Question

我列出了每周每笔儿童的个人最佳时间。对于孩子A，我如何计算前几个星期孩子的平均收入，不包括本周的结果呢？例如，一个孩子的结果可能如下所示：

df.trial <- data.frame(Week= c("w10", "w9", "w9", "w5", "w5", "w5", "w6", "w6", "w3"), Stroke= c("Fly","Free","Breast","Back","Free","Breast","Fly","Back","Free"),  Score = c(5.5,4.5,4.6,5.2, 4.3, 5.7, 4.7,5.5,4.8))

我正在尝试添加一个新列来计算如下平均值：

df.desired <- data.frame(Week= c("w10", "w9", "w9", "w5", "w5", "w5", "w6", "w6", "w3"), Stroke= c("Fly","Free","Breast","Back","Free","Breast","Fly","Back","Free"), Score = c(5.5, 4.5, 4.6, 5.2, 4.3, 5.7, 4.7, 5.5, 4.8), prev.mean = c(4.91, 5.03, 5.03, 4.80, 4.80, 4.80, 5.00, 5.00, NA))

我的数据有很多孩子，所以我需要按每个学生的姓名分组。另外，如果没有以前的几周，那么我想添加一个NA。

我尝试了几种不同的方式，并尝试编写类似rollsum的函数，如下所述：R previous week total

到目前为止没有运气。有什么建议吗？

Answer 1

编辑：新答案

我误解了您的问题，我认为这就是您想要的。请注意，我在您的data.frame中添加了一个kid_id列，以向您显示适用于多个ID的解决方案。

还请注意，我必须将Weeks转换为数字，以便可以安排Weeks，否则安排将不正确。

library(tidyverse)

df.trial <- data.frame(Kid_id = 1, Week= c("w10", "w9", "w9", "w5", "w5", "w5", "w6", "w6", "w3"), Stroke= c("Fly","Free","Breast","Back","Free","Breast","Fly","Back","Free"),  Score = c(5.5,4.5,4.6,5.2, 4.3, 5.7, 4.7,5.5,4.8))

df.trial %>% as_tibble() %>% 
  mutate(Week = as.numeric(gsub("w", "", as.character(Week))), 
         kid_id = 1) %>% 
  group_by(kid_id, Week) %>% 
  arrange(kid_id, Week) %>% 
  summarise(sum_score = sum(Score),
            n_score = n()) %>% 
  mutate(prev.mean = map_dbl(seq_along(sum_score), ~sum(sum_score[1:.x - 1]) / sum(n_score[1:.x - 1])),
         Week = paste0("w", Week)) %>% 
  ungroup() %>% 
  select(kid_id, Week, prev.mean) %>% 
  left_join(df.trial, ., by = "Week")


  Week Stroke Score kid_id prev.mean
1  w10    Fly   5.5      1  4.912500
2   w9   Free   4.5      1  5.033333
3   w9 Breast   4.6      1  5.033333
4   w5   Back   5.2      1  4.800000
5   w5   Free   4.3      1  4.800000
6   w5 Breast   5.7      1  4.800000
7   w6    Fly   4.7      1  5.000000
8   w6   Back   5.5      1  5.000000
9   w3   Free   4.8      1       NaN

在这里您拥有NaN而不是NA，但可以根据需要轻松地替换它们。

先前值的得分的平均值

1 个答案: