I want to create a function that produces a matrix containing several lags of a variable. A simple example that works is
a <- ts(1:10)
cbind(a, lag(a, -1))
To do this for multiple lags, I have
lagger <- function(var, lags) {
### Create list of lags
lagged <- lapply(1:lags, function(x){
lag(var, -x)
})
### Join lags together
do.call(cbind, list(var, lagged))
}
Using the above example gives unexpected results;
lagger(a, 1)
gives a length 20 list with the original time series broken out into separate list slots and the final 10 each being a replication of the lagged series.
Any suggestions to getting this working? Thanks!
This gives a lag of 0 and of 1.
library(zoo)
a <- ts(11:13)
lags <- -(0:1)
a.lag <- as.ts(lag(as.zoo(a), lags))
Now a.lag is this:
> a.lag
Time Series:
Start = 1
End = 4
Frequency = 1
lag0 lag-1
1 11 NA
2 12 11
3 13 12
4 NA 13
If you don't want the NA entries then use: as.ts(na.omit(lag(as.zoo(a), lags))) .
Based on #Joshua Ulrich answer.
I thinkd embed is the correct answer but you get the vectors in the other way around. I mean using embed you'll get the lagged series not in the proper order, see the following
lagged <- embed(a,4)
colnames(lagged) <- paste('t', 3:0, sep='-')
lagged
t-3 t-2 t-1 t-0
[1,] 4 3 2 1
[2,] 5 4 3 2
[3,] 6 5 4 3
[4,] 7 6 5 4
[5,] 8 7 6 5
[6,] 9 8 7 6
[7,] 10 9 8 7
this gives the correct answer to you but not in the correct order, since the lags are in descending order.
But it you reorder just like this:
lagged_OK <- lagged[,ncol(lagged):1]
colnames(lagged_OK) <- paste('t', 0:3, sep='-')
lagged_OK
lag.0 lag.1 lag.2 lag.3
[1,] 1 2 3 4
[2,] 2 3 4 5
[3,] 3 4 5 6
[4,] 4 5 6 7
[5,] 5 6 7 8
[6,] 6 7 8 9
[7,] 7 8 9 10
Then, you get the right lagged matrix.
I add colnames only for explanation purpose, you can just do:
embed(a,4)[ ,4:1]
If you really want a lagger function, try this
lagger <- function(x, lag=1){
lag <- lag+1
Lagged <- embed(x,lag)[ ,lag:1]
colnames(Lagged) <- paste('lag', 0:(lag-1), sep='.')
return(Lagged)
}
lagger(a, 4)
lag.0 lag.1 lag.2 lag.3 lag.4
[1,] 1 2 3 4 5
[2,] 2 3 4 5 6
[3,] 3 4 5 6 7
[4,] 4 5 6 7 8
[5,] 5 6 7 8 9
[6,] 6 7 8 9 10
lagger(a, 1)
lag.0 lag.1
[1,] 1 2
[2,] 2 3
[3,] 3 4
[4,] 4 5
[5,] 5 6
[6,] 6 7
[7,] 7 8
[8,] 8 9
[9,] 9 10
I'm not sure what's wrong with your function, but you can probably use embed instead.
> embed(a,4)
[,1] [,2] [,3] [,4]
[1,] 4 3 2 1
[2,] 5 4 3 2
[3,] 6 5 4 3
[4,] 7 6 5 4
[5,] 8 7 6 5
[6,] 9 8 7 6
[7,] 10 9 8 7
Related
Assume I have to generate 1000 Sample Pairs (Y1,Y2) (from a Normal Distribution with replacement). Each of the pairs should have 20 observations.
y1 <- rep(sample(c(1:10),10, replace = TRUE))
y2 <- rep(sample(c(1:10),10, replace = TRUE))
How would I now generate 1000 of these pairs, so that they are easy to access for further computations.
I had the idea of looping them a 1000 times and saving them in a dataframe, but this may get chaotic.
Is there a simpler/nicer way to do this? A package or a function that I am missing?
Help would be appreciated!
One way is to use replicate, i.e.
replicate(5, rep(sample(c(1:10), 10, replace = TRUE)))
# [,1] [,2] [,3] [,4] [,5]
# [1,] 3 9 2 4 5
# [2,] 4 1 10 8 1
# [3,] 5 6 1 3 7
# [4,] 1 9 9 6 5
# [5,] 5 3 4 7 9
# [6,] 4 5 4 4 5
# [7,] 2 10 9 4 9
# [8,] 3 1 10 5 3
# [9,] 7 3 10 9 10
#[10,] 10 3 10 10 1
Professionals of R, I have a question:
I have the matrix as below, and I want create the criteria: to construct the matrix using only the next strings: 1st string + i, where i=3 so I want to get the new matrix with the first, 5th, 9th strings of the initial matrix, and so the dimension of new matrix has to be 3x3. Maybe is there the special function in R for this procedure or needed to realize this task through the FUN in R?
[,1] [,2] [,3]
[1,] 1 2 15
[2,] 2 3 16
[3,] 3 4 1
[4,] 4 5 2
[5,] 5 6 3
[6,] 6 7 4
[7,] 7 8 5
[8,] 8 9 6
[9,] 9 10 7
Below the desired matrix:
1 2 15
5 6 3
9 10 7
For example suppose I have matrix A
x y z f
1 1 2 A 1005
2 2 4 B 1002
3 3 2 B 1001
4 4 8 C 1001
5 5 10 D 1004
6 6 12 D 1004
7 7 11 E 1005
8 8 14 E 1003
From this matrix I want to find the repeated values like 1001, 1005, D, 2 (in third column) and I also want to find their index (which row, or which position).
I am new to R!
Obviously it is possible to do with simple searching element by element by using a for loop, but I want to know, is there any function available in R for this kind of problem.
Furthermore, I tried using duplicated and unique, both functions are giving me the duplicated row number or column number, they are also giving me how many of them were repeated, but I can not search for whole matrix using both of them!
You can write a rather simple function to get this information. Though note that this solution works with a matrix. It does not work with a data.frame. A similar function could be written for a data.frame using the fact that the data.frame data structure is a subset of a list.
# example data
set.seed(234)
m <- matrix(sample(1:10, size=100, replace=T), 10)
find_matches <- function(mat, value) {
nr <- nrow(mat)
val_match <- which(mat == value)
out <- matrix(NA, nrow= length(val_match), ncol= 2)
out[,2] <- floor(val_match / nr) + 1
out[,1] <- val_match %% nr
return(out)
}
R> m
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 8 6 6 7 6 7 4 10 6 9
[2,] 8 6 6 3 10 4 5 4 6 9
[3,] 1 6 9 2 9 2 3 6 4 2
[4,] 8 6 7 8 3 9 9 4 9 2
[5,] 1 1 5 6 7 1 5 1 10 6
[6,] 7 5 4 7 8 2 4 4 7 10
[7,] 10 4 7 8 3 1 8 6 3 4
[8,] 8 8 2 2 7 5 6 4 10 4
[9,] 10 2 9 6 6 9 7 2 4 7
[10,] 3 9 9 4 2 7 7 2 9 6
R> find_matches(m, 8)
[,1] [,2]
[1,] 1 1
[2,] 2 1
[3,] 4 1
[4,] 8 1
[5,] 8 2
[6,] 4 4
[7,] 7 4
[8,] 6 5
[9,] 7 7
In this function, the row index is output in column 1 and the column index is output in column 2
I have a table which is
[,1] [,2] [,3] [,4] [,5]
[1,] 1 5 10 0.00040803 0.00255277
[2,] 1 11 3 0.01765470 0.01584580
[3,] 1 6 2 0.15514850 0.15509000
[4,] 1 8 14 0.02100531 0.02572320
[5,] 1 9 4 0.04748648 0.00843252
[6,] 2 5 10 0.00040760 0.06782680
[7,] 2 11 3 0.01765480 0.01584580
[8,] 2 6 2 0.15514810 0.15509000
[9,] 2 8 14 0.02100491 0.02572320
[10,] 2 9 4 0.04748608 0.00843252
[11,] 3 5 10 0.00040760 0.06782680
[12,] 3 11 3 0.01765480 0.01584580
[13,] 3 8 14 0.02100391 0.02572320
[14,] 3 9 4 0.04748508 0.00843252
[15,] 4 5 10 0.00040760 0.06782680
[16,] 4 11 3 0.01765480 0.01584580
[17,] 4 8 14 0.02100391 0.02572320
[18,] 4 9 4 0.04748508 0.00843252
[19,] 5 8 14 0.02100391 0.02572320
[20,] 5 9 4 0.04748508 0.00843252
I want to remove duplicates from this table. However, only colums 2,3,4 matter. Example: rows 1,6,11,15 are identical if only columns 2,3,4 are observed. Note for column 4: is it possible to incorporate that it is considered as being the same as long as it is within 10e-5 of the value? So that rows 1 and 6 would be considered as being identical although the value in column 4 differs slightly (within the tolerance I mentioned)?
Then it would be great to get an output which would be like:
column 2 value | column 3 value | column 1 value at which the the pair has been first observed (with the tolerance) (in the example 1) | column 1 value at which the pair has been last observed (with tolerance) (in the example 4) | value of column 4 at first appearance (0.00040803 in the example)
This is a way of thinking about it, but I'm not sure it's what you're looking for. The logic should be able to get you started though.
dat <- YOUR DATA SET
dat
V1 V2 V3 V4 V5
1 1 5 10 0.00040803 0.00255277
2 1 11 3 0.01765470 0.01584580
3 1 6 2 0.15514850 0.15509000
4 1 8 14 0.02100531 0.02572320
5 1 9 4 0.04748648 0.00843252
# TRUNCATED
dat <- dat[, c(2, 3, 4)]
dat$V4 <- round(dat$V4, 5)
unique(dat)
V2 V3 V4
1 5 10 0.00041
2 11 3 0.01765
3 6 2 0.15515
4 8 14 0.02101
5 9 4 0.04749
9 8 14 0.02100
You could do something like this:
# read your data
yy <- read.csv('your-data.csv', header=F)
## V1 V2 V3 V4 V5
## 1 1 5 10 0.00040803 0.00255277
## 2 1 11 3 0.01765470 0.01584580
## 3 1 6 2 0.15514850 0.15509000
## 4 1 8 14 0.02100531 0.02572320
# create a logical matrix indicating value is within tolerance
mat.eq.tol <- sapply(yy$V4, function(x) abs(yy$V4-x) < 1E-5)
# minimum index
eq.min <- apply(mat.eq.tol, 1, function(x) min(which(x)))
# maximum index
eq.max <- apply(mat.eq.tol, 1, function(x) max(which(x)))
# combine result
res <- cbind(yy$V2, yy$V3, yy$V1[eq.min], yy$V1[eq.max], yy$V4[eq.min])
## [,1] [,2] [,3] [,4] [,5]
## [1,] 5 10 1 4 0.00040803
## [2,] 11 3 1 4 0.01765470
## [3,] 6 2 1 2 0.15514850
## [4,] 8 14 1 5 0.02100531
## [5,] 9 4 1 5 0.04748648
## [6,] 5 10 1 4 0.00040803
Using R, I'm trying to construct a dataframe of the row and col numbers of a given matrix. E.g., if
a <- matrix(c(1:15), nrow=5, ncol=3)
then I'm looking to construct a dataframe that gives:
row col
1 1
1 2
1 3
. .
5 1
5 2
5 3
What I've tried:
row <- matrix(row(a), ncol=1, nrow=dim(a)[1]*dim(a)[2], byrow=T)
col <- matrix(col(a), ncol=1, nrow=dim(a)[1]*dim(a)[2], byrow=T)
out <- cbind(row, col)
colnames(out) <- c("row", "col")
results in:
row col
[1,] 1 1
[2,] 2 1
[3,] 3 1
[4,] 4 1
[5,] 5 1
[6,] 1 2
[7,] 2 2
[8,] 3 2
[9,] 4 2
[10,] 5 2
[11,] 1 3
[12,] 2 3
[13,] 3 3
[14,] 4 3
[15,] 5 3
Which isn't what I'm looking for, as the sequence of rows and cols in suddenly reversed, even tough I specified "byrow=T". I don't see if and where I'm making a mistake but would hugely appreciate suggestions to overcome this problem. Thanks in advance!
I'd use expand.grid on the vectors 1:ncol and 1:nrow, then flip the columns with [,2:1] to get them in the order you want:
> expand.grid(seq(ncol(a)),seq(nrow(a)))[,2:1]
Var2 Var1
1 1 1
2 1 2
3 1 3
4 2 1
5 2 2
6 2 3
7 3 1
8 3 2
9 3 3
10 4 1
11 4 2
12 4 3
13 5 1
14 5 2
15 5 3
Use row and col, but more directly manipulate their output ordering since they return corresponding indices in place for the input array. Use t to get the non-default order you want in the end:
data.frame(row = as.vector(t(row(a))), col = as.vector(t(col(a))))
row col
1 1 1
2 1 2
3 1 3
4 2 1
5 2 2
6 2 3
7 3 1
8 3 2
9 3 3
10 4 1
11 4 2
12 4 3
13 5 1
14 5 2
15 5 3
Or, as a matrix not a data.frame:
cbind(as.vector(t(row(a))), as.vector(t(col(a))))
[,1] [,2]
[1,] 1 1
[2,] 1 2
[3,] 1 3
[4,] 2 1
[5,] 2 2
[6,] 2 3
[7,] 3 1
[8,] 3 2
[9,] 3 3
[10,] 4 1
[11,] 4 2
[12,] 4 3
[13,] 5 1
[14,] 5 2
[15,] 5 3
You may want to have a look at ?expand.grid, which does just about exactly what you want to achieve.
Since there are many ways to skin a cat, I'll chip in with yet another variant based on rep:
data.frame(row=rep(seq(nrow(a)), each=ncol(a)), col=rep(seq(ncol(a)), nrow(a)))
...but to announce a "winner", I think you need to time the solutions:
# Make up a huge matrix...
a <- matrix(runif(1e7), 1e4)
system.time( a1<-data.frame(row = as.vector(t(row(a))),
col = as.vector(t(col(a)))) ) # 0.68 secs
system.time( a2<-expand.grid(col = seq(ncol(a)),
row = seq(nrow(a)))[,2:1] ) # 0.49 secs
system.time( a3<-data.frame(row=rep(seq(nrow(a)), each=ncol(a)),
col=rep(seq(ncol(a)), nrow(a))) ) # 0.59 secs
identical(a1, a2) && identical(a1, a3) # TRUE
...so it seems #Spacedman has the speediest solution!