在R语言中计算相邻列差值并插入原数据集

先构造和你一致的示例数据集：

df <- data.frame(
  id = c(1,2,3),
  time.1 = c(2,4,2),
  time.2 = c(3,6,2.1),
  time.3 = c(4.5,7,2.8)
)

方法一：基础R实现 #

不需要额外安装包，直接用原生代码就能搞定：

# 提取所有时间列（排除id列）
time_cols <- df[, -1]
# 计算后一列减前一列的差值
diff_cols <- time_cols[, -1] - time_cols[, -ncol(time_cols)]
# 给差值列命名为diff.后序号.前序号
colnames(diff_cols) <- paste0("diff.", seq(2, ncol(time_cols)), ".", seq(1, ncol(time_cols)-1))

# 构造最终的列顺序：id → time.1 → diff.2.1 → time.2 → diff.3.2 → time.3
new_col_order <- c()
for(i in 1:ncol(time_cols)){
  new_col_order <- c(new_col_order, colnames(time_cols)[i])
  if(i < ncol(time_cols)){
    new_col_order <- c(new_col_order, colnames(diff_cols)[i])
  }
}

# 合并并按顺序排列列
new_df <- cbind(df[, "id", drop=FALSE], time_cols, diff_cols)[, c("id", new_col_order)]

运行后new_df就是你要的结果：

> new_df
  id time.1 diff.2.1 time.2 diff.3.2 time.3
1  1    2.0      1.0    3.0      1.5    4.5
2  2    4.0      2.0    6.0      1.0    7.0
3  3    2.0      0.1    2.1      0.7    2.8

方法二：用tidyverse工具实现 #

如果习惯用tidyverse生态（dplyr+tidyr），代码更简洁，适合列数较多的场景：

library(tidyverse)

new_df_tidy <- df %>%
  # 把时间列转成长格式
  pivot_longer(cols = starts_with("time"), names_to = "time_col", values_to = "value") %>%
  # 提取时间列的序号数字
  mutate(time_num = as.numeric(str_extract(time_col, "\\d+"))) %>%
  group_by(id) %>%
  # 计算当前值与前一行值的差
  mutate(diff = value - lag(value)) %>%
  # 给差值列生成目标名称
  mutate(diff_col = ifelse(!is.na(diff), paste0("diff.", time_num, ".", time_num-1), NA)) %>%
  # 转回宽格式，同时保留原时间列和差值列
  pivot_wider(
    id_cols = id,
    names_from = c(time_col, diff_col),
    values_from = c(value, diff),
    names_sep = ""
  ) %>%
  # 清理列名里的无效后缀
  rename_with(~str_remove(., "_NA")) %>%
  # 调整列顺序到目标格式
  select(id, time.1, diff.2.1, time.2, diff.3.2, time.3)

补充说明 #

• 两种方法都能得到相同结果，基础R适合不想额外装包的场景，tidyverse更灵活，后续要加更多时间列时几乎不用改代码。
• 如果你需要把0.1显示成.1，可以用format()函数处理列值，比如diff_cols[] <- lapply(diff_cols, function(x) sub("^0", "", format(x, nsmall=1)))，不过保留0.1的格式通常更规范。

内容的提问来源于stack exchange，提问作者alex