I have a 1D array with an odd number of rows, 2435 rows. I want to split the array into smaller arrays and each time perform a small test.
Firstly, I want to split the big array into two smaller arrays.
Then I would like to split my array into 4 smaller arrays, then into 8 small arrays and so on.
Can anyone help with that?
An example is the following:
A<-1:2435
A1
1,2,3,4,...,1237
A2
1238, 1239,...,2435
Thanks in advance
Why not simply use the split() function? For example (using an odd length will return warnings but that's fine):
split(x = 1:11, f = 1:2) # to split into 2 distinct list elements
#$`1`
#[1] 1 3 5 7 9 11
#
#$`2`
#[1] 2 4 6 8 10
split(x = 1:11, f = 1:4)
#$`1`
#[1] 1 5 9
#
#$`2`
#[1] 2 6 10
#
#$`3`
#[1] 3 7 11
#
#$`4`
#[1] 4 8
And if you are really keen on splitting to 2, and then again by 2, you can always use the lapply() function which works on each element of a list:
lapply(split(x = 1:11, f = 1:2), split, f = 1:2)
#$`1`
#$`1`$`1`
#[1] 1 5 9
#
#$`1`$`2`
#[1] 3 7 11
#
#
#$`2`
#$`2`$`1`
#[1] 2 6 10
#
#$`2`$`2`
#[1] 4 8
The nested structure is a little bit of a pain but there are other methods for dealing with that, for example:
L <- split(x = 1:11, f = 1:2) # the main (first) split
names(L) <- letters[1:length(L)] # names the main split a and b
LL <- lapply(L, split, f = 1:2) # split the main split
unlist(LL, recursive = F)
#$a.1
#[1] 1 5 9
#
#$a.2
#[1] 3 7 11
#
#$b.1
#[1] 2 6 10
#
#$b.2
#[1] 4 8
If you want to split the data through the middle of the array, you can also use the split function:
a <- 1:2435
divide <- function(x, n = 2)
{
i <- ceiling(length(x)/n)
split(x,x%/%i+1)
}
divide(a)
and with more parts you can use
divide(a, n = 4)
Or in two itterations use
lapply(divide(a,2),function(x) divide(x,2))
With a higher value of n, the sizes will not be equal anymore, due to rounding issues. Which warrants the use of the nested approach.
Related
i have created an array consisting of N values, say
a <- array(dim = c(x, y)).
I want to partition the array into K groups, where array values are randomly assigned to one of K groups. For example if K = 2 and N = 10, one subarray could have 7 values while the other would have 3. The values in all subarrays must sum to N (e.g., 7 + 3 = 10).
I know using split() and sample() is probably the easiest route. I have tried
split(a, sample(a, 2))
but this does not work as it should.
Any ideas?
a <- array(c(1,2,3,4,5,6))
k <- 3
split(sample(a),1:k)
# [1] 2 3
#
# $`2`
# [1] 4 5
#
# $`3`
# [1] 6 1
Hi there i was wondering if is there a way to store a vector into an array or matrix.
for example,
array1<-array(dim=c(1,2))
vector1<-as.vector(1:5)
vector2<-as.vector(6:10)
array1[1,1]<-vector1
array1[1,2]<-vector2
so that when i call for
array1[1,1]
i will receive
[1] 1 2 3 4 5
I've tried doing what i did above and what i get the error
number of items to replace is not a multiple of replacement length
is there a way to get around this?
also, the problem that i face is that i do not know the vector length and that the vectors could have different length as well.
i.e vector 1 can be length of 6 and vector 2 can be a length of 7.
thanks!
Try with a list:
my_list <- list()
my_list[[1]] <- c(1:5)
my_list[[2]] <- c(6:11)
A list allows you to store vectors of varying length. The vectors can be retrieved by addressing the element of the list:
> my_list[[1]]
#[1] 1 2 3 4 5
You can use a matrix of lists:
m <- matrix(list(), 2, 2)
m[1,1][[1]] <- 1:2
m[1,2][[1]] <- 1:3
m[2,1][[1]] <- 1:4
m[2,2][[1]] <- 1:5
m
# [,1] [,2]
#[1,] Integer,2 Integer,3
#[2,] Integer,4 Integer,5
m[1, 2]
#[[1]]
#[1] 1 2 3
m[1, 2][[1]]
#[1] 1 2 3
I right away give an example,
now suppose I have 3 arrays a,b,c such as
a = c(3,5)
b = c(6,1,8,7)
c = c(4,2,9)
I must be able to extract consecutive triplets among them i,e.,
c(1,2,3),c(4,5,6)
But this was just an example, I would be having a larger data set with even more than 10 arrays, hence must be able to find the consecutive series of length ten.
So could anyone provide an algorithm, to generally find the consecutive series of length 'n' among 'n' arrays.
I am actually doing this stuff in R, so its preferable if you give your code in R. Yet algorithm from any language is more than welcomed.
