I need to find which elements in a new vector (vb) have been added to another vector (va). If there for example is unly one "2" in va, but two "2" in vb, then one "2" has been added.
The comment in the code below shows what is sought.
va <- c(1, 2) # Original vector
vb <- c(1, 2) # NA or NULL
vb <- c(2, 2) # 2
vb <- c(1, 1) # 1
vb <- c(1) # NA or NULL
vb <- c(2) # NA or NULL
vb <- c(3, 3) # c(3, 3)
I've tried match, union, intersect, %in%, etc. but can't get it to work to consider also multiple instances. This feels irritatingly simple...
The following reproduces your expected outcome. Just as an honest heads-up, I'm not really happy with my solution, this seems oddly convoluted:
f <- function(a, b) {
a <- as.data.frame(unclass(rle(a)));
b <- as.data.frame(unclass(rle(b)));
t <- merge(a, b, by = "values", all = TRUE);
t$lengths.x[is.na(t$lengths.x)] <- 0;
t$diff <- t$lengths.y - t$lengths.x;
t <- t[!is.na(t$diff) & t$diff > 0, ];
return(rep(t$values, t$diff));
}
va <- c(1, 2);
vb <- c(1, 2) # NA or NULL
f(va, vb);
#numeric(0)
vb <- c(2, 2) # 2
f(va, vb);
#[1] 2
vb <- c(1, 1) # 1
f(va, vb);
#[1] 1
vb <- c(1) # NA or NULL
f(va, vb);
#numeric(0)
vb <- c(2) # NA or NULL
f(va, vb);
#numeric(0)
vb <- c(3, 3) # c(3, 3)
#[1] 3 3
Explanation: I'm making use of rle to compare the lengths (level of duplicity) of different entries in va and vb; then report only those that are not already in va.
Update
Here is a much cleaner method using a recursive function.
f <- function(a, b) {
if (length(a) == 0 | length(b) == 0) return(NULL);
m <- data.frame(idx.a = 1:length(a), idx.b = match(a, b));
m <- m[complete.cases(m), ];
# Here is the recursive call
if (nrow(m) > 0) f(a[-m$idx.a[1]], b[-m$idx.b[1]]) else b;
}
va <- c(1, 2) # Original vector
f(va, c(1, 2));
#NULL
f(va, c(2, 2));
#[1] 2
f(va, c(1, 1));
#[1] 1
f(va, c(1));
#NULL
f(va, c(2));
#NULL
f(va, c(3, 3));
#[1] 3 3
Not the most elegant, but it works for all your cases:
Diff_frequency <- function(va,vb){
df <- merge(as.data.frame(table(va)), as.data.frame(table(vb)), by.x="va", by.y="vb", all=T)
df$Freq.x[is.na(df$Freq.x)] <- 0
df$Dif <- df$Freq.y - df$Freq.x
df$Dif[is.na(df$Dif) | df$Dif < 0] <- 0
return(rep(as.numeric(as.character(df[,1])), df$Dif))
}
Diff_frequency(va,vb)
Examples of output:
va=c(1,1,1,2,2,2,3)
vb=c(1,1,4,4,2,2,5)
Diff_frequency(va,vb)
[1] 4 4 5
va=c(1,1,1,2,2,2,3)
vb=c(1,1,1,1,2,2,2,3,3,5)
Diff_frequency(va,vb)
1] 1 3 5
va=c(1,1,1,2,2,2,3)
vb=c(1,1,2,3)
Diff_frequency(va,vb)
numeric(0)
Related
I am using the lapp functin of {terra} in R and I want to update rast_a with values from rast_b or rast_c (and some other math) depending on the value in each cell of rast_a.
sample data
rast_a <- rast(ncol = 2, nrow = 2)
values(rast_a) <- 1:4
rast_b <- rast(ncol = 2, nrow = 2)
values(rast_b) <- c(2,2,2,2)
rast_c <- rast(ncol = 2, nrow = 2)
values(rast_c) <- c(3,3,3,3)
Problem
This is my (wrong) attempt.
my_update_formula <- function(a, b, c) {
a[a == 1] <- b[a == 1] + 10 + 20 - 30
a[a == 2] <- c[a == 2] + 10 + 50 - 50
return(a)
}
result <- lapp(c(rast_a, rast_b, rast_c),
fun = my_update_formula)
values(result)
lyr1
[1,] 3
[2,] 3
[3,] 3
[4,] 4
The actual result should be 2,3,3,4. But because of the operations inside the formula, the first value gets updated twice. First it is changed from 1 to 2 (correctly) but then it fulfills the condition of the second line of code also, and is changed again (I don't want that to happen).
How can I solve this please?
You can change your formula to
f1 <- function(a, b, c) {
d <- a
d[a == 1] <- b[a == 1]
d[a == 2] <- c[a == 2] + 10
d
}
#or
f2 <- function(a, b, c) {
i <- a == 1
j <- a == 2
a[i] <- b[i]
a[j] <- c[j] + 10
return(a)
}
lapp(c(rast_a, rast_b, rast_c), fun = f1) |> values()
# lyr1
#[1,] 2
#[2,] 13
#[3,] 3
#[4,] 4
lapp(c(rast_a, rast_b, rast_c), fun = f2) |> values()
# lyr1
#[1,] 2
#[2,] 13
#[3,] 3
#[4,] 4
You can get the same result with
x <- ifel(rast_a==1, rast_b,
ifel(rast_a == 2, rast_c + 10, rast_a))
I really need some help to write a recursion in R.
The function that I want changes a certain observation according to a set of comparisons between different rows in a data frame, which I shall call g. One of these comparisons depends on the previous value of this same observation.
Suppose first that I want to update the value of column index, row i in my data df in the following way:
j <- 1:4
g <- (df$dom[i] > 0 &
abs(df$V2009[i] - df$V2009[j]) <= w) |
df$index[i] == df$index[j]
df$index[i] <- ifelse(any(g), which(g)[[1]], df$index[[i]])
The thing is, the object w is actually a list:
w = list(0, 1, 2, df$age[i])
So, as you can see, I want to create a function foo() that updates df$index iteratively. It changes it by looping through w and comparisons depend on updated values.
Here is some data:
df <- data.frame(dom = c(0, 0, 6, 6),
V2009 = c(9, 11, 9, 11),
index = c(1, 2, 1, 2),
age = c(2, 2, 2, 2))
I am not sure if a recursive function is actually needed or if something like reduce or map would do it.
Thank you!
The following function uses a double for loop to change the values of column index according to the condition defining g. It accepts a data.frame as input and returns the updated data.frame.
foo <- function(x){
change_index <- function(x, i, w){
j <- seq_len(nrow(x))
(x$dom[i] > 0 & abs(x$V2009[i] - x$V2009[j]) <= w) |
x$index[i] == x$index[j]
}
for(i in seq_len(nrow(x))){
W <- list(0, 1, 2, x$age[i])
for(w in W){
g <- change_index(x, i, w)
if(any(g)) x$index[i] <- which(g)[1]
}
}
x
}
foo(df)
# dom V2009 index age
#1 0 9 1 2
#2 0 11 2 2
#3 6 9 1 2
#4 6 11 1 2
One can define w inside a function and use lexical scoping (closure).
Using your instructions, the function index_value calculates for any given i the index value.
correct_index_col returns the corrected df.
df <- data.frame(dom = c(0, 0, 6, 6),
V2009 = c(9, 11, 9, 11),
index = c(1, 2, 1, 2),
age = c(2, 2, 2, 2))
index_value <- function(df, i) {
j <- nrow(df)
w <- c(0, 1, 2, df$age[i])
g <- (df$dom[i] > 0 & abs(df$V2009[i] - df$V2009[j]) <= w) |
df$index[i] == df$index[j]
ifelse(any(g), which(g)[[1]], df$index[[i]])
}
correct_index_col <- function(df) {
indexes <- Vectorize(function(i) {
index_value(df, i)
})
df$index <- indexes(1:nrow(df))
df
}
# > correct_index_col(df)
# dom V2009 index age
# 1 0 9 1 2
# 2 0 11 1 2
# 3 6 9 3 2
# 4 6 11 1 2
#
If you want to really update (mutate) your df, then you have to do
df <- correct_index_col(df).
Here is an attempt of my own. I guess I figured out a way to use recursion over mutate:
test <- function(i, df, k){
j <- 1:nrow(df)
w <- list(0, 1, 2, df$age[i])
g <- (df$dom[i] > 0 & abs(df$V2009[i] - df$V2009[j]) <= w[k]) |
df$index[i] == df$index[j]
l <- ifelse(any(g), which(g)[1], df$index[i])
return(l)
}
loop <- function(data,
k = 1) {
data <- data %>%
mutate(index = map_dbl(seq(n()),
~ test(.x, df = cur_data(), k)))
if (k == 4) {
return(data)
} else {
return(loop(data, k + 1))
}
}
df %>% loop()
I welcome any comments in case this is inefficient considering large datasets
Assume one wished to calculate a cumulative sum based on a multivariate condition, all(Z[i] <= x), for all i over a multivariate grid x. One may obviously implement this naively
cSums <- numeric(nrow(x))
for(i in seq(nrow(x))){
for(j in seq(nrow(Z))){
if(all(Z[j, ] <= x[i, ]))
cSums[i] <- cSums[i] + R[j] # <== R is a single vector to be summed
}
}
which would be somewhere around O((n*p)^2), or slightly faster by iteratively subsetting the columns
cSums <- numeric(nrow(x))
for(i in seq(nrow(x))){
indx <- seq(nrow(Z))
for(j in seq(ncol(Z))){
indx <- indx[which(Z[indx, j] <= x[i, j])]
}
cSums[i] <- sum(R[indx])
}
but this still worst-case as slow as the naive-implementation. How could one improve this to achieve faster performance, while still allowing an undefined number of columns to be compared?
Dummy data and Reproducible example
var1 <- c(3,3,3,5,5,5,4,4,4,6)
var2 <- rep(seq(1,5), each = 2)
Z <- cbind(var1, var2)
x <- Z
R <- rep(1, nrow(x))
# Result using either method.
#[1] 2 2 3 4 6 6 5 5 6 10
outer is your friend, just Vectorize your comparison. colSums yields the desired result then. Should be fast.
f <- Vectorize(function(k, l) all(Z[k, ] <= x[l, ]))
res <- colSums(outer(1:nrow(Z), 1:nrow(x), f))
res
# [1] 2 2 3 4 6 6 5 5 6 10
Data
x <- Z <- structure(c(3, 3, 3, 5, 5, 5, 4, 4, 4, 6, 1, 1, 2, 2, 3, 3, 4,
4, 5, 5), .Dim = c(10L, 2L), .Dimnames = list(NULL, c("var1",
"var2")))
We can use apply row-wise and compare every row with every other row and count how many of them satidy the criteria.
apply(Z, 1, function(x) sum(rowSums(Z <= as.list(x)) == length(x)))
#[1] 2 2 3 4 6 6 5 5 6 10
Similar approach can also be performed using sapply + split
sapply(split(Z, seq_len(nrow(Z))), function(x)
sum(rowSums(Z <= as.list(x)) == length(x)))
data
var1 <- c(3,3,3,5,5,5,4,4,4,6)
var2 <- rep(seq(1,5), each = 2)
Z <- data.frame(var1, var2)
I have two matrices, one is generated out of the other by deleting some rows. For example:
m = matrix(1:18, 6, 3)
m1 = m[c(-1, -3, -6),]
Suppose I do not know which rows in m were eliminated to create m1, how should I find it out by comparing the two matrices? The result I want looks like this:
1, 3, 6
The actual matrix I am dealing with is very big. I was wondering if there is any efficient way of conducting it.
Here are some approaches:
1) If we can assume that there are no duplicated rows in m -- this is the case in the example in the question -- then:
which(tail(!duplicated(rbind(m1, m)), nrow(m)))
## [1] 1 3 6
2) Transpose m and m1 giving tm and tm1 since it is more efficient to work on columns than rows.
Define match_indexes(i) which returns a vector r such that each row in m[r, ] matches m1[i, ].
Apply that to each i in 1:n1 and remove the result from 1:n.
n <- nrow(m); n1 <- nrow(m1)
tm <- t(m); tm1 <- t(m1)
match_indexes <- function(i) which(colSums(tm1[, i] == tm) == n1)
setdiff(1:n, unlist(lapply(1:n1, match_indexes)))
## [1] 1 3 6
3) Calculate an interaction vector for each matrix and then use setdiff and finally match to get the indexes:
i <- interaction(as.data.frame(m))
i1 <- interaction(as.data.frame(m1))
match(setdiff(i, i1), i)
## [1] 1 3 6
Added If there can be duplicates in m then (1) and (3) will only return the first of any multiply occurring row in m not in m1.
