2

我正在使用包含大量 NA 的矩阵。我想在一个新矩阵中记录每个 NA 序列的长度。

下面的例子应该更简单。

#Generating a random 5x5 population matrix with 15 NA
M=matrix(sample(1:9,25,T),5)
M[sample(1:length(M),15,F)]=NA
dimnames(M)=list(paste(rep("City",dim(M)[1]),1:dim(M)[1],sep=""),paste(rep("Year",dim(M)[2]),1:dim(M)[2],sep=""))
M

      Year1 Year2 Year3 Year4 Year5
City1     2    NA    NA    NA    NA
City2    NA    NA    NA     6     8
City3     1    NA    NA     6    NA
City4    NA     5    NA    NA     1
City5     8    NA     1    NA     2

所需的输出如下。例如 4 4 4 4 表示 4 个连续 NA 的序列。

          Year1 Year2 Year3 Year4 Year5
City1     0     4     4     4     4
City2     3     3     3     0     0
City3     0     2     2     0     1
City4     1     0     2     2     0
City5     0     1     0     1     0

你知道我该怎么做吗？

4

2 回答 2

2

不是有史以来最有效的代码：

r1=c(1,1,NA,1,1)
r2=c(1,NA,NA,1,1)
r3=c(1,NA,NA,NA,1)
r4=c(NA,NA,1,1,1)
r5=c(1,1,1,NA,NA)
M=rbind(r1,r2,r3,r4,r5)

就像@Pascal 指出的那样，您的方法会将整个矩阵转换为字符，因此您可以将 1 分配给 0 并执行以下操作：

M[M == 1] <- 0

(xx <- t(apply(M, 1, function(x) {
  s <- sum(is.na(x))
  if (is.na(x[1])) x[is.na(x)] <- rep(4, s) else
    if (is.na(tail(x, 1))) x[is.na(x)] <- rep(5, s) else 
    x[is.na(x)] <- s
  x
})))

#    [,1] [,2] [,3] [,4] [,5]
# r1    0    0    1    0    0
# r2    0    2    2    0    0
# r3    0    3    3    3    0
# r4    4    4    0    0    0
# r5    0    0    0    5    5

这是您想要的输出。如果您不相信我，请将 0 转换回 1 并根据整数分配字母

xx[xx > 0] <- letters[xx[xx > 0]]
xx[xx == '0'] <- 1


r1=c(1,1,"a",1,1)
r2=c(1,"b","b",1,1)
r3=c(1,"c","c","c",1)
r4=c("d","d",1,1,1)
r5=c(1,1,1,"e","e")
R=rbind(r1,r2,r3,r4,r5)


identical(R, xx)
# [1] TRUE
于 2015-05-08T05:48:02.283 回答
1

这是将应用于每一行的函数的另一个基础。我试过了，但无法避免 for 循环：

x = c(1,NA,1,NA,NA,1,NA,NA,NA,1,NA,NA,NA,NA)

#Find the Start and End of each sequence of NA's (Vectorized)
(start <- is.na(x) * c(T,!is.na(x[-length(x)])))
#>  [1] 0 1 0 1 0 0 1 0 0 0 1 0 0 0

(end <- is.na(x) * c(!is.na(x[-1]),T))
#>  [1] 0 1 0 0 1 0 0 0 1 0 0 0 0 1

# The difference betweeen the start and end of the sequence +1 is the sequence length
wStart <- which(!!start)
wEnd <- which(!!end)
sequenceLength <- wEnd[i] - wStart[i] + 1

# replace the sequence of NA's with it's class
for(i in seq_along(wStart))
    x[`:`(wStart[i],wEnd[i])] <- letters[sequenceLength] 

x
#> [1] "1" "a" "1" "b" "b" "1" "c" "c" "c" "1" "d" "d" "d" "d"

如：

(xx <- t(apply(M, 1, function(x) {
    wStart <- which(!!(is.na(x) * c(T,!is.na(x[-length(x)]))))
    wEnd <- which(!!is.na(x) * c(!is.na(x[-1]),T))
    sequenceLength <- 
    for(i in seq_along(wStart))
        x[`:`(wStart[i],wEnd[i])] <- letters[wEnd[i] - wStart[i] + 1] 
    return(x)
})))
于 2015-05-08T05:54:25.667 回答