Reorganize the data first into a list containing value and array number.
Sort the list; you'd have smth like:
1-2
2-3
3-1 (i.e. " there' s a three in array 1" )
4-3
5-1
6-2
7-2
8-2
9-3
Then loop the list, check if there are actually n consecutive numbers, then check if these had different array numbers
Here's one approach. This assumes there are no breaks in the sequence of observations in the number of groups. Here the data.
N <- 3
a <- c(3,5)
b <- c(6,1,8,7)
c <- c(4,2,9)
Then i combine them together and order by the observations
dd <- lattice::make.groups(a,b,c)
dd <- dd[order(dd$data),]
Now I look for rows in this table where all three groups are represented
idx <- apply(embed(as.numeric(dd$which),N), 1, function(x) {
length(unique(x))==N
})
Then we can see the triplets with
lapply(which(idx), function(i) {
dd[i:(i+N-1),]
})
# [[1]]
# data which
# b2 1 b
# c2 2 c
# a1 3 a
#
# [[2]]
# data which
# c1 4 c
# a2 5 a
# b1 6 b
Here is a brute force method with expand.grid and three vectors as in the example
# get all combinations
df <- expand.grid(a,b,c)
Using combn to calculate difference for each pairwise combination.
# get all parwise differences
myDiffs <- combn(names(df), 2, FUN=function(x) abs(x[1]-x[2]))
# subset data using `rowSums` and `which`
df[which(rowSums(myDiffs == 1) == ncol(myDiffs)-1), ]
df[which(rowSums(myDiffs == 1) == ncol(myDiffs)-1), ]
Var1 Var2 Var3
2 5 6 4
11 3 1 2
I have hacked together a little recursive function that will find all the consecutive triplets amongst as many vectors as you pass it (need to pass at least three). It is probably a little crude, but seems to work.
The function uses the ellipsis, ..., for passing arguments. Hence it will take however many arguments (i.e. numeric vectors) you provide and put them in the list items. Then the smallest value amongst each passed vector is located, along with its index.
Then the indeces of the vectors corresponding to the smallest triplet are created and iterated through using a for() loop, where the output values are passed to the output vector out. The input vectors in items are pruned and passed again into the function in a recursive fashion.
Only, when all vectors are NA, i.e. there are no more values in the vectors, the function returns the final result.
library(magrittr)
# define function to find the triplets
tripl <- function(...){
items <- list(...)
# find the smallest number in each passed vector, along with its index
# output is a matrix of n-by-2, where n is the number of passed arguments
triplet.id <- lapply(items, function(x){
if(is.na(x) %>% prod) id <- c(NA, NA)
else id <- c(which(x == min(x)), x[which(x == min(x))])
}) %>% unlist %>% matrix(., ncol=2, byrow=T)
# find the smallest triplet from the passed vectors
index <- order(triplet.id[,2])[1:3]
# create empty vector for output
out <- vector()
# go through the smallest triplet's indices
for(i in index){
# .. append the coresponding item from the input vector to the out vector
# .. and remove the value from the input vector
if(length(items[[i]]) == 1) {
out <- append(out, items[[i]])
# .. if the input vector has no value left fill with NA
items[[i]] <- NA
}
else {
out <- append(out, items[[i]][triplet.id[i,1]])
items[[i]] <- items[[i]][-triplet.id[i,1]]
}
}
# recurse until all vectors are empty (NA)
if(!prod(unlist(is.na(items)))) out <- append(list(out),
do.call("tripl", c(items), quote = F))
else(out <- list(out))
# return result
return(out)
}
The function can be called by passing the input vectors as arguments.
# input vectors
a = c(3,5)
b = c(6,1,8,7)
c = c(4,2,9)
# find all the triplets using our function
y <- tripl(a,b,c)
The result is a list, which contains all the neccesary information, albeit unordered.
print(y)
# [[1]]
# [1] 1 2 3
#
# [[2]]
# [1] 4 5 6
#
# [[3]]
# [1] 7 9 NA
#
# [[4]]
# [1] 8 NA NA
Ordering everything can be done using sapply():
# put everything in order
sapply(y, function(x){x[order(x)]}) %>% t
# [,1] [,2] [,3]
# [1,] 1 2 3
# [2,] 4 5 6
# [3,] 7 9 NA
# [4,] 8 NA NA
The thing is, that it will use only one value per vector to find triplets.
It will therefore not find the consecutive triplet c(6,7,8) among e.g. c(6,7,11), c(8,9,13) and c(10,12,14).
