Question

我正在尝试替换向量中的缺失值（NA）。两个相等数字之间的NA被该数字替换。两个不同值之间的NA应保持NA。例如，给定向量＆＃34; a＆＃34;，我希望它是＆＃34; b＆＃34;。

a = c(1, NA, NA, NA, 1, NA, NA, NA, 2, NA, NA, 2, 3, NA, NA, 3)
b = c(1, 1, 1, 1, 1, NA, NA, NA, 2, 2, 2, 2, 3, 3, 3, 3)

如您所见，价值NA和1之间的第二轮2未被替换。

有没有办法对计算进行矢量化？

Answer 1

OP要求提供一个 vecgorized 解决方案，所以这里有一个可能的矢量化基础R解决方案（没有for循环），它也处理领先/滞后的NAs的情况

# Define a vector with Leading/Lagging NAs
a <- c(NA, NA, 1, NA, NA, NA, 1, NA, NA, NA, 2, NA, NA, 2, 3, NA, NA, 3, NA, NA)

# Save the boolean vector as we are going to reuse it a lot
na_vals <- is.na(a)

# Find the NAs location compared to the non-NAs
ind <- findInterval(which(na_vals), which(!na_vals))

# Find the consecutive values that equal
ind2 <- which(!diff(a[!na_vals]))

# Fill only NAs between equal consequtive files
a[na_vals] <- a[!na_vals][ind2[match(ind, ind2)]]
a
# [1] NA NA  1  1  1  1  1 NA NA NA  2  2  2  2  3  3  3  3 NA NA

对大型载体进行一些时间比较

# Create a big vector
set.seed(123)
a <- sample(c(NA, 1:5), 5e7, replace = TRUE)

############################################
##### Cainã Max Couto-Silva

fill_data <- function(vec) {

  for(l in unique(vec[!is.na(vec)])) {

    g <- which(vec %in% l)

    indexes <- list()

    for(i in 1:(length(g) - 1)) {
      indexes[[i]] <- (g[i]+1):(g[i+1]-1)
    }

    for(i in 1:(length(g) - 1)) { 
      if(all(is.na(vec[indexes[[i]]]))) {
        vec[indexes[[i]]] <- l
      }
    }
  }

  return(vec)
}

system.time(res <- fill_data(a))
#   user  system elapsed 
#  81.73    4.41   86.48 

############################################
##### Henrik

system.time({
  a_ap <- na.approx(a, na.rm = FALSE)
  a_locf <- na.locf(a, na.rm = FALSE)
  a[which(a_ap == a_locf)] <- a_ap[which(a_ap == a_locf)]
})
#  user  system elapsed 
# 12.55    3.39   15.98 

# Validate
identical(res, as.integer(a))
# [1] TRUE

############################################
##### David

## Recreate a as it been overridden
set.seed(123)
a <- sample(c(NA, 1:5), 5e7, replace = TRUE)

system.time({
  # Save the boolean vector as we are going to reuse it a lot
  na_vals <- is.na(a)

  # Find the NAs location compaed to the non-NAs
  ind <- findInterval(which(na_vals), which(!na_vals))

  # Find the consecutive values that equl
  ind2 <- which(!diff(a[!na_vals]))

  # Fill only NAs between equal consequtive files
  a[na_vals] <- a[!na_vals][ind2[match(ind, ind2)]]
})
# user  system elapsed 
# 3.39    0.71    4.13 

# Validate
identical(res, a)
# [1] TRUE

Answer 2

您可以制作类似的功能：

{
    "message": "Must authenticate to access this API.",
    "documentation_url": "https://developer.github.com/enterprise/2.11/v3"
}

跑步功能：

fill_data <- function(vec) {

  for(l in unique(vec[!is.na(vec)])) {

    g <- which(vec %in% l)

    indexes <- list()

    for(i in 1:(length(g) - 1)) {
      indexes[[i]] <- (g[i]+1):(g[i+1]-1)
    }

    for(i in 1:(length(g) - 1)) { 
      if(all(is.na(vec[indexes[[i]]]))) {
        vec[indexes[[i]]] <- l
      }
    }
  }

  return(vec)
}

如果你有一个值在不同位置的向量，它也可以工作：

a = c(1, NA, NA, NA, 1, NA, NA, NA, 2, NA, NA, 2, 3, NA, NA, 3)

fill_data(a)
[1]  1  1  1  1  1 NA NA NA  2  2  2  2  3  3  3  3

说明：

首先，您会找到唯一的非NA值。

然后它获取每个唯一非NA值的索引并获取它们之间的值;

然后测试这些值是否都是NA，如果是，则用等级值替换它们。

Answer 3

您可以使用zoo包中的便捷功能。在这里，我们替换原始向量中的NA，其中插值（由na.approx创建）等于'最后一次观察结果'（由na.locf创建）：

library(zoo)
a_ap <- na.approx(a)
a_locf <- na.locf(a)
a[which(a_ap == a_locf)] <- a_ap[which(a_ap == a_locf)]
a
# [1]  1  1  1  1  1 NA NA NA  2  2  2  2  3  3  3  3

要考虑领先和尾随NA，请添加na.rm = FALSE：

a <- c(NA, 1, NA, NA, NA, 1, NA, NA, NA, 2, NA, NA, 2, 3, NA, NA, 3, NA)

a_ap <- na.approx(a, na.rm = FALSE)
a_locf <- na.locf(a, na.rm = FALSE)
a[which(a_ap == a_locf)] <- a_ap[which(a_ap == a_locf)]
a
# [1] NA  1  1  1  1  1 NA NA NA  2  2  2  2  3  3  3  3 NA

根据周围的非缺失值有条件地替换缺失值

3 个答案: