Question

我有一个要分组的数据框，并以一致的单位获得diff（滞后差）的中位数。与下面的示例非常相似。如您所见，通过运行下面的代码，我遇到了问题，因为diff的summarise函数没有考虑到units属性

library(tidyverse)

# Initialise random data
t = Sys.time()
rnd <- sample(1:10000,10,replace=F)
add <- rnd[order(rnd)]

# Create 2 dtaaframes
time1 <- data.frame(datetime = t + add)
time2 <- data.frame(datetime = t + add * 1000)

# Bind dataframe together
mydata <- bind_rows(time1, time2, .id = "group")

# Trying to get a summary table
mydata %>% group_by(group) %>% summarise(elapsed = median(diff(datetime[order(datetime)])))

# These are the values that I should get in my summary table
median(diff(time1$datetime))
median(diff(time2$datetime))

Answer 1

这里是一个选项，它将在几秒钟内显示所有结果。如果更合适，请使用dminutes(1)或dhours(1)或ddays(1)。

mydata %>% 
  group_by(group) %>% 
  summarise(elapsed = median(diff(datetime[order(datetime)])) / lubridate::dseconds(1))

Answer 2

如何使用difftime并设置单位？

mydata %>% 
  group_by(group) %>%
  summarise(elapsed = median(difftime(datetime, lag(datetime), units = "mins"), na.rm = TRUE))

汇总数据框以获得diff（滞后差）

2 个答案: