我还是data.table的新手。我的问题类似于this one和this one。区别在于我想按组计算多个变量的加权平均值，但每个均值使用多个权重。
考虑以下data.table（实际情况要大得多）：
library(data.table)
set.seed(123456)
mydata <- data.table(CLID = rep("CNK", 10),
ITNUM = rep(c("First", "Second", "First", "First", "Second"), 2),
SATS = rep(c("Always", "Amost always", "Sometimes", "Never", "Always"), 2),
ASSETS = rep(c("0-10", "11-25", "26-100", "101-200", "MORE THAN 200"), 2),
AVGVALUE1 = rnorm(10, 10, 2),
AVGVALUE2 = rnorm(10, 10, 2),
WGT1 = rnorm(10, 3, 1),
WGT2 = rnorm(10, 3, 1),
WGT3 = rnorm(10, 3, 1))
#I set the key of the table to the variables I want to group by,
#so the output is sorted
setkeyv(mydata, c("CLID", "ITNUM", "SATS", "ASSETS"))
我想要实现的是按AVGVALUE1，AVGVALUE2，{{定义的组计算ITNUM和SATS（以及可能更多变量）的加权平均值1}}使用每个权重变量ASSETS，WGT1，WGT2（可能还有更多）。因此，对于我想要计算加权平均值的每个变量，我将有三个加权平均值（或任何权重数）。
我可以分别为每个变量执行此操作，例如：
WGT3我在all.weights <- c("WGT1", "WGT2", "WGT3")
avg.var <- "AVGVALUE1"
split.vars <- c("ITNUM", "SATS", "ASSETS")
mydata[ , Map(f = weighted.mean, x = .(get(avg.var)), w = mget(all.weights),
na.rm = TRUE), by = c(key(mydata)[1], split.vars)]
中添加了第一个键变量，虽然它是一个常量，因为我希望将它作为输出中的列。我得到了：
by然而，对于实际的 CLID ITNUM SATS ASSETS V1 V2 V3
1: CNK First Always 0-10 11.66824 11.66819 11.66829
2: CNK First Never 101-200 11.37378 12.21008 11.60182
3: CNK First Sometimes 26-100 12.43004 13.13450 12.01330
4: CNK Second Always MORE THAN 200 12.32265 11.81613 12.56786
5: CNK Second Amost always 11-25 10.76556 11.34669 10.52458
，我有更多的列来计算加权平均值（以及更多要使用的权重），一个接一个地执行它将是相当麻烦的。我想象的是一个函数，其中每个变量（data.table，AVGVALUE1等）的均值用每个权重变量（AVGVALUE2，WGT1计算，WGT2等等，并将计算加权平均值的每个变量的输出添加到列表中。我猜这个列表是最好的选择，因为如果所有的估计都在同一个输出中，那么列的数量可能是无穷无尽的。所以像这样：
WGT3使用[[1]]
CLID ITNUM SATS ASSETS V1 V2 V3
1: CNK First Always 0-10 11.66824 11.66819 11.66829
2: CNK First Never 101-200 11.37378 12.21008 11.60182
3: CNK First Sometimes 26-100 12.43004 13.13450 12.01330
4: CNK Second Always MORE THAN 200 12.32265 11.81613 12.56786
5: CNK Second Amost always 11-25 10.76556 11.34669 10.52458
[[2]]
CLID ITNUM SATS ASSETS V1 V2 V3
1: CNK First Always 0-10 9.132899 9.060045 9.197005
2: CNK First Never 101-200 12.896584 13.278680 13.000772
3: CNK First Sometimes 26-100 10.972260 11.215390 10.828431
4: CNK Second Always MORE THAN 200 11.704404 11.611072 11.749586
5: CNK Second Amost always 11-25 8.086409 8.225030 8.028928
lapply使用all.weights <- c("WGT1", "WGT2", "WGT3")
avg.vars <- c("AVGVALUE1", "AVGVALUE2")
split.vars <- c("ITNUM", "SATS", "ASSETS")
lapply(mydata, function(i) {
mydata[ , Map(f = weighted.mean, x = mget(avg.vars)[i], w = mget(all.weights),
na.rm = TRUE), by = c(key(mydata)[1], split.vars)]
})
Error in weighted.mean.default(x = dots[[1L]][[1L]], w = dots[[2L]][[1L]], :
