求 numeric 列的 3 个最低价格的平均值

Question

我如何找到数字列 (Country_1) 的 3 个最低价格的平均值？假设我有数千个值？

d<-structure(list(Subarea = c("SA_1", "SA_2", "SA_3", "SA_4", "SA_5", 
"SA_6", "SA_7", "SA_8", "SA_10", "SA_9"), Country_1 = c(101.37519256645, 
105.268942332558, 100.49933368058, 104.531597221684, NA, 83.4404308144341, 
86.2833044714836, 81.808967345926, 79.6786979951661, 77.6863475527052
)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-10L))

Answer 1

按值升序对向量进行排序，取前 3 个值，然后计算平均值。

mean(head(sort(d$Country_1), 3))
# [1] 79.72467

如果要对多列执行此操作，请使用 sapply 或 dplyr::across：

sapply(df[, your_columns], \(x) mean(head(sort(x), 3)))

# or

library(dplyr)
d %>%
   mutate(across(your_columns, ~ mean(head(sort(.x), 3)))

Answer 2

如果你只关心最小的3个值，而且数据量很大，使用sort()和partial = 1:3效率更高。

mean(sort(sample(d$Country_1), partial = 1:3)[1:3])

Answer 3

选项slice_min

library(dplyr)
d %>% 
 slice_min(n = 3, order_by = Country_1) %>% 
 summarise(Mean = mean(Country_1))
# A tibble: 1 × 1
   Mean
  <dbl>
1  79.7

求 numeric 列的 3 个最低价格的平均值

Find the average of 3 minimum prices of numeric column

r

conditional-formatting

mean

dataframe