The vector (1 2 3 4 5 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8 5 6 7 8 9)
seq() and rep() maybe can not deliver parameters.
I read the help doc but fail to find the way.
You could try
(1:5) + rep(0:4,each=5)
#[1] 1 2 3 4 5 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8 5 6 7 8 9
NOTE: (1:5) and 0:4 can be replaced by seq(1,5) and seq(0,4)
Another one:
as.vector(outer(1:5,0:4,"+"))
Related
I am looking at the point pattern data set in spatstat anemones which has 231 points with marks attached to them which define the diameter.. I want to delete the marks and the points within the point pattern when the diameter is equal to 2
Here is the data:
>
[1] 6 4 4 6 3 3 5 3 5 4 4 6 5 3 4 7 4 6 6 5 4 4 5 3 3 6 4 5 4 4 5 3 3 5
3 4 5 8 5 4 6 5 6 4 5 3 3 4 5 6 4 4 3 4 4 6 5 4 3 6 5 3 [63] 3 6 5 3
3 2 5 7 4 4 4 3 3 4 3 6 2 6 6 3 4 3 7 6 3 4 2 7 4 5 4 4 4 6 4 3 3 3 3
6 7 3 7 3 2 4 3 5 2 3 4 4 3 3 3 6 3 4 5 3 6 3 [125] 7 5 3 3 4 4 5 4 4
6 5 3 3 3 5 3 6 5 5 4 4 3 4 4 4 4 3 4 7 4 6 5 7 6 3 6 5 4 6 4 5 4 5 3
6 3 3 6 4 6 4 4 6 3 5 3 4 6 5 5 4 5 [187] 4 3 3 4 4 4 4 5 4 5 5 5 4 6
4 4 5 3 5 4 3 4 4 4 3 4 5 5 3 3 5 3 4 5 6 2 5 2 3 2 3 3 7 5 4
thanks!
Another solution is to use the generic R command subset:
X <- subset(anemones, marks != 2)
From the question it is not quite clear whether you want to get rid of all the marks after deleting these points. In that case use unmark:
X <- unmark(X)
Correct me if I'm wrong but I'm reading this as meaning you wish to remove observations when anemones$marks is equal to 2.
If so this should do it:
updated_anemones <- anemones[!anemones$marks == 2,]
My current dataset look like this
Order V1
1 7
2 5
3 8
4 5
5 8
6 3
7 4
8 2
1 8
2 6
3 3
4 4
5 5
6 7
7 3
8 6
I want to create a new variable called "V2" based on the variables "Order" and "V1". For every 8 items in the "Order" variable, I want to assign a value of "0" in "V2" if the varialbe "Order" has observation equals to 1; otherwise, "V2" takes the value of previous item in "V1".
This is the dataset that I want
Order V1 V2
1 7 0
2 5 7
3 8 5
4 5 8
5 8 5
6 3 8
7 4 3
8 2 4
1 8 0
2 6 8
3 3 6
4 4 3
5 5 4
6 7 5
7 3 7
8 6 3
Since my actual dataset is very large, I'm trying to use for loop with if statement to generate "V2". But my code keeps failing. I appreciate if anyone can help me on this, and I'm open to other statements. Thank you!
(Up front: I am assuming that the order of Order is perfectly controlled.)
You need simply ifelse and lag:
df <- read.table(text="Order V1
1 7
2 5
3 8
4 5
5 8
6 3
7 4
8 2
1 8
2 6
3 3
4 4
5 5
6 7
7 3
8 6 ", header=T)
df$V2 <- ifelse(df$Order==1, 0, lag(df$V1))
df
# Order V1 V2
# 1 1 7 0
# 2 2 5 7
# 3 3 8 5
# 4 4 5 8
# 5 5 8 5
# 6 6 3 8
# 7 7 4 3
# 8 8 2 4
# 9 1 8 0
# 10 2 6 8
# 11 3 3 6
# 12 4 4 3
# 13 5 5 4
# 14 6 7 5
# 15 7 3 7
# 16 8 6 3
with(dat,{V2<-c(0,head(V1,-1));V2[Order==1]<-0;dat$V2<-V2;dat})
Order V1 V2
1 1 7 0
2 2 5 7
3 3 8 5
4 4 5 8
5 5 8 5
6 6 3 8
7 7 4 3
8 8 2 4
9 1 8 0
10 2 6 8
11 3 3 6
12 4 4 3
13 5 5 4
14 6 7 5
15 7 3 7
16 8 6 3
How to create a vector sequence of:
2 3 4 5 6 7 8 3 4 5 6 7 8 4 5 6 7 8 5 6 7 8 6 7 8 7 8
I tried to use:
2:8+rep(0:6,each=6)
but the result is:
2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 .... 12 13 14
Please help. Thanks.
