Question

我想在阵列的开始处对NA观测值进行估算，使用以下两个非NA观测值的线性近似来推断缺失值。然后使用前两个非NA观测值对阵列末端的NA观测值进行相同的处理。

我的df的可重现的例子：

M=matrix(sample(1:9,10*10,T),10);M[sample(1:length(M),0.5*length(M),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 Year6 Year7 Year8 Year9 Year10
City1     NA     4     5    NA     3    NA    NA    NA     5     NA
City2      6    NA     3     3    NA     4     6    NA    NA      7
City3     NA     7    NA     8     8    NA    NA     8    NA      5
City4      3     5     3    NA    NA     3     5     9     8      7
City5      4     6     6    NA    NA     8    NA     7     1     NA
City6     NA    NA    NA    NA     4    NA     8     3     6      7
City7      9     3    NA    NA    NA    NA    NA     4    NA     NA
City8      5     6     9     8     5    NA    NA     1     4     NA
City9     NA    NA     6    NA     3     3     8    NA     7     NA
City10    NA    NA    NA    NA    NA    NA    NA    NA    NA      1

idx=rowSums(!is.na(M))>=2 # Index of rows with 2 or more non-NA to run na.approx

library(zoo)
M[idx,]=t(na.approx(t(M[idx,]),rule=1,method="linear")) # I'm using t as na.approx works on columns

       Year1 Year2 Year3 Year4    Year5 Year6 Year7 Year8 Year9 Year10
City1     NA   4.0     5   4.0 3.000000  3.50   4.0   4.5     5     NA
City2    6.0   5.5     3   3.0 5.500000  4.00   6.0   6.0     6      7
City3    4.5   7.0     3   8.0 8.000000  3.50   5.5   8.0     7      5
City4    3.0   5.0     3   8.0 6.666667  3.00   5.0   9.0     8      7
City5    4.0   6.0     6   8.0 5.333333  8.00   6.5   7.0     1      7
City6    6.5   4.5     7   8.0 4.000000  6.75   8.0   3.0     6      7
City7    9.0   3.0     8   8.0 4.500000  5.50   8.0   4.0     5     NA
City8    5.0   6.0     9   8.0 5.000000  4.25   8.0   1.0     4     NA
City9     NA    NA     6   4.5 3.000000  3.00   8.0   7.5     7     NA
City10    NA    NA    NA    NA       NA    NA    NA    NA    NA      1

我想根据前两个观察结果使用线性近似来推断边界（City1和City9）。例如，M[1,1]应为3，M[1,10]应为5,5。

你知道我怎么能这样做吗？

Answer 1

这为您提供了第一列，其中填充了NA的线性外推值。你可以适应最后一栏。

firstNAfill <- function(x) {
  ans <- ifelse(!is.na(x[1]),
                x[1],
                ifelse(sum(!is.na(x))<2, NA,
                       2*x[which(!is.na(x[1, ]))[1]] - x[which(!is.na(x[1, ]))[2]]
                )
  )
  return(ans)
}


dat$Year1 <- unlist(lapply(seq(1:nrow(dat)), function(x) {firstNAfill(dat[x, ])}))

<强>结果：

       Year1 Year2 Year3 Year4    Year5 Year6 Year7 Year8 Year9 Year10
City1    3.0   4.0     5   4.0 3.000000  3.50   4.0   4.5     5     NA
City2    6.0   5.5     3   3.0 5.500000  4.00   6.0   6.0     6      7
City3    4.5   7.0     3   8.0 8.000000  3.50   5.5   8.0     7      5
City4    3.0   5.0     3   8.0 6.666667  3.00   5.0   9.0     8      7
City5    4.0   6.0     6   8.0 5.333333  8.00   6.5   7.0     1      7
City6    6.5   4.5     7   8.0 4.000000  6.75   8.0   3.0     6      7
City7    9.0   3.0     8   8.0 4.500000  5.50   8.0   4.0     5     NA
City8    5.0   6.0     9   8.0 5.000000  4.25   8.0   1.0     4     NA
City9    7.5    NA     6   4.5 3.000000  3.00   8.0   7.5     7     NA
City10    NA    NA    NA    NA       NA    NA    NA    NA    NA      1

