I have a vector containing 0 and 1. I want to return the maximum value of number of times 1 appears consecutively. For e.g. if x is the input vector
x <-c(0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1)
Expected Output: 3
My attempt:
I'm using function rle to do this job. Here is my sample code:
x<-c(0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1)
y<-rle(x)
max_repeat <-max(y$lengths)
In this scenario, I get output as 4 (corresponding to 0 instead of 1). I tried to use tapply to access the complete output of rle, but I am not able to extract out the maximum repeat corresponding to value 1.
out <-tapply(y$lengths, y$values, max)
This is what I get for out:
0 1
4 3
When I look at the structure of out, it is " int [1:2(1d)] 4 3". I do not have enough experience with dealing this type of variables. I need to extract the value corresponding to 1 i.e. 3. Any help will be appreciated!
Thanks
You can try this:
x<-c(0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1)
y<-rle(x)
max(y$lengths[y$values==1])
# 3
If you want informations about any object, here y you can use the function str, which will return informations about what contains any object.
I found it like this =>
x <-c(0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1)
y=rle(x)
max(y$lengths[y$values==1])
hope it meets your expectations.
Related
I am doing the next task.
Suppose that I have the next vector.
(1,1,0,0,0,1,1,1,1,0,0,1,1,1,0)
I need to extract the next info.
the maximum number of sets of consecutive zeros
the mean number of consecutive zeros.
FOr instance in the previous vector
the maximum is: 3, because I have 000 00 0
Then the mean number of zeros is 2.
I am thinking in this idea because I need to do the same but with several observations. I think to implement this inside an apply function.
We could use rle for this. As there are only binary values, we could just apply the rle on the entire vector, then extract the lengths that correspond to 0 (!values - returns TRUE for 0 and FALSE others)
out <- with(rle(v1), lengths[!values])
And get the length and the mean from the output
> length(out)
[1] 3
> mean(out)
[1] 2
data
v1 <- c(1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0)
You can try another option using regmatches
> v <- c(1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0)
> s <- paste0(v, collapse = "")
> zeros <- unlist(regmatches(s, gregexpr("0+", s)))
> length(zeros)
[1] 3
> mean(nchar(zeros))
[1] 2
Suppose I have a vector
vec <- c(0, 1, 0, 0, 0, 1, 1, 1, 1, 2)
How do I random sample a nonzero element and turn other elements into 0?
Suppose the element sampled was vec[2], then the resulting vector would be
vec <- c(0, 1, 0, 0, 0, 0, 0, 0, 0, 0)
I know that I can sample the indice of one nonzero element by sample(which(vec != 0), 1), but I am not sure how to proceed from that. Thanks!
You can try the code below
> replace(0 * vec, sample(which(vec != 0), 1), 1)
[1] 0 0 0 0 0 0 0 1 0 0
where
which returns the indices of non-zero values
sample gives a random index
replace replaces the value to 1 at the specific index
Watch out for sample's behavior if which returns only 1 value:
> vec <- c(rep(0, 9), 1)
> sample(which(vec != 0), 1)
[1] 4
This preserves the vector value (instead of turning it to 1) and guards against vectors with only one nonzero value using rep to guarantee sample gets a vector with more than one element:
vec[-sample(rep(which(vec != 0), 2), 1)] <- 0
In this tutorial, I tried to use another method for converting categorical variables to factor.
In the article, the following method is used.
library(MASS)
library(rpart)
cols <- c('low', 'race', 'smoke', 'ht', 'ui')
birthwt[cols] <- lapply(birthwt[cols], as.factor)
and I replaced the last line by
birthwt[cols] <- as.factor((birthwt[cols]))
but the result is NA all
What is wrong with that?
as.factor((birthwt[cols])) is calling as.factor on a list of 5 vectors. If you do that R will interpret each of those 5 vectors as the levels, and the column headers as the labels, of a factor variable, which is clearly not what you want:
> as.factor(birthwt[cols])
low race smoke ht ui
<NA> <NA> <NA> <NA> <NA>
5 Levels: c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) ...
> labels(as.factor(birthwt[cols]))
[1] "low" "race" "smoke" "ht" "ui"
lapply iterates over a list, calling the function as.factor on each of the vectors separately in that list. You need to do this to convert each variable separately into a factor, rather than attempting to convert the entire list into a single factor, which is what as.factor(birthwt[cols]) does.
Is there an efficient way to calculate the length of portions of a vector that repeat a specified value?
For instance, I want to calculate the length of rainless periods along a vector of daily rainfall values:
daily_rainfall=c(15, 2, 0, 0, 0, 3, 3, 0, 0, 10)
Besides using the obvious but clunky approach of looping through the vector, what cleaner way can I get to the desired answer of
rainless_period_length=c(3, 2)
given the vector above?
R has a built-in function rle: "run-length encoding":
daily_rainfall <- c(15, 2, 0, 0, 0, 3, 3, 0, 0, 10)
runs <- rle(daily_rainfall)
rainless_period_length <- runs$lengths[runs$values == 0]
rainless_period_length
output:
[1] 3 2
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I am trying out R studio myself and have a question.
I have a vector
vec <- c(1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1)
I want to make a function to do the following: if the distance between two subsequences of 1's less then 5, then it is going to show 0. But if it is more than 5 it will show 1.
So, if looking at
vec <- c(1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1)
the output is going to be:
0 0 1
I understand how I can find a position of 1:
function_start_of_seq <- function(x) {
one_pos<-which(rle(x)$values==1 %in% TRUE)
And I know that I need to use diff function and cumsum, but I don't know how...
Perhaps an approach regarding rather the 0s than the 1s is more appropriate. In the next line you can check the lengths of the rle() output which distance (i.e. number of 0s between the 1s) exceeds the 5. Just convert it into 0-1 with as.numeric()at the end.
fun1 <- function(x) {
null_pos <- which(rle(x)$values == 0)
tf <- rle(x)$lengths[null_pos] > 5
return(as.numeric(tf))
}
> fun1(vec)
[1] 0 0 1
Does that make sense?
In case you want a one-liner, just do
> as.numeric(rle(vec)$lengths[which(rle(vec)$values == 0)] > 5)
[1] 0 0 1
The part which(rle(vec)$values == 0) selects the positions with distance between 1s sequences (i.e. the output of rle() regarding the 0s) is greater than 5.
as.numeric() then "translates" the output into the 0-1 - form you desire.
An uncool, non-obfuscated, only-calling-rle-once, no-use-of-which answer:
vec <- c(1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1)
r <- rle(vec)
r
## Run Length Encoding
## lengths: int [1:7] 1 2 2 4 1 6 1
## values : num [1:7] 1 0 1 0 1 0 1
So it seems the distance between the 1 sequences is what you're after. We'll assume you know you always have 0's and 1's.
r$values == 0 will return a vector with TRUE or FALSE for the result of each positional evalution. We can use that directly in r$lengths.
rl <- r$lengths[r$values == 0]
rl
## [1] 2 4 6
Since it's just 0 and 1, we don't need a double. integers will do just fine:
as.integer(rl > 5)
## [1] 0 0 1