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行是不同的.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,这可能导致需要进行计算很多很多行.
这是一个"为每行问题选择特定列集"的小工作示例.
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
那又怎样？到目前为止我写了很多错误......
我没有在数据表示例中找到任何示例,其中用于计算每行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,我正在征求关于如何开始解决这个问题的建议.
首先,当使用诸如sample(每次运行时设置随机种子)等功能创建可重现的示例时,您应该使用set.seed.
其次,不是循环遍历每一行,而是可以循环遍历lookup列表,该列表总是小于数据(多次显着缩小)并将其组合rowMeans.你也可以使用base R来做,但是你要求一个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转换为NAs,但你也可以尝试将其概括为你的特定情况)
所以使用set.seed(123),你的功能给出
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应用程序只在lookup列表for上运行(列表中的循环在R btw中非常有效,请参见此处)
## 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,这可以改善利用set,但我想不出目前的好方法.
PS
正如@Arun建议,请看看他自己写的短文这里以熟悉的:=操作,.SD,with = FALSE,等.