如果不是NA，则函数返回第一列的当前值，如果没有两个值可以推断，则返回NA，否则返回外推值。

Answer 2

在extrap中，nlead是输入向量x中的主要NA的数量。 non.na是x的元素子集，不是NA。如果没有前导NA元素或者少于2个非NA元素，则返回输入。 m是前两个非NA的斜率。用推断替换nlead的第一个x元素。最后，我们使用extrap将M应用于MM[] <-的每一行，以便保留列名，然后反转每一行，重复和反向：

library(zoo)

extrap <- function(x) {
    nlead <- which.min(x * 0) - 1
    non.na <- na.omit(x)
    if (length(nlead) == 0 || nlead == 0) || length(non.na) < 2) return(x)
    m <- diff(head(non.na, 2))      
    replace(x, seq_len(nlead), non.na[1] - nlead:1 * m)
}

nc <- ncol(M)

naApprox <- function(x) if (length(na.omit(x)) < 2) x else na.approx(x, na.rm = FALSE)
MM <- M
MM[] <- t(apply(MM, 1, naApprox))

MM[] <- t(apply(MM, 1, extrap)) # extraploate to fill leading NAs
MM[] <- t(apply(MM[, nc:1], 1, extrap))[, nc:1] # extrapolate to fill trailing NAs

，并提供：

> MM
       Year1 Year2    Year3    Year4    Year5    Year6    Year7    Year8    Year9    Year10
City1    3.0   4.0 5.000000 4.000000 3.000000 3.500000 4.000000 4.500000 5.000000  5.500000
City2    6.0   4.5 3.000000 3.000000 3.500000 4.000000 6.000000 6.333333 6.666667  7.000000
City3    6.5   7.0 7.500000 8.000000 8.000000 8.000000 8.000000 8.000000 6.500000  5.000000
City4    3.0   5.0 3.000000 3.000000 3.000000 3.000000 5.000000 9.000000 8.000000  7.000000
City5    4.0   6.0 6.000000 6.666667 7.333333 8.000000 7.500000 7.000000 1.000000 -5.000000
City6   -4.0  -2.0 0.000000 2.000000 4.000000 6.000000 8.000000 3.000000 6.000000  7.000000
City7    9.0   3.0 3.166667 3.333333 3.500000 3.666667 3.833333 4.000000 4.166667  4.333333
City8    5.0   6.0 9.000000 8.000000 5.000000 3.666667 2.333333 1.000000 4.000000  7.000000
City9    9.0   7.5 6.000000 4.500000 3.000000 3.000000 8.000000 7.500000 7.000000  6.500000
City10    NA    NA       NA       NA       NA       NA       NA       NA       NA  1.000000

注意我们将其用作M：

M <- structure(c(NA, 6L, NA, 3L, 4L, NA, 9L, 5L, NA, NA, 4L, NA, 7L, 
5L, 6L, NA, 3L, 6L, NA, NA, 5L, 3L, NA, 3L, 6L, NA, NA, 9L, 6L, 
NA, NA, 3L, 8L, NA, NA, NA, NA, 8L, NA, NA, 3L, NA, 8L, NA, NA, 
4L, NA, 5L, 3L, NA, NA, 4L, NA, 3L, 8L, NA, NA, NA, 3L, NA, NA, 
6L, NA, 5L, NA, 8L, NA, NA, 8L, NA, NA, NA, 8L, 9L, 7L, 3L, 4L, 
1L, NA, NA, 5L, NA, NA, 8L, 1L, 6L, NA, 4L, 7L, NA, NA, 7L, 5L, 
7L, NA, 7L, NA, NA, NA, 1L), .Dim = c(10L, 10L), .Dimnames = list(
    c("City1", "City2", "City3", "City4", "City5", "City6", "City7", 
    "City8", "City9", "City10"), c("Year1", "Year2", "Year3", 
    "Year4", "Year5", "Year6", "Year7", "Year8", "Year9", "Year10"
    )))

更新：已修复。

使用线性近似来限制NA观测的估算

2 个答案: