Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
660 views
in Technique[技术] by (71.8m points)

dataframe - R subtract value for the same ID (from the first ID that shows)

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

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 

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

Thank you for your help!

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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)]
}))

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

57.0k users

...