如何在R数据框中仅将相同非NA值之间的NA替换为前一个非NA值

实现思路 #

• 对每一列数值列单独处理，先记录所有非NA值的位置和对应取值
• 检查相邻两个非NA值是否相等：如果相等，就将两个值中间的所有NA填充为前一个非NA的取值；如果不相等则不填充

代码实现 #

tidyverse 版本（简洁易读） #

library(dplyr)
library(tidyr)

# 读入示例数据
df <- structure(list(data_gas = structure(c(12779, 12780, 12781, 12786, 
12787, 12788, 12789, 12793, 12794, 12795, 12800, 12801, 12802, 
12807, 12808, 12809, 12814, 12815, 12816, 12822), class = "Date"), 
    `1` = c(NA, 2.299, NA, NA, NA, 2.299, NA, NA, 2.299, NA, 
    NA, 2.299, NA, NA, NA, 2.299, 2.299, NA, NA, NA), `3` = c(NA, 
    2.349, NA, NA, NA, NA, NA, NA, 2.349, NA, NA, NA, NA, NA, 
    2.369, NA, NA, NA, NA, NA), `4` = c(NA, 2.348, NA, NA, NA, 
    NA, NA, NA, 2.348, NA, NA, NA, NA, NA, 2.368, NA, NA, NA, 
    NA, NA)), row.names = c(NA, 20L), class = "data.frame")

# 重命名列名和示例匹配，可根据自己实际数据集调整
colnames(df) <- c("date", "1", "2", "3")

# 自定义符合要求的填充函数
fill_between_same <- function(x) {
  # 提取所有非NA值的位置和取值
  non_na_pos <- which(!is.na(x))
  non_na_val <- x[non_na_pos]
  # 遍历相邻非NA值对
  for (i in seq_len(length(non_na_val) - 1)) {
    if (non_na_val[i] == non_na_val[i + 1]) {
      start <- non_na_pos[i] + 1
      end <- non_na_pos[i + 1] - 1
      if (start <= end) {
      # 相同值中间的NA全部填充
        x[start:end] <- non_na_val[i]
      }
    }
  }
  return(x)
}

# 对所有数值列应用填充函数
df_result <- df %>%
  mutate(across(-date, fill_between_same))

# 查看输出结果
print(df_result, n = 20)

基础R版本（无需加载额外包） #

# 读入示例数据
df <- structure(list(data_gas = structure(c(12779, 12780, 12781, 12786, 
12787, 12788, 12789, 12793, 12794, 12795, 12800, 12801, 12802, 
12807, 12808, 12809, 12814, 12815, 12816, 12822), class = "Date"), 
    `1` = c(NA, 2.299, NA, NA, NA, 2.299, NA, NA, 2.299, NA, 
    NA, 2.299, NA, NA, NA, 2.299, 2.299, NA, NA, NA), `3` = c(NA, 
    2.349, NA, NA, NA, NA, NA, NA, 2.349, NA, NA, NA, NA, NA, 
    2.369, NA, NA, NA, NA, NA), `4` = c(NA, 2.348, NA, NA, NA, 
    NA, NA, NA, 2.348, NA, NA, NA, NA, NA, 2.368, NA, NA, NA, 
    NA, NA)), row.names = c(NA, 20L), class = "data.frame")
colnames(df) <- c("date", "1", "2", "3")

fill_between_same <- function(x) {
  non_na_pos <- which(!is.na(x))
  non_na_val <- x[non_na_pos]
  for (i in seq_len(length(non_na_val) - 1)) {
    if (non_na_val[i] == non_na_val[i + 1]) {
      start <- non_na_pos[i] + 1
      end <- non_na_pos[i + 1] - 1
      if (start <= end) {
        x[start:end] <- non_na_val[i]
      }
    }
  }
  return(x)
}

# 对除日期列外的所有列应用函数
df[, -1] <- lapply(df[, -1], fill_between_same)

以上代码运行后输出结果和你给出的理想输出完全一致。

内容的提问来源于stack exchange，提问作者moreirasd