Remove isolated elements of a vector
I have a vector of integers and I want to filter it by eliminating the components that are "isolated".
What do I mean by "isolated"? those components that does not lie in an 4-neighbourhood of other component.
The components in the vector are ordered increasingly, and there are no repetitions.
For example if I have c(1,2,3,8,15,16,17) then I need to eliminate 8 because is not in a 4-neighbourhood of other element.
I've tried applying
for (p in 1:(length(index)-2))
if((index[p+1]>3+index[p])&(index[p+2]>3+index[p+1])){index[p+1]<-0}
index<-index[index!=0]
where index is my vector of interest, but there's some problem with the logical condition.
Could you please give me some hints?
Thanks in advance.
r if-statement vector
asked Feb 25 at 8:42
RScrlliRScrlli
13511
add a comment |
I have a vector of integers and I want to filter it by eliminating the components that are "isolated".
What do I mean by "isolated"? those components that does not lie in an 4-neighbourhood of other component.
The components in the vector are ordered increasingly, and there are no repetitions.
For example if I have c(1,2,3,8,15,16,17) then I need to eliminate 8 because is not in a 4-neighbourhood of other element.
I've tried applying
for (p in 1:(length(index)-2))
if((index[p+1]>3+index[p])&(index[p+2]>3+index[p+1])){index[p+1]<-0}
index<-index[index!=0]
where index is my vector of interest, but there's some problem with the logical condition.
Could you please give me some hints?
Thanks in advance.
r if-statement vector
asked Feb 25 at 8:42
RScrlliRScrlli
13511
add a comment |
I have a vector of integers and I want to filter it by eliminating the components that are "isolated".
What do I mean by "isolated"? those components that does not lie in an 4-neighbourhood of other component.
The components in the vector are ordered increasingly, and there are no repetitions.
For example if I have c(1,2,3,8,15,16,17) then I need to eliminate 8 because is not in a 4-neighbourhood of other element.
I've tried applying
for (p in 1:(length(index)-2))
if((index[p+1]>3+index[p])&(index[p+2]>3+index[p+1])){index[p+1]<-0}
index<-index[index!=0]
where index is my vector of interest, but there's some problem with the logical condition.
Could you please give me some hints?
Thanks in advance.
r if-statement vector
asked Feb 25 at 8:42
RScrlliRScrlli
13511
I have a vector of integers and I want to filter it by eliminating the components that are "isolated".
What do I mean by "isolated"? those components that does not lie in an 4-neighbourhood of other component.
The components in the vector are ordered increasingly, and there are no repetitions.
For example if I have c(1,2,3,8,15,16,17) then I need to eliminate 8 because is not in a 4-neighbourhood of other element.
I've tried applying
for (p in 1:(length(index)-2))
if((index[p+1]>3+index[p])&(index[p+2]>3+index[p+1])){index[p+1]<-0}
index<-index[index!=0]
where index is my vector of interest, but there's some problem with the logical condition.
Could you please give me some hints?
Thanks in advance.
r if-statement vector
r if-statement vector
asked Feb 25 at 8:42
RScrlliRScrlli
13511
asked Feb 25 at 8:42
RScrlliRScrlli
13511
edited Feb 25 at 8:57
RScrlli
asked Feb 25 at 8:42
RScrlliRScrlli
13511
asked Feb 25 at 8:42
RScrlliRScrlli
13511
asked Feb 25 at 8:42
RScrlliRScrlli
13511
13511
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You can achieve it with a combination of outer and colSums, i.e.
x[colSums(abs(outer(x, x, `-`)) >= 4) == length(x)-1]
#[1] 8
To eliminate the values, we can do,
i1 <- colSums(outer(x, x, FUN = function(i, j) abs(i - j) >= 4)) == length(x) - 1
x[!i1]
#[1] 1 2 3 15 16 17
where,
x <- c(1,2,3,8,15,16,17)
answered Feb 25 at 8:52
SotosSotos
31.1k51741
add a comment |
We keep values where preceding or next difference is lower or equal to 4 :
v <- c(1,2,3,8,15,16,17)
v[c(FALSE, diff(v) <= 4) | c(diff(v) <= 4, FALSE)]
answered Feb 25 at 9:01
ClemsangClemsang
802414
2
Neat, but since the vector is assumed to be ordered, you can even remove theabs:)
– niko
Feb 25 at 13:23
@nate Done thanks !
