需要计算R中向量中特定转换的数量

Question

I am programming a sampler in R, which basically is a big for loop, and for every Iterations I have to count the number of transitions in a vector. I have a vector called k, which contains zeros and ones, with 1000 entries in the vector.

I have used the following, horribly slow, code:

#we determine the number of transitions n00,n01,n10,n11
  n00=n01=n10=n11=0 #reset number of transitions between states from last time
  for(j in 1:(1000-1)){
    if(k[j+1]==1 && k[j]==0) {n01<-n01+1}
    else { if(k[j+1]==1 && k[j]==1) {n11<-n11+1}
           else { if(k[j+1]==0 && k[j]==1) {n10<-n10+1}
                  else{n00<-n00+1}
           }

    }
  }

So for every time the loop goes, the variables n00,n01,n10,n11 counts the transitions in the vector. For example, n00 counts number of times a 0 is followed by another 0. And so on...

This is very slow, and I am very new to R, so I am kind of desperate here. I do not understand how to use grep, if that even is possible.

Thank you for your help

Answer 1

Try something like this:

x <- sample(0:1,20,replace = TRUE)
> table(paste0(head(x,-1),tail(x,-1)))

00 01 10 11 
 4  3  4  8 

The head and tail return portions of the vector x : all but the last element, and then all but the first element. This means that the corresponding elements are the consecutive pairs from x .

Then paste0 just converts each one to a character vector and pastes the first elements, the second element, etc. The result is a character vector with elements like "00", "01", etc. Then table just counts up how many of each there are.

You can assign the result to a new variable like so:

T <- table(paste0(head(x,-1),tail(x,-1)))

Experiment yourself with each piece of the code to see how it works. Run just head(x,-1) , etc. to see what each piece does.

To address the comment below, to ensure that all types appear with counts when you run table , convert it to a factor first:

x1 <- factor(paste0(head(x,-1),tail(x,-1)),levels = c('00','01','10','11'))
table(x1)

Answer 2

x <- sample(0:1, 10, replace = TRUE)
# my sample: [1] 0 0 0 0 0 1 0 1 1 0

rl <- rle(x)

zero_to_zero <- sum(rl$len[rl$val == 0 & rl$len > 1] - 1)
one_to_one <- sum(rl$len[rl$val == 1 & rl$len > 1] - 1)

zero_to_one <- sum(diff(rl$val) == -1)
one_to_zero <- sum(diff(rl$val) == 1)

x
# [1] 0 0 0 0 0 1 0 1 1 0
zero_to_zero
# [1] 4

one_to_one
# [1] 1

zero_to_one
# [1] 2

one_to_zero
# [1] 2

@joran's answer is faaaar cleaner though...Still, I thought I just as well could finish the stroll I started down (the dirty) trail, and share the result.

Answer 3

If we don't care about distinguishing the n00 and n11 cases, then this becomes much simpler:

x <- sample(0:1,20,replace = TRUE)
#  [1] 0 0 0 0 0 1 0 1 1 0 0 0 1 0 0 1 0 0 0 0

table(diff(x))      
# -1  0  1 
#  4 11  4 

Since the question says that you're primarily interested in the transitions, this may be acceptable, otherwise one of the other answers would be preferable.