5

我有一个包含 3 组和 3 天的数据框：

set.seed(10)
dat <- data.frame(group=rep(c("g1","g2","g3"),each=3), day=rep(c(0,2,4),3), value=runif(9))
#   group day    value
# 1    g1   0 0.507478
# 2    g1   2 0.306769
# 3    g1   4 0.426908
# 4    g2   0 0.693102
# 5    g2   2 0.085136
# 6    g2   4 0.225437
# 7    g3   0 0.274531
# 8    g3   2 0.272305
# 9    g3   4 0.615829

我想取 log2 并将每个值除以每组中的第 0 天值。我现在这样做的方式是在中间步骤中计算每一天的组：

day_0 <- dat[dat$day==0, "value"]
day_2 <- dat[dat$day==2, "value"]
day_4 <- dat[dat$day==4, "value"]
res <- cbind(0, log2(day_2/day_0), log2(day_4/day_0))
rownames(res) <- c("g1","g2","g3")
colnames(res) <- c("day_0","log_ratio_day_2_day_0","log_ratio_day_4_day_0")
#    day_0 log_ratio_day_2_day_0 log_ratio_day_4_day_0
# g1     0            -0.7261955             -0.249422
# g2     0            -3.0252272             -1.620346
# g3     0            -0.0117427              1.165564

res没有中间步骤的正确计算方法是什么？

4

4 回答 4

5

data.table编码优雅和内存效率的解决方案

library(data.table)

DT <- data.table(dat)

# assign within DT by reference

DT[, new_value := log2(value / value[day == 0]), by = group]

或者你可以使用joinsand keysandby-without-by

DTb <- data.table(dat)

setkey(DTb, group)

# val0 contains just those records for day 0
val0 <- DTb[day==0]

 # the i.value refers to value from the i argument 
 # which is in this case `val0` and thus the value for 
 # day = 0 
 DTb[val0, value := log2(value / i.value)]

这两种解决方案都不需要您进行排序 day以确保value将是第一个（或任何特定）元素。

编辑

i.语法文档

    **********************************************
    **                                          **
    **   CHANGES IN DATA.TABLE VERSION 1.7.10   **
    **                                          **
    **********************************************
     NEW FEATURES

o   New function setcolorder() reorders the columns by name
    or by number, by reference with no copy. This is (almost)
    infinitely faster than DT[,neworder,with=FALSE].
    
o   The prefix i. can now be used in j to refer to join inherited
    columns of i that are otherwise masked by columns in x with
    the same name.
于 2012-11-04T22:34:26.290 回答
4

您的朋友ddply来自plyr包裹：

require(plyr)
> ddply(dat, .(group), mutate, new_value = log2(value / value[1]))
  group day      value   new_value
1    g1   0 0.50747820  0.00000000
2    g1   2 0.30676851 -0.72619548
3    g1   4 0.42690767 -0.24942179
4    g2   0 0.69310208  0.00000000
5    g2   2 0.08513597 -3.02522716
6    g2   4 0.22543662 -1.62034599
7    g3   0 0.27453052  0.00000000
8    g3   2 0.27230507 -0.01174274
9    g3   4 0.61582931  1.16556397
于 2012-11-04T16:32:29.483 回答
3

基础解决方案：

> res <- do.call(rbind,by(dat,dat$group,function(x) log2(x$value/x$value[x$day==0])))
> res

   [,1]       [,2]       [,3]
g1    0 -1.6496538 -2.3673937
g2    0  0.3549090  0.4537402
g3    0 -0.9423506  1.4603706

> colnames(res) <- c("day_0","log_ratio_day_2_day_0","log_ratio_day_4_day_0")
> res

   day_0 log_ratio_day_2_day_0 log_ratio_day_4_day_0
g1     0            -1.6496538            -2.3673937
g2     0             0.3549090             0.4537402
g3     0            -0.9423506             1.4603706
于 2012-11-04T20:57:26.267 回答
1

这ave在 R 的核心中使用：

transform(dat, value0 = ave(value, group, FUN = function(x) log2(x / x[1])))
于 2012-11-05T12:42:10.843 回答