– Clemsang
Feb 25 at 14:04
add a comment |
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2 Answers
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oldest
votes
2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
You can achieve it with a combination of outer and colSums, i.e.
x[colSums(abs(outer(x, x, `-`)) >= 4) == length(x)-1]
#[1] 8
To eliminate the values, we can do,
i1 <- colSums(outer(x, x, FUN = function(i, j) abs(i - j) >= 4)) == length(x) - 1
x[!i1]
#[1] 1 2 3 15 16 17
where,
x <- c(1,2,3,8,15,16,17)
answered Feb 25 at 8:52
SotosSotos
31.1k51741
add a comment |
You can achieve it with a combination of outer and colSums, i.e.
x[colSums(abs(outer(x, x, `-`)) >= 4) == length(x)-1]
#[1] 8
To eliminate the values, we can do,
i1 <- colSums(outer(x, x, FUN = function(i, j) abs(i - j) >= 4)) == length(x) - 1
x[!i1]
#[1] 1 2 3 15 16 17
where,
x <- c(1,2,3,8,15,16,17)
answered Feb 25 at 8:52
SotosSotos
31.1k51741
add a comment |
You can achieve it with a combination of outer and colSums, i.e.
x[colSums(abs(outer(x, x, `-`)) >= 4) == length(x)-1]
#[1] 8
To eliminate the values, we can do,
i1 <- colSums(outer(x, x, FUN = function(i, j) abs(i - j) >= 4)) == length(x) - 1
x[!i1]
#[1] 1 2 3 15 16 17
where,
x <- c(1,2,3,8,15,16,17)
answered Feb 25 at 8:52
SotosSotos
31.1k51741
You can achieve it with a combination of outer and colSums, i.e.
x[colSums(abs(outer(x, x, `-`)) >= 4) == length(x)-1]
#[1] 8
To eliminate the values, we can do,
i1 <- colSums(outer(x, x, FUN = function(i, j) abs(i - j) >= 4)) == length(x) - 1
x[!i1]
#[1] 1 2 3 15 16 17
where,
x <- c(1,2,3,8,15,16,17)
answered Feb 25 at 8:52
SotosSotos
31.1k51741
edited Feb 25 at 8:58
answered Feb 25 at 8:52
SotosSotos
31.1k51741
answered Feb 25 at 8:52
SotosSotos
31.1k51741
answered Feb 25 at 8:52
SotosSotos
31.1k51741
31.1k51741
add a comment |
add a comment |
We keep values where preceding or next difference is lower or equal to 4 :
v <- c(1,2,3,8,15,16,17)
v[c(FALSE, diff(v) <= 4) | c(diff(v) <= 4, FALSE)]
answered Feb 25 at 9:01
ClemsangClemsang
802414
2
Neat, but since the vector is assumed to be ordered, you can even remove theabs:)
– niko
Feb 25 at 13:23
@nate Done thanks !
– Clemsang
Feb 25 at 14:04
add a comment |
We keep values where preceding or next difference is lower or equal to 4 :
v <- c(1,2,3,8,15,16,17)
v[c(FALSE, diff(v) <= 4) | c(diff(v) <= 4, FALSE)]
answered Feb 25 at 9:01
ClemsangClemsang
802414
2
Neat, but since the vector is assumed to be ordered, you can even remove theabs:)
– niko
Feb 25 at 13:23
@nate Done thanks !
– Clemsang
Feb 25 at 14:04
add a comment |
We keep values where preceding or next difference is lower or equal to 4 :
v <- c(1,2,3,8,15,16,17)
v[c(FALSE, diff(v) <= 4) | c(diff(v) <= 4, FALSE)]
answered Feb 25 at 9:01
ClemsangClemsang
802414
We keep values where preceding or next difference is lower or equal to 4 :
v <- c(1,2,3,8,15,16,17)
v[c(FALSE, diff(v) <= 4) | c(diff(v) <= 4, FALSE)]
answered Feb 25 at 9:01
ClemsangClemsang
802414
edited Feb 25 at 14:04
answered Feb 25 at 9:01
ClemsangClemsang
802414
answered Feb 25 at 9:01
ClemsangClemsang
802414
answered Feb 25 at 9:01
ClemsangClemsang
802414
802414
2
Neat, but since the vector is assumed to be ordered, you can even remove theabs:)
– niko
Feb 25 at 13:23
@nate Done thanks !
– Clemsang
Feb 25 at 14:04
add a comment |
2
Neat, but since the vector is assumed to be ordered, you can even remove theabs:)
– niko
Feb 25 at 13:23
@nate Done thanks !
– Clemsang
Feb 25 at 14:04
2
2
Neat, but since the vector is assumed to be ordered, you can even remove the
abs :)– niko
Feb 25 at 13:23
Neat, but since the vector is assumed to be ordered, you can even remove the
abs :)– niko
Feb 25 at 13:23
@nate Done thanks !
– Clemsang
Feb 25 at 14:04
@nate Done thanks !
– Clemsang
Feb 25 at 14:04
add a comment |
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