vector v1, excluding element 13, up to and including the final element (currently 17). v1<-c(v1[1],v1[2],v1[3],v1[4],v1[5],v1[6],v1[7],v1[8],v1[9],v1[10],v1[11],v1[12],v1[14],v1[15],v1[16],v1[17]) v1 #4. Using v1, create a new vector v2 that contains the 1st to 5th and 9th to 11th elements of v1. # Concatenate elements 1 to 5 with 9 to 11 from v1, and use these to create v2. v2<-c(v1[1:5],v1[9:11]) v2 #5. Calculate v1*v2 and display the results. In your comments, explain how R adds two vectors of different lengths. # Use the multiply operation (*) to multiply the new vectors and create a new vector ans1 to contain the answer. Display ans1. ans1<-v1*v2 ans1 # When adding, or multiplying two vectors of differing lengths, R reaches the and off the shorter of the two vectors, then returns to the start of it, whilst continuing along the longer vector. # I.e When adding a vector of length 16 another of length 8; element 9 of the longer vector is added/multiplied to/with element 1 of the shorter vector. Element 10 with element 2 and so on. #6. Create a new vector v3 by selecting all even elements of v1 using a filter. # Filter v1 so that all elements, which when modulated to 2 equal 0 (are even), are used to create v3. v3<-v1[v1%%2==0] v3 #7. Using a loop, create a new vector v4 the same length as v1, that contains the value TRUE if the corresponding element of v1 has an integer square root and the value FALSE otherwise. #HINT: You will need to use the sqrt() and round() functions. #Create v4 and set it to NULL v4<-NULL # Main Loop # Using seq to go through each element in v1. for (i in seq(v1)){ # Calculate square root of v1[i], check to see whether it is an integer, if it is return TRUE, otherwise return FALSE. #For each result; concatenate V4 as it stands, together with the latest result. v4<-c(v4,ifelse(sqrt(v1[i])==round(sqrt(v1[i]), digits=0), TRUE,FALSE)) } #Display v4 v4 #8. Create a matrix m1 with 4 rows from vector v1. # Use the matrix operation to create the matrix. # Use v1 as the values to be included in the matrix. # Set the number of columns as 4. m1<-matrix(v1,nrow=4) m1 #9. Create a matrix m2 with 2 columns from vector v2. # Use the matrix operation to create the matrix. # Use v2 as the values to be included in the matrix. # Set the number of rows as 4=2. m2<-matrix(v2,ncol=2) m2 #10. Swap all of the elements in row 3 of m1 with the elements in column 2 of m1. # Create a vector h1, which contains the elements within row 3 of m1. h1<-m1[3,] # Create a vector h1, which contains the elements within row 3 of m1. h2<-m1[,2] # Insert the vector h2 into row 3 of m1. m1[3,]<-h2 # Display m1 following the initial replacement. m1 # Insert the vector h1 into column 2 of m1. m1[,2]<-h1 # This method ensures that the elements which are subsequently entered into the corresponding row/column is as it was at the starting position. # The "swap" that has been requested in the question cannot be completed without overwriting a particular element. # In this case, as we have first replaced row 3 with elements from column 2; then replacing column 2 overwrites one change initially made to row 3. # In effect m1[3,2] changes from 10 to 9 and finally to 14. # Display m1 following the subsequent replacement. m1 #11. Create and display a new matrix m3 from m1 and m2 where m3 contains all of the columns of both source matrices. # Use the column bind function to bind the matrix m2 to the right hand side of m1, and subsequently create m3. m3<-cbind(m1,m2) # Display m3. m3 #12. Replace all values in m3 that are in (column 1 or column 3) and in (row 2 or row 3) with double their value. # Using a loop throughout all rows of the matrix:- #For each column within each row:- #If column = 1, elements are replaced with double their value. #If column = 3, elements are replaced with double their value. #If row = 2, elements are replaced with double their value. #If row = 3, elements are replaced with double their value. #If none of the conditions above are satisfied; the initial value is brought back into the matrix. for (i in seq(m3[,1])) { for(j in seq(m3[1,])){ ifelse(j==1,m3[i,j]<-m3[i,j]*2, ifelse(j==3,m3[i,j]<-m3[i,j]*2, ifelse(i==2,m3[i,j]<-m3[i,j]*2, ifelse(i==3,m3[i,j]<-m3[i,j]*2, m3[i,j]<-m3[i,j] )))) } } #Display m3 m3 #13. Remove row 2 of m3. # Use the "-" sign to drop the second row using reassignment as follows:- m3<-m3[-2,] #Display m3 m3 #14. Label the columns of m3 as "c1", "c2", "c3", "c4", "c5", and "c6". #Use the colnames function to rename the columns:- colnames(m3)<-c("c1", "c2", "c3", "c4", "c5","c6") #Display m3 m3 #15. Create a vector mmeans that contains the means of each column. #Use the apply function to generate the means of each column, and enter them into the new vector "mmeans". #"Mean" is a function which exists in R already,m and so there is no need to create this first. mmeans<-apply(m3,2,mean) #Display mmeans mmeans #16. Create a barplot of mmeans. #The barplot function is used, and the matrix "mmeans" is called to act as the values and x axis names. #Colour the bars green. #Label the x axis "Column". #Label the y axis "Mean". barplot(mmeans, col = "Green", xlab = "Column", ylab = "Mean")