Assignment10

Question 1: Using for loops to determine # zeros in a numerical vector

Vector <- c(0,0,1,1,0,1,0,1,1,1)
head(Vector)

## [1] 0 0 1 1 0 1

Vec0<-vector(mode="logical")
for (i in 1:length(Vector)) {
  if (Vector[i] == 0) {
    Vec0[i] <- 1 
     }
  else {
    Vec0[i]<- 0
  }
}
  sum(Vec0)

## [1] 4

Question 2: Using subsetting rather than for loops

Vector <- c(0,0,1,1,0,1,0,1,1,1)

length((Vec0) <= subset(Vector, Vector == 0))

## Warning in (Vec0) <= subset(Vector, Vector == 0): longer object length is
## not a multiple of shorter object length

## [1] 10

Question 3: Maximum difference between all pairs

From this point forward, script courtesy of Alex Looi, annnotations done by Erin L. Keller

X = rnorm(20, mean = 0, sd = 10) # sampling 20 random numbers with mean of 0

max_diff = function(X){
  
  pairs = expand.grid(X,X) # identifying all pairwise comparisons in X
  diffs = abs(pairs$Var1 - pairs$Var2) # calculating the absolute difference between all pairs
  
  max_difference = max(diffs, na.rm = T) # determining max difference by using the max function to determine the largest diference present in the diffs vector
  
  return(max_difference) # returns the max difference
  
}
max_diff(X)

## [1] 41.14262

Question 4: Modifying output of Question 3 to contain 3 elements: 1st is the pair of values that give the greatest difference, 2nd is the position of these elements, and 3rd is the maximum distance from this pair

X = rnorm(20, mean = 0, sd = 10)

max_diff = function(X= rnorm(20, mean = 0, sd = 10)){
  
  pairs = as.matrix(expand.grid(X,X))
  diffs = abs(pairs[,1] - pairs[,2]) # up until this point in the script is the same as Q3
  
  vec_max = which.max(diffs) # this is selecting the maximum difference from all pairs
  
  pair_max = pairs[vec_max,] # this is identify which pair of values generated largest difference
  
  max_difference = max(diffs, na.rm = T) # This is pulling out the maximum difference between two values
  
  var1_vec = which(pair_max[1] == X) # this is determining the position of the first value of the pair
  var2_vec = which(pair_max[2] == X) # this is determinig the position of the second value of the pair
  pair_vec = c(var1_vec, var2_vec) # this is concatenating the above output and contains the position of these values
  
  output = list(pair_max, pair_vec, max_difference) # ouput will be a list containing the 3 (really 4) values
  
  return(output)
  
}
max_diff()

## [[1]]
##      Var1      Var2 
## -24.57302  17.65702 
## 
## [[2]]
## [1] 13  4
## 
## [[3]]
## [1] 42.23004

Question 5: Keep only maximum difference but cycle through all possibilities

X = rnorm(20, mean = 0, sd = 10)

max_diff = function(X= rnorm(20, mean = 0, sd = 10)){
  
  pairs = expand.grid(X,X) # making pairwise comparisons
  diffs = abs(pairs$Var1 - pairs$Var2) # calculating absolute difference between every pairwise comparison
  temp_max = 0 # creating a temporary maximum vector containing what will be the largest difference
  for(v in 1:length(diffs)){ # for loop in which each pairwise comparison is being evaluated (length(diffs) means that every difference value in the vector will be evaluated)
    
    if(temp_max <= diffs[v]){ # if statement saying that if the difference value being evaluated is less than or equal to what is already in the temp_max file
      temp_max = diffs[v] # so if this is true (new diff is the greater than what was previously stored) then we replace what is in temp_max with the new maximum difference
    }
  }
  return(temp_max)
}
max_diff(X)

## [1] 42.31205

Question 6: Multiplying two matrices together

M1 = matrix(c(1:6),2,3, byrow = T) # generating two matrices
M2 = matrix(c(7:12),3,2, byrow = T)

Mat_mult = function(M1, M2){ # creatig function that will take the two matrices as inputs and multiply them together
  # create a new matrix table
  M3 = matrix(0, nrow(M1), ncol(M2)) # making empty matrix with rows = # rows in M1 and columns = # columns in M2
  if (nrow(M1) != ncol(M2)){ # if the row and column lengths are not equal, give the following warning:
    cat("input matrices are not correct dimensions", "\n")
  } else { # if they are the same length, do the following:
    for (nr in 1:nrow(M1)){ # for the number of rows in 1:length of M1 and for the number of columns in 1:length of M2, take the product of the values at the corresponding position in the matrix
      for (nc in 1:ncol(M2)){
        cell_val = sum(M1[nr,]*M2[,nc])
        M3[nr,nc] = cell_val # output will be a matrix containing the product of the two matrices
      }
    }
  }
  return(M3)
}

Mat_mult(M1, M2)

##      [,1] [,2]
## [1,]   58   64
## [2,]  139  154

# Using R funtion formultiplying matrices:
M1%*%M2 # returns same result!

##      [,1] [,2]
## [1,]   58   64
## [2,]  139  154

Question 7:

Write a function that takes as input two integers representing the number of rows and columns in a matrix. The output is a matrix of these dimensions in which each element is the product of the row number x the column number

M1 = matrix(c(1:6),2,3, byrow = T)

Mat_thing = function(nr, nc){ # number of rows and number of columns are the two input variables
  M = matrix(0, nr, nc) # constructing a matrix filled with 0's that has the dimensions specified by nr and nc
  for (r in 1:nr){ # for each value r from 1:nr (number of rows)...
    for (c in 1:nc){ # for each value r from 1:nc *(number of columns)...
      M[r,c] = r*c # multiply the number of the row by the number of the column and put it into matrix M at the position corresponding to the row/column
    } 
  }
  return(M)
}
Mat_thing(4, 6)

##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    2    3    4    5    6
## [2,]    2    4    6    8   10   12
## [3,]    3    6    9   12   15   18
## [4,]    4    8   12   16   20   24