Microbial Informatics

• Homework is due Friday
• We will not be meeting on Friday
• Read The Art of R Programming (Chapter 7)

• Lists
• Data frames
• Factors

• Understand how to direct flow through a program using loops and conditionals

• Much of what we've been doing so far follows a very linear path. "Do this, then that, and then do this." In more complicated programs, you will want
  the program to have some intelligence and make decisions. That is where control statements come in to play.
• Today we'll discuss two forms of control statements:
  • loops
  • conditionals

• Recall that many things can be vectorized within R. So while you could write a loop to calcualte the square root of every number between 1 and 100, it's far more efficient to create a vector and run sqrt on the vector.
• There are a set of "apply" functions that we'll cover later that largely allow us to eliminate the need for explicit loops
  

for(i in 1:10){
#   do something clever
}

• This is essentially saying we'll do something with i=1, then repeat it with i=2, ... i=10. Then we'll stop.
• i is called an index varaible and can be a number or string. the numbers need not be sequential think of indexing over different start codon positions in a genome or indexing over all possible codons
• Recall that 1:10 is a vector

x <- seq(2,100,2)
for(i in x){
#   do something clever...
}

or

x <- c("red", "green", "blue")
for(i in x){
#   do something clever...
}

for(i in 10:1){
    i
}
print("blast off!")

## [1] "blast off!"

Hmmm. What went wrong?

Within a loop (or any function) you have to explicitly tell R to output the value of a variable

for(i in 10:1){
    print(i)
}
print("blast off!")

## [1] 10
## [1] 9
## [1] 8
## [1] 7
## [1] 6
## [1] 5
## [1] 4
## [1] 3
## [1] 2
## [1] 1
## [1] "blast off!"

for(i in 1:10){
    squares[i] <- i^2
}

## Error in squares[i] <- i^2: object 'squares' not found

squares

## Error in eval(expr, envir, enclos): object 'squares' not found

What's wrong?

Have to create vector before starting loop:

squares <- vector()
for(i in 1:10){
    squares[i] <- i^2
}
squares

##  [1]   1   4   9  16  25  36  49  64  81 100

Have to create vector before starting loop:

squares <- rep(NA, 10)
#squares <- rep(0, 10)
for(i in 1:9){
    squares[i] <- i^2
}
squares

##  [1]  1  4  9 16 25 36 49 64 81 NA

    x<-c(5,12,13)
    for(i in x) {   print(x^2)  }
    for(i in x) {   print(i^2)  }

    x<-c(5,12,13)
    for(i in x) {   print(x^2)  }

## [1]  25 144 169
## [1]  25 144 169
## [1]  25 144 169

    for(i in x) {   print(i^2)  }

## [1] 25
## [1] 144
## [1] 169

i<-1
while(i <= 10){
    print(i)
}

Notice anything wrong with this statement?

You have to modify the index within the loop because i will always be less than or equal to 10:

i<-1
while(i <= 10){
    i<-i+3
    print(i)
}

• The break command breaks you out of the current loop
• Can be used with for and repeat loops

x<-0;
while(TRUE){
  x<-rnorm(1);
  print(x)
  if(x<0){  break   }
}

x<-0;
repeat {
  x<-rnorm(1);
  print(x)
  if(x<0){  break   }
}

## [1] 1.258434
## [1] 0.1496421
## [1] -0.6827691

• the next command allows us to skip the remainder of the iteration and return to the top of the loop block...

counter <- 0
while(counter < 10){
    x<-rnorm(1);
    if(x<0){    next    }
    print(x)
    counter <- counter+1
}

## [1] 0.3141362
## [1] 0.4966642
## [1] 0.9285402
## [1] 2.262056
## [1] 0.3757543
## [1] 1.150501
## [1] 0.09672476
## [1] 1.104058
## [1] 1.085318
## [1] 0.5036225

• R objects are immutable / unchangeable.
• To assign a value to a variable, R operations reassign values to given objects
• To do this they have to come up with new space in RAM to store information
• Recall the following...

    squared[i] <- i^2

• The i^2 clearly costs some unit of time, but look at the "squared[i] <-" part.
• Each cycle through the loop is doing this

    squared <- "[<-"(squared,i,value=i^2)

• The <- part is a function!

• A copy of squared is being made, element i is changed to i^2 and the resulting vector is reassigned to squared. The end result is to re-assign the entire vector because you changed one value.
• When you vectorize an operation you are really removing a number of these steps. Consider this:

squared <- (1:10)^2

• There is only one assignment operation

• We've already seen conditionals throughout the course when we filtered the metadata table
• Sometimes we'd like to have a program bifurcate at a given step based on user input or the behavior of the data.
• For example, if a start and stop codon are less than 300 bp apart, then we'll ignore it. Otherwise, let's translate it to amino acids

• Familiar with ifelse from Excel? It's the same in R...

ifelse(LOGICALTEST, Do this if TRUE, Do this if FALSE)

• An example:

x <- seq(1, 100, 5)
ifelse(x>40, "old", "young")

##  [1] "young" "young" "young" "young" "young" "young" "young" "young"
##  [9] "old"   "old"   "old"   "old"   "old"   "old"   "old"   "old"  
## [17] "old"   "old"   "old"   "old"

x <- seq(1, 100, 5)
ifelse(x<10, "kid", ifelse(x<20, "adolscent", "adult"))

##  [1] "kid"       "kid"       "adolscent" "adolscent" "adult"    
##  [6] "adult"     "adult"     "adult"     "adult"     "adult"    
## [11] "adult"     "adult"     "adult"     "adult"     "adult"    
## [16] "adult"     "adult"     "adult"     "adult"     "adult"

for(age in x){
        if(age < 10){       #age has to be an atomic varaible
            print("kid")
        } else if(age<20) {
            print("adolescent")
        } else {
            print("ancient")
        }
    }

## [1] "kid"
## [1] "kid"
## [1] "adolescent"
## [1] "adolescent"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"
## [1] "ancient"