获取存储在单个列中的连续日期之间的时间 [英] Get the time between consecutive dates stored in a single column

问题描述

当事件被存储为数据框中的日期列时，我试图弄清楚如何获取连续事件之间的时间.

I am trying to figure out how to get the time between consecutive events when events are stored as a column of dates in a dataframe.

sampledf=structure(list(cust = c(1L, 1L, 1L, 1L), date = structure(c(9862, 
9879, 10075, 10207), class = "Date")), .Names = c("cust", "date"
), row.names = c(NA, -4L), class = "data.frame")

我可以得到答案

as.numeric(rev(rev(difftime(c(sampledf$date[-1],0),sampledf$date))[-1]))
# [1]  17 196 132

但是真的很丑.除此之外，我只知道如何排除向量中的第一项，而不是最后一项，所以我必须 rev() 两次才能删除最后一个值.

but it is really ugly. Among other things, I only know how to exclude the first item in a vector, but not the last so I have to rev() twice to drop the last value.

有没有更好的方法?

顺便说一下，我将使用 ddply 对每个客户 ID 的更大数据集执行此操作，因此该解决方案需要使用 ddply.

By the way, I will use ddply to do this to a larger set of data for each cust id, so the solution would need to work with ddply.

library(plyr)
ddply(sampledf, 
              c("cust"), 
              summarize,
              daysBetween = as.numeric(rev(rev(difftime(c(date[-1],0),date))[-1]))
)

谢谢！

推荐答案

你在找这个吗?

as.numeric(diff(sampledf$date))
# [1]  17 196 132

要删除最后一个元素，请使用 head:

To remove the last element, use head:

head(as.numeric(diff(sampledf$date)), -1)
# [1]  17 196

require(plyr)
ddply(sampledf, .(cust), summarise, daysBetween = as.numeric(diff(date)))

#   cust daysBetween
# 1    1          17
# 2    1         196
# 3    1         132

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