R - 沿矩阵对角线填充长度为 'n' 或更短的数据间隙
R -Fill in data gaps with length of 'n' or less along a matrix diagonal
我正在处理一些大矩阵,其值沿对角线类似于以下内容。
ontrack <- matrix(c(
runif(1),NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
runif(1),NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,runif(1),NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,runif(1),runif(1),NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,runif(1),NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,runif(1)),
nrow=14, byrow=T
)
我想填充长度为 'n' 或更短的数据间隙以连接对角线段。例如使用上面的矩阵并填充 3 或更少的数据间隙,我想从这个开始:
diag_indx <- which(!is.na(ontrack), arr.ind=T)
这给出了
row col
[1,] 1 1
[2,] 2 1
[3,] 3 3
[4,] 7 5
[5,] 7 6
[6,] 9 8
[7,] 14 13
至此
row col
1 1
2 1
newV 3 2
3 3
new 4 4
new 5 4
new 6 4
7 5
7 6
new 8 7
9 8
14 13
对于 newV
这样的实例,结果可以是 (2,2) 或 (3,2)。我的后续代码使用 diag_indx
矩阵,但如果效率更高,可以直接在 ontrack
矩阵中填充数据间隙(使用任何值都可以)。
在尝试制定解决方案时,我设想使用此 sequence length equation
在 diag_indx
矩阵中找到数据间隙
seqle <- function(x, incr=1) {
if(!is.integer(x)) x <- as.integer(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)])
}
然后使用 seq()
填补数据空白。我只是不确定如何有效地将它们组合在一起。谢谢你的帮助。
经过反复试验,我想出了一个(不太漂亮)只需要基本 R 函数的解决方案。
diagFillSeq <- function(diag_indx, fillgap=1){
repeat{
for(cols in 1:2){
diag_indx <- diag_indx[order(diag_indx[, cols]), ] #Sort by selected column
repeat{
diffs <- diff(diag_indx[, cols])
#Find breaks in sequence with differences >1 (diffs==1 are in sequence) and less than or equal to fillgap
gap_indx <- which(diffs > 1 & diffs <= (fillgap +1)) #need +1 because fencepost error: 3rd & 7th post diffs=4 but fillgap=3)
if(length(gap_indx) == 0){break}
insert_indx <- gap_indx[1]
seq_length <- diffs[gap_indx[1]] - 1 #need -1 because fencepost error
#Subset diag_indx and insert filling sequence
diag_indx <- rbind(diag_indx[1:insert_indx, ],
cbind(
as.integer( seq(from=diag_indx[insert_indx, 1] +1, to=diag_indx[insert_indx+1, 1] -1, length.out=seq_length) ),
as.integer( seq(from=diag_indx[insert_indx, 2] +1, to=diag_indx[insert_indx+1, 2] -1, length.out=seq_length) )
),
diag_indx[(insert_indx+1):nrow(diag_indx), ])
}
}
#Recheck first column to see if any new sequence gaps were created
diffs <- diff(diag_indx[, 1])
gap_indx <- which(diffs > 1 & diffs <= (fillgap +1))
if(length(gap_indx) == 0){return(unname(diag_indx))}
}
}
以及上面diag_indx
的测试
whatIwant <- matrix(as.integer(c(1,2,3,3,4,5,6,7,7,8,9,14, 1,1,2,3,4,4,4,5,6,7,8,13)), ncol=2)
whatIwant
# [,1] [,2]
# [1,] 1 1
# [2,] 2 1
# [3,] 3 2
# [4,] 3 3
# [5,] 4 4
# [6,] 5 4
# [7,] 6 4
# [8,] 7 5
# [9,] 7 6
#[10,] 8 7
#[11,] 9 8
#[12,] 14 13
identical(diagFillSeq(diag_indx, fillgap=3), whatIwant)
#TRUE
我正在处理一些大矩阵,其值沿对角线类似于以下内容。
ontrack <- matrix(c(
runif(1),NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
runif(1),NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,runif(1),NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,runif(1),runif(1),NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,runif(1),NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,
NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,runif(1)),
nrow=14, byrow=T
)
我想填充长度为 'n' 或更短的数据间隙以连接对角线段。例如使用上面的矩阵并填充 3 或更少的数据间隙,我想从这个开始:
diag_indx <- which(!is.na(ontrack), arr.ind=T)
这给出了
row col
[1,] 1 1
[2,] 2 1
[3,] 3 3
[4,] 7 5
[5,] 7 6
[6,] 9 8
[7,] 14 13
至此
row col
1 1
2 1
newV 3 2
3 3
new 4 4
new 5 4
new 6 4
7 5
7 6
new 8 7
9 8
14 13
对于 newV
这样的实例,结果可以是 (2,2) 或 (3,2)。我的后续代码使用 diag_indx
矩阵,但如果效率更高,可以直接在 ontrack
矩阵中填充数据间隙(使用任何值都可以)。
在尝试制定解决方案时,我设想使用此 sequence length equation
在diag_indx
矩阵中找到数据间隙
seqle <- function(x, incr=1) {
if(!is.integer(x)) x <- as.integer(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)])
}
然后使用 seq()
填补数据空白。我只是不确定如何有效地将它们组合在一起。谢谢你的帮助。
经过反复试验,我想出了一个(不太漂亮)只需要基本 R 函数的解决方案。
diagFillSeq <- function(diag_indx, fillgap=1){
repeat{
for(cols in 1:2){
diag_indx <- diag_indx[order(diag_indx[, cols]), ] #Sort by selected column
repeat{
diffs <- diff(diag_indx[, cols])
#Find breaks in sequence with differences >1 (diffs==1 are in sequence) and less than or equal to fillgap
gap_indx <- which(diffs > 1 & diffs <= (fillgap +1)) #need +1 because fencepost error: 3rd & 7th post diffs=4 but fillgap=3)
if(length(gap_indx) == 0){break}
insert_indx <- gap_indx[1]
seq_length <- diffs[gap_indx[1]] - 1 #need -1 because fencepost error
#Subset diag_indx and insert filling sequence
diag_indx <- rbind(diag_indx[1:insert_indx, ],
cbind(
as.integer( seq(from=diag_indx[insert_indx, 1] +1, to=diag_indx[insert_indx+1, 1] -1, length.out=seq_length) ),
as.integer( seq(from=diag_indx[insert_indx, 2] +1, to=diag_indx[insert_indx+1, 2] -1, length.out=seq_length) )
),
diag_indx[(insert_indx+1):nrow(diag_indx), ])
}
}
#Recheck first column to see if any new sequence gaps were created
diffs <- diff(diag_indx[, 1])
gap_indx <- which(diffs > 1 & diffs <= (fillgap +1))
if(length(gap_indx) == 0){return(unname(diag_indx))}
}
}
以及上面diag_indx
的测试
whatIwant <- matrix(as.integer(c(1,2,3,3,4,5,6,7,7,8,9,14, 1,1,2,3,4,4,4,5,6,7,8,13)), ncol=2)
whatIwant
# [,1] [,2]
# [1,] 1 1
# [2,] 2 1
# [3,] 3 2
# [4,] 3 3
# [5,] 4 4
# [6,] 5 4
# [7,] 6 4
# [8,] 7 5
# [9,] 7 6
#[10,] 8 7
#[11,] 9 8
#[12,] 14 13
identical(diagFillSeq(diag_indx, fillgap=3), whatIwant)
#TRUE