![print(paste('Iterations exceed the maximum allowed number of', maxcount))
break }
}
[1] "Solution reached after 77 observations"
x
[1] 16 14 12 18 5 19 14 20 20 20 12
print('sum')
[1] "sum"
tibble(x)
A tibble: 11 x 1
x
<int>
1 16
2 14
3 12
4 18
5 5
6 19
7 14
8 20
9 20
10 20
11 12
print('count')
[1] "count"
counter
[1] 77
What if R cannot give you the required numbers? Break it up!
Here I chose some funky numbers on purpose to show how a break statement works:
• k <- 1000 (constant)
• m <- 0 (lower limit)
• n <- 0 (upper limit)
• x <- 1 (initialize the sequence)
• y <- 50 (size of data set)
• maxcount <- 80000 (maximum amount of iterations)
• counter <- 0](https://image.slidesharecdn.com/controllingtheflow-191014194402/85/Understanding-control-flow-2-320.jpg)
![k <- 1000 constant
m <- 0 lower limit
n <- 0 upper limit
x <- 1 initialize the sequence
y <- 50 size of data set
maxcount <- 80000 maximum amount of iterations
counter <- 0
while(sum(x) != k | counter < maxcount){
counter <- counter+1
if (sum(x)== k){
print(paste('Solution reached after', counter, 'observations'))
break
} else
x <- sample(m:n, y, replace = T)
if (counter> maxcount){
print(paste('Iterations exceed the maximum allowed number of', maxcount))
break }
}
[1] "Iterations exceed the maximum allowed number of 80000"
print('sum')
[1] "sum"
tibble(x)
A tibble: 50 x 1
x
<int>
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 0
10 0
... with 40 more rows
print('count')
[1] "count"
counter
[1] 80001](https://image.slidesharecdn.com/controllingtheflow-191014194402/85/Understanding-control-flow-3-320.jpg)
Understanding control-flow
- 1. Controlling the flow Control flow statements • Have you ever needed to quickly come up with numbers that add up to a different number? • Below is a code chunk that generates a sequence of numbers that add up to a constant. • It reads as follows: Give me a “y” amounts of numbers between “m” and “n” that add up to “k” • So this loop will repeat until it gets y different numbers between m and n that add up to k. • I controlled it by limiting the number of iterations to 80000 but you can change it if you feel so inclined by changing the value of maxcount after you have copied the code into your Rscript. • k <- 170 (constant) • m <- 5 (lower limit) • n <- 20 (upper limit) • x <- 1 (initialize the sequence) • y <- 11 (size of data set) • maxcount <- 80000 (maximum amount of iterations) • counter <- 0 k <- 170 constant m <- 5 lower limit n <- 20 upper limit x <- 1 initialize the sequence y <- 11 size of data set maxcount <- 80000 maximum amount of iterations counter <- 0 while(sum(x) != k | counter < maxcount){ counter <- counter+1 if (sum(x)== k){ print(paste('Solution reached after', counter, 'observations')) break } else x <- sample(m:n, y, replace = T) if (counter> maxcount){
- 2. print(paste('Iterations exceed the maximum allowed number of', maxcount)) break } } [1] "Solution reached after 77 observations" x [1] 16 14 12 18 5 19 14 20 20 20 12 print('sum') [1] "sum" tibble(x) A tibble: 11 x 1 x <int> 1 16 2 14 3 12 4 18 5 5 6 19 7 14 8 20 9 20 10 20 11 12 print('count') [1] "count" counter [1] 77 What if R cannot give you the required numbers? Break it up! Here I chose some funky numbers on purpose to show how a break statement works: • k <- 1000 (constant) • m <- 0 (lower limit) • n <- 0 (upper limit) • x <- 1 (initialize the sequence) • y <- 50 (size of data set) • maxcount <- 80000 (maximum amount of iterations) • counter <- 0
- 3. k <- 1000 constant m <- 0 lower limit n <- 0 upper limit x <- 1 initialize the sequence y <- 50 size of data set maxcount <- 80000 maximum amount of iterations counter <- 0 while(sum(x) != k | counter < maxcount){ counter <- counter+1 if (sum(x)== k){ print(paste('Solution reached after', counter, 'observations')) break } else x <- sample(m:n, y, replace = T) if (counter> maxcount){ print(paste('Iterations exceed the maximum allowed number of', maxcount)) break } } [1] "Iterations exceed the maximum allowed number of 80000" print('sum') [1] "sum" tibble(x) A tibble: 50 x 1 x <int> 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 10 0 ... with 40 more rows print('count') [1] "count" counter [1] 80001