多列的 R 时间聚合

Question

我添加了一个数据框，其中包含一次和 6 个数据列，如下所示...

df <- data.frame(structure(list(Time = c(100, 100.1, 100.2, 100.2, 100.3, 100.3,100.3, 100.4, 100.4, 100.5, 100.5, 100.6, 100.6, 100.7),
               x = c(4,NA, 7, NA, 3, 7, NA, 9, NA, 7, NA, 3, NA, 7),
               y = c(NA, 7, NA,9, NA, 9, 7, NA, NA, NA, 9, NA, 5, NA), 
               a = c(7, NA, 3, 3, NA,NA, 7, NA, NA, 7, 7, NA, NA, 9),
               b = c(8, NA, 4, NA, 5, 4, NA,9, NA, 1, NA, 7, NA, 2),
               j = c(NA, 4, NA, 6, NA, 6, 4, NA, NA, NA, 6, NA, 2, NA), 
               k = c(1, NA, 5, 5, NA, NA, 1, NA, NA, 2, 2,NA, NA, 6)), 
          .Names = c("Time", "x", "a", "j", "y", "b", "k"), 
          class = c("tbl_df","tbl", "data.frame"), row.names = c(NA, -14L)))


Time    x   y   a   b   j   k
100     4   NA  7   8   NA  1
100.1   NA  7   NA  NA  4   NA
100.2   7   NA  3   4   NA  5
100.2   NA  9   3   NA  6   5
100.3   3   NA  NA  5   NA  NA
100.3   7   9   NA  4   6   NA
100.3   NA  7   7   NA  4   1
100.4   9   NA  NA  9   NA  NA
100.4   NA  NA  NA  NA  NA  NA
100.5   7   NA  7   1   NA  2
100.5   NA  9   7   NA  6   2
100.6   3   NA  NA  7   NA  NA
100.6   NA  5   NA  NA  2   NA
100.7   7   NA  9   2   NA  6

我想使用时间列进行聚合。必须在 x 和 y、a 和 b、j 和 k 之间计算时间平均值。输出应该是这样的..

Time    xy_mean ab_mean jk_mean
100         
100.1           
100.2           
100.3           
100.4           
100.5           
100.6           
100.7           

请帮忙...

（如果问题不清楚也请评论）

Answer 1

编辑

根据@Marijn Stevering 的评论，这种方法会更有效：

 df_final <- df %>% 
    group_by(Time) %>% 
   summarize(av_xy = mean(c(x,y), na.rm = TRUE), 
   av_ab = mean(c(a,b), na.rm = TRUE), 
   av_jk = mean(c(j,k), na.rm = TRUE))


df_final
## A tibble: 8 x 4
#   Time av_xy av_ab av_jk
#  <dbl> <dbl> <dbl> <dbl>
#1 100.0  6.00   NaN   4.0
#2 100.1   NaN   5.5   NaN
#3 100.2  5.50   7.5   4.0
#4 100.3  4.75   6.5   4.0
#5 100.4  9.00   NaN   NaN
#6 100.5  4.00   7.5   4.5
#7 100.6  5.00   3.5   NaN
#8 100.7  4.50   NaN   7.5

---

原始答案

我知道必须有更直接的东西，但这里有一个dplyr 方法，其中包含一些步骤：

library(dplyr)

df_xy <- df %>%
      group_by(Time) %>%
      summarise(av_xy = mean(c(x,y), na.rm = TRUE))

df_ab <- df %>%
      group_by(Time) %>%
      summarise(av_ab = mean(c(a,b), na.rm = TRUE))

df_jk <- df %>%
      group_by(Time) %>%
      summarise(av_jk = mean(c(j,k), na.rm = TRUE))
      
df_final <- df_xy %>%
  left_join(df_ab) %>%
  left_join(df_jk)

 df_final
## A tibble: 8 x 4
#   Time av_xy av_ab av_jk
#  <dbl> <dbl> <dbl> <dbl>
#1 100.0  6.00   NaN   4.0
#2 100.1   NaN   5.5   NaN
#3 100.2  5.50   7.5   4.0
#4 100.3  4.75   6.5   4.0
#5 100.4  9.00   NaN   NaN
#6 100.5  4.00   7.5   4.5
#7 100.6  5.00   3.5   NaN
#8 100.7  4.50   NaN   7.5

Answer 2

以下代码可以满足您的需求。它不漂亮，它使用split 将data.frame 拆分为Time 的子df，然后连续使用*apply 来计算结果。
如果要删除NA 值，请在代码开头设置NA.RM <- TRUE。

fun <- function(x, y, na.rm = FALSE){
    mean(c(x, y), na.rm = na.rm)
}

NA.RM <- FALSE

inx <- seq_along(names(df2))[2:4]
res <- lapply(split(df, df2$Time), function(DF)
            sapply(inx, function(i) fun(DF[[i]], DF[[i + 3]], NA.RM)))
res <- do.call(rbind, res)
res <- cbind.data.frame(row.names(res), as.data.frame(res))
row.names(res) <- NULL
names(res)[1] <- names(df2)[1]
names(res)[2:4] <- sapply(inx, function(i) paste0(names(df2)[i], names(df2)[i + 1]))
names(res)[2:4] <- paste(names(res)[2:4], "mean", sep = "_")
res
#   Time xy_mean ya_mean ab_mean
#1   100     6.0      NA     4.0
#2 100.1      NA     5.5      NA
#3 100.2      NA      NA     4.0
#4 100.3      NA      NA      NA
#5 100.4      NA      NA      NA
#6 100.5      NA      NA     4.5
#7 100.6      NA      NA      NA
#8 100.7     4.5      NA     7.5