In this instance it would return c(6,8,10) (see below).
a<-c(6,7,11)
b<-c(8,9,13)
c<-c(10,12,14)
y <- tripl(a,b,c)
sapply(y, function(x){x[order(x)]}) %>% t
# [,1] [,2] [,3]
# [1,] 6 8 10
# [2,] 7 9 12
# [3,] 11 13 14
I'm slightly confused as to whether lapply works on a list or on a vector. See two examples below
Here, the mean function is applied over an array of numbers, ie, 1 to 5
x = list(a=1:5, b=rnorm(10))
x
#$a
#[1] 1 2 3 4 5
#
#$b
#[1] -0.57544290 0.51035240 0.43143241 -0.97971957 -0.99845378
#[6] 0.77389008 -0.08464382 0.68420547 1.64793617 -0.39688809
lapply(x,mean)
#$a
#[1] 3
#
#$b
#[1] 0.1012668
But here, the runif function is applied over each individual element of the array
x = 1:4
lapply(x,runif)
#[[1]]
#[1] 0.5914268
#[[2]]
#[1] 0.6762355 0.3072287
#[[3]]
#[1] 0.8439318 0.8488374 0.1158645
#[[4]]
#[1] 0.8519037 0.8384169 0.2894639 0.4066553
My question is, what exactly does lapply work on? an array or an individual element? And how does it choose it correctly depending on the function?
lapply will work on whatever is the highest level which defines the structure of the R object.
If I have 4 individual integers, lapply will work on each integer:
x <- 1:4
lapply(x, identity)
#[[1]]
#[1] 1
#
#[[2]]
#[1] 2
#
#[[3]]
#[1] 3
#
#[[4]]
#[1] 4
If however I have a list of length==2 each containing 2 values, lapply will work on each list object.
x <- list(1:2,3:4)
lapply(x, identity)
#[[1]]
#[1] 1 2
#
#[[2]]
#[1] 3 4
I have two arrays with different lengths
value <- c(1,1,1,4,4,4,1,1,1)
time <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)
How can I resize the value array to make it same length as the time array, saving it's approximate values ?
approx() function tells that lengths are differ.
I want to get value array to be like
value <- c(1,1,1,1,1,4,4,4,4,4,4,1,1,1,1)
time <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)
so lengths are equal
UPD
Okay, the main goal is to calculate correlation of v1 from v2, where
v1 inside of data.frame v1,t1 , and v2 inside of data.frame v2,t2.
the v1,t1 and v2,t2 data frames have different lengths, but we know that t1 and t2 is for equal time period so we can overlay them.
for t1 we have 1,3,5,7,9 and for t2 we have 1,2,3,4,5,6,7,8,9,10.
The problem is that two data frames are recorded separately but simultaneusly so I need to scale one of them to overlay another data.frame. And then I can calculate correlation of how v1 affects on v2.
That why I need to scale v1 to t2 length.
I'm sorry guys, I dont know how to write the goal correctly in english.
You may use the xout argument in approx
"xout: an optional set of numeric values specifying where interpolation is to take place.".
# create some fake data, which I _think_ may resemble the data you described in edit.
set.seed(123)
# "for t1 we have 1,3,5,7,9"
df1 <- data.frame(time = c(1, 3, 5, 7, 9), value = sample(1:10, 5))
df1
# "for t2 we have 1,2,3,4,5,6,7,8,9,10", the 'full time series'.
df2 <- data.frame(time = 1:10, value = sample(1:10))
# interpolate using approx and the xout argument
# The time values for 'full time series', df2$time, is used as `xout`.
# default values of arguments (e.g. linear interpolation, no extrapolation)
interpol1 <- with(df1, approx(x = time, y = value, xout = df2$time))
# some arguments you may wish to check
# extrapolation rules
interpol2 <- with(df1, approx(x = time, y = value, xout = df2$time,
rule = 2))
# interpolation method ('last observation carried forward")
interpol3 <- with(df1, approx(x = time, y = value, xout = df2$time,
rule = 2, method = "constant"))
df1
# time value
# 1 1 3
# 2 3 8
# 3 5 4
# 4 7 7
# 5 9 6
interpol1
# $x
# [1] 1 2 3 4 5 6 7 8 9 10
#
# $y
# [1] 3.0 5.5 8.0 6.0 4.0 5.5 7.0 6.5 6.0 NA
interpol3
# $x
# [1] 1 2 3 4 5 6 7 8 9 10
#
# $y
# [1] 3 3 8 8 4 4 7 7 6 6
# correlation between a vector of inter-(extra-)polated values
# and the 'full' time series
cor.test(interpol3$y, df2$value)
This little function tries to pad the values in the shorter vector out as evenly as possible and is generalisable. Haven't thought too much about edge cases, and I am sure there are many that break it. Plus it seems like it could be simplified, but is this what you are looking to do...
pad <- function(x,y){
fill <- length(y) - length(x)
run <- rle(x)
add <- fill %/% length(run$lengths)
pad <- diff( c( 0 , as.integer( seq( add , fill , length.out = length(run$lengths) ) ) ) )
rep(run$values , times = run$lengths+pad)
}
pad(value,time)
[1] 1 1 1 1 1 4 4 4 4 4 1 1 1 1 1
Or e.g.
value <- 1:2
time <- 1:10
pad(value,time)
[1] 1 1 1 1 1 2 2 2 2 2