Simple question: I've got two vectors of 0's and 1's, a and b. The b vector has as many entries as there are 1's in a. I would like to replace the 1's in a with the entries from b. Of course I can do this in a for loop, but is there a nice vectorized way to do this?
From
a <- c(0, 1, 1, 0, 1)
b <- c(1, 0, 1)
create
c <- c(0, 1, 0, 0, 1)
This is pretty simple: a[a == 1] <- b
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
I have recently started using R. When working on carriage of problematic bacteria, I encountered one problem that I hope somebody could help solve. Apologies if the question is on the easy side.
I want to calculate the cumulative proportion of people who get colonized by the problem bug at various time points (a, b, c) as shown in the dataset below "df". "0" means negative test, "1" means positive test for resistant bug, "NA" means test was not done at the time point. The result should be as described in "x", i.e. if the person ever tests positive on either time point (a,b,c) he should have the value "1" in x. If all his tests were negative he should have value "0", and if he never had a test done, the value should be "NA". Is there a good way to calculate this "x" automatically?
a <- c(0, 0, 1, 0, 0, 1, 0, 0, NA, NA)
b <- c(0, 0, 1, 0, 1, NA, 0, 0, NA, 0)
c <- c(NA, 1, 0, 0, 0, 1, 1, 0, NA, 0)
df <- cbind(a, b, c)
df
x <- c(0, 1, 1, 0, 1, 1, 1, 0,NA,0)
df <- cbind(df, x)
df
I tried to create the x-variable using ifelse, but get problems with missing values. For instance, using the following expression:
y <- ifelse(a==1 | b==1 | c==1, 1, ifelse(a==0 | b==0 | c==0, 0, NA))
df <- cbind(df, y)
df
... the resultant column erroneously get "NA" in row 1 and 10, i.e. when there is a combination of 0 and NA, the result should be 0, not NA.
You can use rowSums :
cols <- c('a', 'b', 'c')
+(rowSums(df[, cols], na.rm = TRUE) > 0) * NA^+(rowSums(!is.na(df[, cols])) == 0)
#[1] 0 1 1 0 1 1 1 0 NA 0
This gives similar result as x shown however, might be difficult to understand.
Here is a simple alternative using apply :
apply(df[, cols], 1, function(x) if(all(is.na(x))) NA else +(any(x == 1, na.rm = TRUE)))
#[1] 0 1 1 0 1 1 1 0 NA 0
This returns NA if all the values in the row are NA else checks if any value has 1 in it.
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
I have these matrices.
matr <- list()
matr[[i]] <- c(0, NA, 3, 4, 4,
0, 0, 3, 4, 1,
0, 0, 0, NA, 1,
0, 0, NA, 0, 3,
0, 0, 0, 0, 0)
matr[[i]] <- matrix(matr[[i]], 5, 5)
I want to set NA to zero using the following code:
x <- lapply(matr,function(x) x[is.na(x) <- 0])
Then I got this result:
> x
[[1]]
numeric(0)
[[2]]
numeric(0)
[[3]]
numeric(0)
Why it does not return the matrices? Is my code correct? any help please?
Since lapply works on lists and return lists I think that isn't what you want.
I think using apply here fits better.
Try x <- apply(matr[[1]], 2, function(x){
x[is.na(x)] <- 0
x
})
The number 2 here indicates that you want to operate column-wise instead of row-wise (1st margin are rows, and 2nd margin are columns).
Also notice that you had put the <- operator within the brackets which was a wrong sintax.
EDIT:
It seems that I have misunderstood your question.
Here follows a code that works for an entire list:
lapply(matr, function(x){
apply(x, 2, function(y){
y[is.na(y)] <- 0
y
})
})
Let's say I want to find the longest length of consecutive numbers (excluding 0) in a sequence in R.
Example: (0,2,3,0,5) in this case it should return 2 .
The solution I came up with is as follows:
A1 <- c(1, 1, 0,1,1,1)
length =NULL
B<-rle(A1==0)
C<-B$lengths
D<-B$values
for(i in 1:length(C)){
if(D[i]==FALSE){length[i]=C[i]}
}
length <- length [!is.na(length )]
max(length)
[1] 3
How can I find the longest sequence of non-zero numbers in a vector in R?
We could use rle. A==0 output a logical index vector, rle computes the lengths and runs of values of adjacent elements that are the same for logical vector. Extract the lengths of values that are not '0' and get the max after removing the first and last elements to account for the maximum lengths of non-zero elements at the start or end of vector.
max(with(rle(A==0), lengths[-c(1, length(lengths))][
!values[-c(1, length(values))]]))
#[1] 2
Another example
A1 <- c(1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0,0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1)
max(with(rle(A1==0), lengths[-c(1, length(lengths))][
!values[-c(1, length(values))]]))
#[1] 4
Or
indx <- A1==0
max(with(rle(A1[which(indx)[1L] : tail(which(indx),1)]==0),
lengths[!values]))
#[1] 4
Update
Based on the new info, may be you can try,
A1 <- c(1, 1, 0,1,1,1)
max(with(rle(A1==0), lengths[!values]))
#[1] 3