Lecture 4 demo code

Remove element (slide 4)

Loop approach

Let’s try using a loop first. A preliminary idea is as follows: 1. Construct a result vector result 2. Loop through the input v - if v[i] does not equal target, add it to result - otherwise, do not add it to result (do nothing) 3. Return result

Here is this idea implemented:

remove_elt <- function(v, target) {
  result <- numeric(0)
  for (i in seq_along(v)) {
    if(v[i] != target) {
      result <- c(result, v[i])
    }
  }
  return(result)
}
remove_elt(c(14, 14, 7, 7, 14, 10), 14)

## [1]  7  7 10

Vectorized approach

Instead of a loop, we can do a vectorized comparison and then filter for the desired values of v.

remove_elt <- function(v, target) {
  v[v != target]
}
remove_elt(c(14, 14, 7, 7, 14, 10), 14)

## [1]  7  7 10

In the above, v != target is a vector of logicals which we use to subset v itself. Please make sure you are comfortable with this operation.

Print Square (slide 5)

Matrix approach

The first approach is clever: just print a matrix.

print_square <- function(n) {
  m <- matrix("*", nrow = n, ncol = n)
  print(m)
}

print_square(5)

##      [,1] [,2] [,3] [,4] [,5]
## [1,] "*"  "*"  "*"  "*"  "*" 
## [2,] "*"  "*"  "*"  "*"  "*" 
## [3,] "*"  "*"  "*"  "*"  "*" 
## [4,] "*"  "*"  "*"  "*"  "*" 
## [5,] "*"  "*"  "*"  "*"  "*"

Think about how recycling occurs so that a single * becomes n^2 *s.

Loop approach

Let’s build a vector of n *s and then use a loop to print it n times.

print_square <- function(n) {
  asts <- rep("*", n)
  for (i in seq_len(n)) {
    print(asts)
  }
}
print_square(4)

## [1] "*" "*" "*" "*"
## [1] "*" "*" "*" "*"
## [1] "*" "*" "*" "*"
## [1] "*" "*" "*" "*"

Print triangle (slide 6)

A matrix approach is less straightforward in this case. But we can easily adapt the above loop. Now, the number of asterisks depends on the loop index.

print_triangle <- function(n) {
  for (i in seq_len(n)) {
    asts <- rep("*", i)
    print(asts)
  }
}
print_triangle(4)

## [1] "*"
## [1] "*" "*"
## [1] "*" "*" "*"
## [1] "*" "*" "*" "*"

Max vector (slide 7)

Loop with comparison

We want to compare nums1 and nums2 element-wise, so it makes sense to loop over them. The vectors are guaranteed to be of length n, so we can use this.

max_vec <- function(nums1, nums2, n) {
  result <- vector(length = n) # this initializes an "empty" vector with length n
  for (i in seq_len(n)) {
    if (nums1[i] < nums2[i]) {
      result[i] <- nums2[i] # Since nums2[i] is larger, place it into the result
    } else {
      result[i] <- nums1[i] # Since nums1[i] is larger, place it into the result
    }
    # Note that if they are equal, it does not matter which we put in.
  }
  return(result)
}
max_vec(1:4, c(0, 27, -5, 19), 4)

## [1]  1 27  3 19

Loop with builtin max

It is always good to check if R already implements some basic operation. In this case, max(a, b) returns the larger of a and b.

max_vec <- function(nums1, nums2, n) {
  result <- vector(length = n)
  for (i in seq_len(n)) {
    result[i] <- max(nums1[i], nums2[i]) # use the builtin max function
  }
  return(result)
}
max_vec(1:4, c(0, 27, -5, 19), 4)

## [1]  1 27  3 19

The above saves us from explicitly writing a branch.

pmax

A even better way is to use Google to find the pmax function, which does everything for us.

pmax(1:4, c(0, 27, -5, 19))

## [1]  1 27  3 19

Small Count Simple (slide 8)

It is possible to do this with a loop. But let’s use a vectorized approach:

small_count_simple <- function(v, target) {
  sum(v < target)
}
small_count_simple(c(10, 15, -2, 5), 11)

## [1] 3

v < target results in a logical vector. Summing up the result gives the number of TRUE elements.

Small Count (slide 9)

Loop over a result vector and fill in the values using the approach from the previous problem.

small_count <- function(v) {
  result <- vector(length = length(v))
  for (i in seq_along(v)) {
    result[i] <- sum(v < v[i])
  }
  return(result)
  small_count(c(12, 100, -3, -12))
}

Two Sum (slide 10)

For this problem, we must be able to enumerate pairs of indices going from 1 to length(v). In other words, if v has 4 elements, we need to check the following pairs:

(i, j) <= (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)

Then if we look at v[i] + v[j], we will see the sum of all pairs of elements of v. Think: why were (2, 1), (4, 2), ... not included?

Loop approach

The idea is to use two loops: i going from 1 to length(v) and j going from 1 to i-1. This ensures we go over each pair exactly once (why?).

two_sum <- function(v, target) {
  i <- 0
  j <- 0
  for (i in seq_along(v)) {
    for (j in seq_len(i-1)) {
      if (v[i] + v[j] == target) {
        return(c(i, j))
      }
    }
  }
}
two_sum(c(2, 7, 11, 15), 9)

## [1] 2 1

First Even (slide 11)

The expression (v %% 2) == 0 gives a vector of logicals indicating TRUE if the entry of v is even and FALSE otherwise. Calling which on this logical vector gives all the indices of TRUE, which are all the indices of even numbers of v. Then index with [1] to find the first such index.

first_even <- function(v) {
  w_idx <- which((v %% 2) == 0)[1]
  v[w_idx]
}
first_even(c(-1, -4, 0, 2))

## [1] -4

It may be clearer to do the indexing after subsetting v: Below, w_idx is now a vector of indices, which we use to subset v.

first_even <- function(v) {
  w_idx <- which((v %% 2) == 0)
  (v[w_idx])[1]
}
first_even(c(-1, -4, 0, 2))

## [1] -4

Last Negative (slide 12)

This is an exercise; please do this on your own. If you are stuck, talk to me.

Any and all (slide 13)

This is also an exercise; please do this on your own. If you are stuck, talk to me.