m <- matrix(1:18, 6, 3)
m1 <- m[c(2, 4, 5),]
m <- rbind(m, m[1:2, ])
# 1
which(tail(!duplicated(rbind(m1, m)), nrow(m)))
## 1 3 6
# 2
n <- nrow(m); n1 <- nrow(m1)
tm <- t(m); tm1 <- t(m1)
match_indexes <- function(i) which(colSums(tm1[, i] == tm) == n1)
setdiff(1:n, unlist(lapply(1:n1, match_indexes)))
## 1 3 6 7
# 3
i <- interaction(as.data.frame(m))
i1 <- interaction(as.data.frame(m1))
match(setdiff(i, i1), i)
## 1 3 6
A possible way is to represent each row as a string:
x1 <- apply(m, 1, paste0, collapse = ';')
x2 <- apply(m1, 1, paste0, collapse = ';')
which(!x1 %in% x2)
# [1] 1 3 6
Some benchmark with a large matrix using my solution and G. Grothendieck's solutions:
set.seed(123)
m <- matrix(rnorm(20000 * 5000), nrow = 20000)
m1 <- m[-sample.int(20000, 1000), ]
system.time({
which(tail(!duplicated(rbind(m1, m)), nrow(m)))
})
# user system elapsed
# 339.888 2.368 342.204
system.time({
x1 <- apply(m, 1, paste0, collapse = ';')
x2 <- apply(m1, 1, paste0, collapse = ';')
which(!x1 %in% x2)
})
# user system elapsed
# 395.428 0.568 395.955
system({
n <- nrow(m); n1 <- nrow(m1)
tm <- t(m); tm1 <- t(m1)
match_indexes <- function(i) which(colSums(tm1[, i] == tm) == n1)
setdiff(1:n, unlist(lapply(1:n1, match_indexes)))
})
# > 15 min, not finish
system({
i <- interaction(as.data.frame(m))
i1 <- interaction(as.data.frame(m1))
match(setdiff(i, i1), i)
})
# run out of memory. My 32G RAM machine crashed.
We can also use do.call
which(!do.call(paste, as.data.frame(m)) %in% do.call(paste, as.data.frame(m1)))
#[1] 1 3 6
Suppose that my vector numbers contains c(1,2,3,5,7,8), and I wish to find if it contains 3 consecutive numbers, which in this case, are 1,2,3.
numbers = c(1,2,3,5,7,8)
difference = diff(numbers) //The difference output would be 1,1,2,2,1
To verify that there are 3 consecutive integers in my numbers vector, I've tried the following with little reward.
rep(1,2)%in%difference
The above code works in this case, but if my difference vector = (1,2,2,2,1), it would still return TRUE even though the "1"s are not consecutive.
Using diff and rle, something like this should work:
result <- rle(diff(numbers))
any(result$lengths>=2 & result$values==1)
# [1] TRUE
In response to the comments below, my previous answer was specifically only testing for runs of length==3 excluding longer lengths. Changing the == to >= fixes this. It also works for runs involving negative numbers:
> numbers4 <- c(-2, -1, 0, 5, 7, 8)
> result <- rle(diff(numbers4))
> any(result$lengths>=2 & result$values==1)
[1] TRUE
Benchmarks!
I am including a couple functions of mine. Feel free to add yours. To qualify, you need to write a general function that tells if a vector x contains n or more consecutive numbers. I provide a unit test function below.
