R:不同组之间的“差异"函数 [英] R: Function “diff” over various groups

问题描述

在寻找问题的解决方案时，我发现了以下线程: Function" ; diff在R 中的各个组上.我有一个非常类似的问题，所以我将在那里处理示例.

While searching for a solution to my problem I found this thread: Function "diff" over various groups in R. I've got a very similar question so I'll just work with the example there.

这是我期望的输出应为:

This is what my desired output should look like:

name class year diff
1    a    c1   2009  NA      
2    a    c1   2010   67
3    b    c1   2009  NA
4    b    c1   2010   20

我有两个构成子组的变量-类和名称.因此，我只想比较具有相同名称和类的值.我还希望获得2009年到2010年之间的差异.如果没有2008年，则diff 2009应该返回NA(因为它无法计算出差异).

I have two variables which form subgroups - class and name. So I want to compare only the values which have the same name and class. I also want to have the differences from 2009 to 2010. If there is no 2008, diff 2009 should return NA (since it can't calculate a difference).

我确定它的工作原理与其他线程非常相似，但我无法使其正常工作.我也使用了这段代码(并通过对数据进行不同的排序来简​​单地解决了升序的一年)，但是以某种方式R仍然设法计算出差异并且不返回NA.

I'm sure it works very similarly to the other thread but I just can't make it work. I used this code too (and simply solved the ascending year by sorting the data differently), but somehow R still manages to calculate a difference and does not return NA.

ddply(df, .(class, name), summarize, year=head(year, -1), value=diff(value))

推荐答案

使用dplyr

  df %>% 
  filter(year!=2008)%>% 
  arrange(name, class, year)%>%
  group_by(class, name)%>%
  mutate(diff=c(NA,diff(value)))
  # Source: local data frame [12 x 5]
  #  Groups: class, name

  #     name class year value diff
  #  1     a    c1 2009    33   NA
  #  2     a    c1 2010   100   67
  #  3     a    c2 2009    80   NA
  #  4     a    c2 2010    90   10
  #  5     a    c3 2009    90   NA
  #  6     a    c3 2010   100   10
  #  7     b    c1 2009    80   NA
  #  8     b    c1 2010    90   10
  #  9     b    c2 2009    90   NA
  #  10    b    c2 2010   100   10
  #  11    b    c3 2009    80   NA
  #  12    b    c3 2010    99   19

更新:

有相对差异

Update:

With relative difference

 df %>%
 filter(year!=2008)%>% 
 arrange(name, class, year)%>%
 group_by(class, name)%>%
 mutate(diff1=c(NA,diff(value)), rel_diff=round(diff1/value[row_number()-1],2))

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