# sample(x,k) generates a random permutation of k objects from the vector x. That is, all k choices are different
# Note every element is chosen at most once.
# x is a vector with 5 elements.
> x = 1:5
> x
[1] 1 2 3 4 5
# randomly sample 3 of the elements of x.
> sample(x,3)
[1] 2 5 1
> sample(x,3)
[1] 3 1 4
> sample(x,3)
[1] 1 2 5
# randomly sampling 5 elements of x is a permutation of all 5 elements.
> sample(x,5)
[1] 1 3 2 5 4
> sample(x,5)
[1] 2 3 1 4 5
# You can't pick more than 5 different elements from a set of 5 things, so R gives an error.
> sample(x,6)
Error in sample.int(length(x), size, replace, prob) :
cannot take a sample larger than the population when 'replace = FALSE'
Allowing repeated elements
# Sometimes you want to allow repeats. For example when we roll a die repeatedly we expect to see numbers repeating. We can think of this as picking a random element from a set and then putting it back, i.e. replacing it, so it can be drawn again.
# In R we do this by setting the optional argument replace=TRUE
# Note that we get can get repeated values
> sample(x,3,replace=TRUE)
[1] 3 1 4
> sample(x,3,replace=TRUE)
[1] 4 5 4
> sample(x,3,replace=TRUE)
[1] 2 1 5
> sample(x,5,replace=TRUE)
[1] 5 1 5 1 4
> sample(x,5,replace=TRUE)
[1] 3 5 5 2 1
> sample(x,5,replace=TRUE)
[1] 4 2 3 2 1
# Now there is no problem asking for more than 5 things from a set of 5 elements.
> sample(x,12,replace=TRUE)
[1] 1 1 2 1 4 2 3 4 1 2 4 5
> sample(x,12,replace=TRUE)
[1] 3 2 5 1 4 2 4 1 4 5 3 1
# To generate an m x n array of random values we can use the function sample function followed by the matrix function.
# Let's simulate rolling a die.
# We use 1:6 to make vector (1,2,3,4,5,6).
# Generate a 3 x 4 array of random dice rolls.
> y = sample(1:6, 12, replace=TRUE)
> matrix(y,3,4)
[,1] [,2] [,3] [,4]
[1,] 1 3 6 1
[2,] 2 2 3 6
[3,] 2 1 4 3
# Or we could make it a 2 x 6 array.
> matrix(y,2,6)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 2 2 6 4 6
[2,] 2 3 1 3 1 3
# First let's simulate rolling 1 die 3 times and checking if one of the rolls was a 6.
# First we use sample to generate 3 samples from 1:6
> x = sample(1:6,3, replace=TRUE)
> x
[1] 6 4 1
# To check if any of the 3 rolls are 6 we use the following command. It returns a vector of TRUE or FALSE depending on whether that entry of x is 6 or not. Compare the result with the value of x above.
> x == 6
[1] TRUE FALSE FALSE
# We can also see which elements of x are less than 6
> x < 6
[1] FALSE TRUE TRUE
# Now let's roll the die 5000 times and see what fraction of the rolls give 6. We expect about 1/6 of them will.
# Simulate 1000 rolls
> x = sample(1:6,1000, replace=TRUE)
# x == 6 gives a vector of TRUE or FALSE. R is clever, when we sum the vector TRUE counts as 1 and FALSE counts as 0. So the sum is the number of 6's. In this case 168.
> sum(x == 6)
[1] 168
# Divide by 1000 to get the fraction of 6's.
> sum(x == 6)/1000
[1] 0.168
# Compare that with the theoretical value of 1/6.
> 1/6
[1] 0.1666667
# Not bad!
# Now let's estimate the probability of getting a 6 in 4 rolls.
# Goal: estimate the probability of getting at least one 6 in 4 rolls.
# Experiment: roll 1 die 4 times and check for a 6.
# Repeat the experiment 10 times and see what fraction of times this happens.
# So you can see all the commands at once, we'll show them all and then explain them later. For commenting, we'll put a command number before each '>'
1. > x = matrix(sample(1:6, 4*10, replace=TRUE), nrow=4, ncol=10)
2. > x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 5 6 1 1 3 6 1 4 6 1
[2,] 3 6 6 4 3 2 2 1 6 5
[3,] 2 3 5 2 5 3 4 6 2 5
[4,] 2 6 6 5 5 3 1 4 5 3
3. > y = (x==6)
4. > y
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] FALSE TRUE FALSE FALSE FALSE TRUE FALSE FALSE TRUE FALSE
[2,] FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE
[3,] FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE
[4,] FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
5. > z = colSums(y)
6. > z
[1] 0 3 2 0 0 1 0 1 2 0
7. > z > 0
[1] FALSE TRUE TRUE FALSE FALSE TRUE FALSE TRUE TRUE FALSE
8. > sum(z > 0)
[1] 5
9. > mean(z > 0)
[1] 0.5
# Command 1: Generate a 4 by 10 random array. Each column represents one experimental trial of 4 rolls. The 10 columns represent the 10 trials.
# Command 3: See which of the rolls are 6's. Note the double equals to check which rolls were 6. The result y is an array of TRUE or FALSE. You can check that it picks out the 6's.
# Command 5: Sum each of the columns. The result is then number of 6's in each
trial (column).
# Command 7: If the column sum is greater than 0 there was at least one TRUE in the column, that is at least one 6. This command returns a vector of TRUE or FALSE representing whether or not the experiment yielded a 6.
# command 8: sum the vector in command 7 to get the number of trials that yielded a 6. We see that 5 out of 10 trials did.
# command 9: The mean function is just the sum divided by the number of trials. This is just 5/10 = .5. Half the trials yielded a 10.
# Let's repeat this but with 1000 trials instead of 10. Without all the comments it's pretty short.
> x = matrix(sample(1:6, 4*1000, replace=TRUE), nrow=4, ncol=1000)
> y = (x==6)
> z = colSums(y)
> sum(z > 0)
[1] 525
> mean(z>0)
[1] 0.525
Our estimate of the probability of at least one 6 in 4 rolls is .525. This is pretty close to the theoretical value of .518.
The dim() function
# We can always check that x has 4 rows and 1000 columns using the dim() function.
> dim(x)
[1] 4 1000
# Goal: estimate the probability of getting a sum of 7 when rolling two dice. 1. > x = matrix(sample(1:6,20000,replace=TRUE), 2,10000) 2. > y = colSums(x) 3. > mean(y == 7) [1] 0.1658 # Command 1: we create 1000 columns with 2 rows. That is, we run 10000 experiments of rolling 2 dice. # Command 2: we sum each of the columns, i.e., we sum the two dice. # Command 3: we find the fraction of sums that are 7. # Note, this is essentially the exact answer of 1/6. Exercise: Try to estimate the probability of two sixes when rolling two dice.