The contenders:
flodel.filter <- function(x, n, incr = 1L) {
if (n > length(x)) return(FALSE)
x <- as.integer(x)
is.cons <- tail(x, -1L) == head(x, -1L) + incr
any(filter(is.cons, rep(1L, n-1L), sides = 1, method = "convolution") == n-1L,
na.rm = TRUE)
}
flodel.which <- function(x, n, incr = 1L) {
is.cons <- tail(x, -1L) == head(x, -1L) + incr
any(diff(c(0L, which(!is.cons), length(x))) >= n)
}
thelatemail.rle <- function(x, n, incr = 1L) {
result <- rle(diff(x))
any(result$lengths >= n-1L & result$values == incr)
}
improved.rle <- function(x, n, incr = 1L) {
result <- rle(diff(as.integer(x)) == incr)
any(result$lengths >= n-1L & result$values)
}
carl.seqle <- function(x, n, incr = 1) {
if(!is.numeric(x)) x <- as.numeric(x)
z <- length(x)
y <- x[-1L] != x[-z] + incr
i <- c(which(y | is.na(y)), z)
any(diff(c(0L, i)) >= n)
}
Unit tests:
check.fun <- function(fun)
stopifnot(
fun(c(1,2,3), 3),
!fun(c(1,2), 3),
!fun(c(1), 3),
!fun(c(1,1,1,1), 3),
!fun(c(1,1,2,2), 3),
fun(c(1,1,2,3), 3)
)
check.fun(flodel.filter)
check.fun(flodel.which)
check.fun(thelatemail.rle)
check.fun(improved.rle)
check.fun(carl.seqle)
Benchmarks:
x <- sample(1:10, 1000000, replace = TRUE)
library(microbenchmark)
microbenchmark(
flodel.filter(x, 6),
flodel.which(x, 6),
thelatemail.rle(x, 6),
improved.rle(x, 6),
carl.seqle(x, 6),
times = 10)
# Unit: milliseconds
# expr min lq median uq max neval
# flodel.filter(x, 6) 96.03966 102.1383 144.9404 160.9698 177.7937 10
# flodel.which(x, 6) 131.69193 137.7081 140.5211 185.3061 189.1644 10
# thelatemail.rle(x, 6) 347.79586 353.1015 361.5744 378.3878 469.5869 10
# improved.rle(x, 6) 199.35402 200.7455 205.2737 246.9670 252.4958 10
# carl.seqle(x, 6) 213.72756 240.6023 245.2652 254.1725 259.2275 10
After diff you can check for any consecutive 1s -
numbers = c(1,2,3,5,7,8)
difference = diff(numbers) == 1
## [1] TRUE TRUE FALSE FALSE TRUE
## find alteast one consecutive TRUE
any(tail(difference, -1) &
head(difference, -1))
## [1] TRUE
It's nice to see home-grown solutions here.
Fellow Stack Overflow user Carl Witthoft posted a function he named seqle() and shared it here.
The function looks like this:
seqle <- function(x,incr=1) {
if(!is.numeric(x)) x <- as.numeric(x)
n <- length(x)
y <- x[-1L] != x[-n] + incr
i <- c(which(y|is.na(y)),n)
list(lengths = diff(c(0L,i)),
values = x[head(c(0L,i)+1L,-1L)])
}
Let's see it in action. First, some data:
numbers1 <- c(1, 2, 3, 5, 7, 8)
numbers2 <- c(-2, 2, 3, 5, 6, 7, 8)
numbers3 <- c(1, 2, 2, 2, 1, 2, 3)
Now, the output:
seqle(numbers1)
# $lengths
# [1] 3 1 2
#
# $values
# [1] 1 5 7
#
seqle(numbers2)
# $lengths
# [1] 1 2 4
#
# $values
# [1] -2 2 5
#
seqle(numbers3)
# $lengths
# [1] 2 1 1 3
#
# $values
# [1] 1 2 2 1
#
Of particular interest to you is the "lengths" in the result.
Another interesting point is the incr argument. Here we can set the increment to, say, "2" and look for sequences where the difference between the numbers are two. So, for the first vector, we would expect the sequence of 3, 5, and 7 to be detected.
Let's try:
> seqle(numbers1, incr = 2)
$lengths
[1] 1 1 3 1
$values
[1] 1 2 3 8
So, we can see that we have a sequence of 1 (1), 1 (2), 3 (3, 5, 7), and 1 (8) if we set incr = 2.
How does it work with ECII's second challenge? Seems OK!
> numbers4 <- c(-2, -1, 0, 5, 7, 8)
> seqle(numbers4)
$lengths
[1] 3 1 2
$values
[1] -2 5 7
Simple but works
numbers = c(-2,2,3,4,5,10,6,7,8)
x1<-c(diff(numbers),0)
x2<-c(0,diff(numbers[-1]),0)
x3<-c(0,diff(numbers[c(-1,-2)]),0,0)
rbind(x1,x2,x3)
colSums(rbind(x1,x2,x3) )==3 #Returns TRUE or FALSE where in the vector the consecutive intervals triplet takes place
[1] FALSE TRUE TRUE FALSE FALSE FALSE TRUE FALSE FALSE
sum(colSums(rbind(x1,x2,x3) )==3) #How many triplets of consecutive intervals occur in the vector
[1] 3
which(colSums(rbind(x1,x2,x3) )==3) #Returns the location of the triplets consecutive integers
[1] 2 3 7
Note that this will not work for consecutive negative intervals c(-2,-1,0) because of how diff() works