对于数据集中的每个学生，可能已经收集了一组特定的分数。我们想要计算每个学生的平均值，但只使用与该学生密切相关的列中的分数。
计算中所需的列对于每一行都是不同的。我已经想过如何使用常用工具在R中编写这个，但我试图用data.table重写，部分是为了好玩，但也部分是为了预期这个小项目的成功可能导致需要制作许多行的计算。
以下是＆＃34;为每个行问题选择特定列集的一个小工作示例。＆＃34;
set.seed(123234)
## Suppose these are 10 students in various grades
dat <- data.frame(id = 1:10, grade = rep(3:7, by = 2),
A = sample(c(1:5, 9), 10, replace = TRUE),
B = sample(c(1:5, 9), 10, replace = TRUE),
C = sample(c(1:5, 9), 10, replace = TRUE),
D = sample(c(1:5, 9), 10, replace = TRUE))
## 9 is a marker for missing value, there might also be
## NAs in real data, and those are supposed to be regarded
## differently in some exercises
## Students in various grades are administered different
## tests. A data structure gives the grade to test linkage.
## The letters are column names in dat
lookup <- list("3" = c("A", "B"),
"4" = c("A", "C"),
"5" = c("B", "C", "D"),
"6" = c("A", "B", "C", "D"),
"7" = c("C", "D"),
"8" = c("C"))
## wrapper around that lookup because I kept getting confused
getLookup <- function(grade){
lookup[[as.character(grade)]]
}
## Function that receives one row (named vector)
## from data frame and chooses columns and makes calculation
getMean <- function(arow, lookup){
scores <- arow[getLookup(arow["grade"])]
mean(scores[scores != 9], na.rm = TRUE)
}
stuscores <- apply(dat, 1, function(x) getMean(x, lookup))
result <- data.frame(dat, stuscores)
result
## If the data is 1000s of thousands of rows,
## I will wish I could use data.table to do that.
## Client will want students sorted by state, district, classroom,
## etc.
## However, am stumped on how to specify the adjustable
## column-name chooser
library(data.table)
DT <- data.table(dat)
## How to write call to getMean correctly?
## Want to do this for each participant (no grouping)
setkey(DT, id)
所需的输出是相应列的学生平均值，如下所示：
> result
id grade A B C D stuscores
1 1 3 9 9 1 4 NaN
2 2 4 5 4 1 5 3.0
3 3 5 1 3 5 9 4.0
4 4 6 5 2 4 5 4.0
5 5 7 9 1 1 3 2.0
6 6 3 3 3 4 3 3.0
7 7 4 9 2 9 2 NaN
8 8 5 3 9 2 9 2.0
9 9 6 2 3 2 5 3.0
10 10 7 3 2 4 1 2.5
我没有在数据表示例中找到任何示例，其中每行的计算中使用的列本身就是一个变量，我感谢您的建议。
我没有要求任何人为我编写代码，我就如何开始解决这个问题寻求建议。
答案 0 :(得分：6)
首先，在使用sample（每次运行时设置随机种子）等功能创建可重现的示例时，应使用set.seed。
其次，您可以循环遍历lookup列表，而不是循环遍历每一行，它始终小于数据（多次显着缩小）并将其与rowMeans结合使用。你也可以用基数R来做，但你要求data.table解决方案，所以这里（为了这个解决方案的目的，我已经将所有9个转换为NA，但你可以尝试将此概括为您的具体案例）
所以使用set.seed(123)，你的函数给出了
apply(dat, 1, function(x) getMean(x, lookup))
# [1] 2.000000 5.000000 4.666667 4.500000 2.500000 1.000000 4.000000 2.333333 2.500000 1.500000
这里有一个可能的data.table应用程序，只运行lookup列表（for循环列表在R btw中非常有效，请参阅here）
## convert all 9 values to NAs
is.na(dat) <- dat == 9L
## convert your original data to `data.table`,
## there is no need in additional copy of the data if the data is huge
setDT(dat)
## loop only over the list
for(i in names(lookup)) {
dat[grade == i, res := rowMeans(as.matrix(.SD[, lookup[[i]], with = FALSE]), na.rm = TRUE)]
}
dat
# id grade A B C D res
# 1: 1 3 2 NA NA NA 2.000000
# 2: 2 4 5 3 5 NA 5.000000
# 3: 3 5 3 5 4 5 4.666667
# 4: 4 6 NA 4 NA 5 4.500000
# 5: 5 7 NA 1 4 1 2.500000
# 6: 6 3 1 NA 5 3 1.000000
# 7: 7 4 4 2 4 5 4.000000
# 8: 8 5 NA 1 4 2 2.333333
# 9: NA 6 4 2 2 2 2.500000
# 10: 10 7 3 NA 1 2 1.500000
可能使用set可以改善这一点，但我目前无法想到一个好方法。
<强> P.S。
正如@Arun所建议的那样，请看一下他自己撰写的小插曲here，以便熟悉:=运算符，.SD，with = FALSE，等
答案 1 :(得分：1)
以下是使用data.table（需要melt.data.table 1.9.5+）的另一种data.table方法，然后在data.table之间加入：
DT_m <- setkey(melt.data.table(DT, c("id", "grade"), value.name = "score"), grade, variable)
lookup_dt <- data.table(grade = rep(as.integer(names(lookup)), lengths(lookup)),
variable = unlist(lookup), key = "grade,variable")
score_summary <- setkey(DT_m[lookup_dt, nomatch = 0L,
.(res = mean(score[score != 9], na.rm = TRUE)), by = id], id)
setkey(DT, id)[score_summary, res := res]
# id grade A B C D mean_score
# 1: 1 3 9 9 1 4 NaN
# 2: 2 4 5 4 1 5 3.0
# 3: 3 5 1 3 5 9 4.0
# 4: 4 6 5 2 4 5 4.0
# 5: 5 7 9 1 1 3 2.0
# 6: 6 3 3 3 4 3 3.0
# 7: 7 4 9 2 9 2 NaN
# 8: 8 5 3 9 2 9 2.0
# 9: 9 6 2 3 2 5 3.0
#10: 10 7 3 2 4 1 2.5
它更冗长，但速度只有两倍：
microbenchmark(da_method(), nk_method(), times = 1000)
#Unit: milliseconds
# expr min lq mean median uq max neval
# da_method() 17.465893 17.845689 19.249615 18.079206 18.337346 181.76369 1000
# nk_method() 7.047405 7.282276 7.757005 7.489351 7.667614 20.30658 1000