n 个不同的最高/最低值的最小值、最大值、平均值，并将它们与时间序列数据一起绘制在 R 的同一图表上

Question

I am dealing with a large time series data set (with almost 100K records) with unix timestamp. I need min, mean, max, avg_of_lowest_n, avg_of_top_n from the value column. I can get min, mean, max as follows:

tapply(df$value, df$pattern, min)
tapply(df$value, df$pattern, mean)
tapply(df$value, df$pattern, max)

Now, I need to get the mean of the lowest n distinct values and top n distinct values in two other columns for each pattern (group). I can get the mean of the lowest and top n (say 5 values) from the following, but I think the mean is not calculated by distinct 5 values for each group (pattern), and here I need to know, how can I do that.

setDT(df_stat) #requires Data.table
df_n[order(value)][, list(mean_of_low_5=mean(value[1:5])), by=pattern]
df_n[order(-value)][, list(mean_of_top_5=mean(value[1:5])), by=pattern]

Any simple way of doing this is highly appreciated.

Sample data-

df <- structure(list(pattern = c(462L, 462L, 462L, 462L, 462L, 462L, 
462L, 462L, 462L, 462L, 462L, 463L, 463L, 463L, 463L, 463L, 463L, 
463L, 463L, 463L, 463L, 463L, 463L, 463L, 463L, 464L, 464L, 464L, 
464L, 464L, 464L, 464L, 464L, 464L, 464L, 464L, 464L, 464L, 465L, 
465L, 465L, 465L, 465L, 466L, 466L, 466L, 466L, 466L, 466L, 466L, 
466L, 466L, 466L, 466L, 466L, 961L, 961L, 961L, 961L, 961L, 961L, 
961L), value = c(5.8e+10, 4.35e+10, 3.96e+10, 3.6e+10, 3.48e+10, 
3.3e+10, 3.3e+10, 3.3e+10, 3.3e+10, 3.3e+10, 3.3e+10, 1e+09, 
1e+09, 1e+09, 1e+09, 1e+09, 1e+09, 1e+09, 1e+09, 1e+09, 1e+09, 
1e+09, 1e+09, 1e+09, 1e+09, 3.3e+10, 3.3e+10, 3.3e+10, 3.3e+10, 
3.3e+10, 3.3e+10, 3.3e+10, 3.3e+10, 3.3e+10, 3.3e+10, 3.3e+10, 
3.3e+10, 3.3e+10, 3e+10, 3e+10, 3e+10, 3e+10, 3e+10, 3.3e+10, 
3.3e+10, 3.3e+10, 3.3e+10, 3.3e+10, 3.3e+10, 3.3e+10, 3.3e+10, 
3.3e+10, 3.2e+10, 3.2e+10, 3.2e+10, 2.6e+10, 2.6e+10, 2.6e+10, 
2.6e+10, 2.6e+10, 2.6e+10, 2.6e+10), timestamp = c(1590604157L, 
1590604157L, 1590604157L, 1590604157L, 1590604157L, 1590604157L, 
1590604157L, 1590604157L, 1590604157L, 1590604157L, 1590604157L, 
1590604170L, 1590604170L, 1590604170L, 1590604170L, 1590604170L, 
1590604170L, 1590604170L, 1590604170L, 1590604170L, 1590604170L, 
1590604170L, 1590604170L, 1590604170L, 1590604170L, 1590604213L, 
1590604213L, 1590604213L, 1590604213L, 1590604213L, 1590604213L, 
1590604213L, 1590604213L, 1590604213L, 1590604213L, 1590604213L, 
1590604213L, 1590604213L, 1590604226L, 1590604226L, 1590604226L, 
1590604226L, 1590604226L, 1590604239L, 1590604239L, 1590604239L, 
1590604239L, 1590604239L, 1590604239L, 1590604239L, 1590604239L, 
1590604239L, 1590604239L, 1590604239L, 1590604239L, 1590610895L, 
1590610895L, 1590610895L, 1590610895L, 1590610895L, 1590610895L, 
1590610895L)), class = "data.frame", row.names = c(NA, -62L))

Answer 1

You can do all the calculation in one pipe using dplyr :

library(dplyr)

df %>%
  group_by(pattern) %>%
  summarise(min_val = min(value), 
            max_val = max(value), 
            mean_val = mean(value), 
            lowest_n_val = mean(head(unique(sort(value)), 5)),
            highest_n_val = mean(tail(unique(sort(value)), 5)))

You can add na.rm. =TRUE na.rm. =TRUE in all the above functions if you have NA in your data.