lab 1 · loops (R) — source
A. Stamm 
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# ============================================================
# Lab 1: Loops, conditionals & the CLT by simulation
# The R Statistical Language · Session 3
#
# Work through the exercises. Comment your thought process
# as you go: a reproducible script is one a stranger can read.
# ============================================================
## Part A: for loops ---------------------------------
# A1. Print the squares of 1:10 with a for() loop.
# A2. Use a loop to sum the integers 1:1000. Confirm the result
# equals sum(1:1000).
# A3. Pre-allocate: compute the first 20 Fibonacci numbers in a
# pre-allocated vector fib (hint: fib[1] <- 1; fib[2] <- 1;
# then for i in 3:20, fib[i] <- fib[i-1] + fib[i-2]).
## Part B: while loops ---------------------------------
# B1. Start x <- 1 and keep doubling while x < 1000. How many
# doublings before it stops? (Add a counter.)
## Part C: conditionals ---------------------------------
# C1. Write an if/else/else-if block that grades a numeric score:
# >= 90 "A", >= 80 "B", >= 70 "C", else "D". Test it.
# C2. Use ifelse() on temps <- c(-5, 4, 12, 25, 31) to label each
# element as "freezing" (<=0), "cold" (<=15), "mild" (<=25),
# else "hot". (Nest ifelse() calls.)
## Part D: the Central Limit Theorem by simulation ---------------------------------
# D1. Simulate the sampling distribution of the mean of rexp(n=30):
# out <- numeric(R) with R <- 2000
# for (i in 1:R) out[i] <- mean(rexp(30))
# Plot a histogram of out. It should look bell-shaped even though
# the source is exponential. Why?
# D2. Now do the same for rbinom(30, 1, 0.2) (a very skewed source).
# Does the histogram still look bell-shaped? What if you instead
# take n=5? (Less averaging -> less bell-shaped.)
# D3. Confirm that the standard deviation of out is close to
# sd_of_source / sqrt(30). (sd of rexp is 1, sd of rbinom(1,0.2)
# is sqrt(0.2*0.8).)
## Part E: STRETCH ---------------------------------
# E1. Time a for() loop that grows a vector vs one that pre-allocates:
# system.time({ x <- c(); for (i in 1:1e4) x <- c(x, i) })
# vs
# system.time({ x <- numeric(1e4); for (i in 1:1e4) x[i] <- i })
# Note the difference; that's why we pre-allocate.