R减去相同ID的值（从显示的第一个ID中减去）

本文介绍了R减去相同ID的值（从显示的第一个ID中减去）的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个与这里发现的问题类似的问题： R，从上一行中减去值，然后按分组（稍作修改；请参见下文）：

I have a similar question to the question found here: R, subtract value from previous row, group by (slight modification; see below):

在R中，可以说我有此数据。 / p>

In R, lets say I have this data.

Data id date value 2380 10/30/12 21.01 2380 10/31/12 22.04 2380 11/1/12 22.65 2380 11/2/12 23.11 20100 10/30/12 35.21 20100 10/31/12 37.07 20100 11/1/12 38.17 20100 11/2/12 38.97 20103 10/30/12 57.98 20103 10/31/12 60.83

我想从相同ID的ID值中减去该值。我希望这是有道理的。见下文:)

And I want to subtract the value from the value of the ID of the same ID. I hope this makes sense. See below :)

id date value diff 2380 10/30/12 21.01 0 2380 10/31/12 22.04 1.03 2380 11/1/12 22.65 1.64 2380 11/2/12 23.11 2.10 20100 10/30/12 35.21 0 20100 10/31/12 37.07 1.86 20100 11/1/12 38.17 2.96 20100 11/2/12 38.97 3.76 20103 10/30/12 57.98 0 20103 10/31/12 60.83 2.85

感谢您的帮助！

推荐答案

假设日期已经排序。我可能会检索每个id的第一个值，然后使用它来计算diff功能。 这样的事情。

Let's assume that dates are already sorted. I would probably retrieve the first value for each id and then use this to compute the diff feature. Something like this.

my.df <- data.frame( id = c(2380, 2380, 2380, 2380, 20100,20100,20100, 20100, 20103, 20103), date = c("10/30/12", "10/31/12", "11/1/12", "11/2/12", "10/30/12", "10/31/12", "11/1/12", "11/2/12", "10/30/12", "10/31/12"), value = c(21.01, 22.04, 22.65, 23.11, 35.21, 37.07, 38.17, 38.97, 57.98, 60.83), stringsAsFactors = F) # # get ids my.ids <- unique(my.df$id) # or levels(my.df$id) # get first val (assuming sorting by date) id.val0 <- sapply(my.ids, (function(id){ my.df$value[my.df$id == id][1] })) names(id.val0) <- my.ids # do operation my.df$diff <- sapply(1:nrow(my.df), (function(i){ tmp.id <- my.df$id[i] my.df$value[i] - id.val0[as.character(tmp.id)] }))