This should accomplish what you're looking for:
x = 2
VecSeq = c(x:8)
while (x < 7) {
x = x + 1
calc = c(x:8)
VecSeq = c(VecSeq, calc)
}
VecSeq # Your desired vector
you could do this:
library(purrr)
unlist(map(2:7, ~.x:8))
# [1] 2 3 4 5 6 7 8 3 4 5 6 7 8 4 5 6 7 8 5 6 7 8 6 7 8 7 8
and a little function in base R:
funky_vec <- function(from,to){unlist(sapply(from:(to-1),`:`,to))}
funky_vec(2,8)
# [1] 2 3 4 5 6 7 8 3 4 5 6 7 8 4 5 6 7 8 5 6 7 8 6 7 8 7 8
This is made really easy with sequence (since R 4.0.0):
sequence(7:2, 2:7)
# [1] 2 3 4 5 6 7 8 3 4 5 6 7 8 4 5 6 7 8 5 6 7 8 6 7 8 7 8
I have a numeric element z as below:
> sort(z)
[1] 1 5 5 5 6 6 7 7 7 7 7 9 9
I would like to sequentially reorganize this element so to have
> z
[1] 1 2 2 2 3 3 4 4 4 4 4 5 5
I guess converting z to a factor and use it as an index should be the way.
You answered it yourself really:
as.integer(factor(sort(z)))
I know this has been accepted already but I decided to look inside factor() to see how it's done there. It more or less comes down to this:
x <- sort(z)
match(x, unique(x))
Which is an extra line I suppose but it should be faster if that matters.
This should do the trick
z = sort(sample(1:10, 100, replace = TRUE))
cumsum(diff(z)) + 1
[1] 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3
[26] 3 3 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6
[51] 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8
[76] 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10 10 10 10 10
Note that diff omits the first element of the series. So to compensate:
c(1, cumsum(diff(z)) + 1)
Alternative using rle:
z = sort(sample(1:10, 100, replace = TRUE))
rle_result = rle(sort(z))
rep(rle_result$values, rle_result$lengths)
> rep(rle_result$values, rle_result$lengths)
[1] 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3
[26] 3 3 3 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 6 6 6
[51] 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8
[76] 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10 10 10 10 10
rep(seq_along(rle(x)$l), rle(x)$l)
I am looking for the Matlab way of doing the following:
> merge(2:4,3:7)
x y
1 2 3
2 3 3
3 4 3
4 2 4
5 3 4
6 4 4
7 2 5
8 3 5
9 4 5
10 2 6
11 3 6
12 4 6
13 2 7
14 3 7
15 4 7
> expand.grid(2:4,3:7)
Var1 Var2
1 2 3
2 3 3
3 4 3
4 2 4
5 3 4
6 4 4
7 2 5
8 3 5
9 4 5
10 2 6
11 3 6
12 4 6
13 2 7
14 3 7
15 4 7
I usually do it with meshgrid:
>> [x y] = meshgrid(2:4, 3:7);
>> [x(:) y(:)]
ans =
2 3
2 4
2 5
2 6
2 7
3 3
3 4
3 5
3 6
3 7
4 3
4 4
4 5
4 6
4 7
Use ndgrid for n variables (2 and more). For example (4-D space)
[X,Y,Z,T] = ndgrid(2:4, 3:7, 1:2, 1:10);