'x' and 'w' must have the same length
mapply我尝试将myfun <- function(data, spl.v, avg.v, wgts) {
data[ , Map(f = weighted.mean, x = mget(avg.v), w = mget(all.weights),
na.rm = TRUE), by = c(key(data)[1], spl.v)]
}
mapply(FUN = myfun, data = mydata, spl.v = split.vars, avg.v = avg.vars,
wgts = all.weights)
Error: value for ‘AVGVALUE2’ not found
包装为列表 - mget(avg.v)，但后来又出现了另一个错误：
.(mget(avg.v))有人可以帮忙吗？
答案 0 :(得分：2)
我们可以使用outer（对两个输入向量中的值的所有组合执行函数）在向量化加权平均函数上运算。通过在数据表范围内定义outer使用的函数，我们可以get计算data.table列：
wmeans = mydata[, {
f = function(X,Y) weighted.mean(get(X), get(Y));
vf = Vectorize(f);
outer(avg.var, all.weights, vf)},
by = split.vars]
这将所有手段放入一个列（即“长”格式）。我们还可以添加几列来指定每个引用的值/权重组合：
wmeans[, mean.v := expand.grid(avg.var, all.weights)[,1]]
wmeans[, mean.w := expand.grid(avg.var, all.weights)[,2]]
head(wmeans)
# ITNUM SATS ASSETS V1 mean.v mean.w
# 1: First Always 0-10 11.668243 AVGVALUE1 WGT1
# 2: First Always 0-10 9.132899 AVGVALUE2 WGT1
# 3: First Always 0-10 11.668192 AVGVALUE1 WGT2
# 4: First Always 0-10 9.060045 AVGVALUE2 WGT2
# 5: First Always 0-10 11.668287 AVGVALUE1 WGT3
# 6: First Always 0-10 9.197005 AVGVALUE2 WGT3
我们可以使用dcast将其重新整形为一个在avg.var中很长但在all.weights中很宽的data.table：
wide.wmeans = dcast(wmeans, mean.v+ITNUM+SATS+ASSETS ~ mean.w, value.var = "V1")
# mean.v ITNUM SATS ASSETS WGT1 WGT2 WGT3
# 1: AVGVALUE1 First Always 0-10 11.668243 11.668192 11.668287
# 2: AVGVALUE1 First Never 101-200 11.373780 12.210083 11.601819
# 3: AVGVALUE1 First Sometimes 26-100 12.430039 13.134499 12.013299
# 4: AVGVALUE1 Second Always MORE THAN 200 12.322651 11.816135 12.567860
# 5: AVGVALUE1 Second Amost always 11-25 10.765557 11.346688 10.524583
# 6: AVGVALUE2 First Always 0-10 9.132899 9.060045 9.197005
# 7: AVGVALUE2 First Never 101-200 12.896584 13.278680 13.000772
# 8: AVGVALUE2 First Sometimes 26-100 10.972260 11.215390 10.828431
# 9: AVGVALUE2 Second Always MORE THAN 200 11.704404 11.611072 11.749586
#10: AVGVALUE2 Second Amost always 11-25 8.086409 8.225030 8.028928
如果您需要将其作为列表而不是data.table，则可以使用
将其拆分lapply(avg.var, function(x) wide.wmeans[mean.v == x])
# [[1]]
# mean.v ITNUM SATS ASSETS WGT1 WGT2 WGT3
# 1: AVGVALUE1 First Always 0-10 11.66824 11.66819 11.66829
# 2: AVGVALUE1 First Never 101-200 11.37378 12.21008 11.60182
# 3: AVGVALUE1 First Sometimes 26-100 12.43004 13.13450 12.01330
# 4: AVGVALUE1 Second Always MORE THAN 200 12.32265 11.81613 12.56786
# 5: AVGVALUE1 Second Amost always 11-25 10.76556 11.34669 10.52458
#
# [[2]]
# mean.v ITNUM SATS ASSETS WGT1 WGT2 WGT3
# 1: AVGVALUE2 First Always 0-10 9.132899 9.060045 9.197005
# 2: AVGVALUE2 First Never 101-200 12.896584 13.278680 13.000772
# 3: AVGVALUE2 First Sometimes 26-100 10.972260 11.215390 10.828431
# 4: AVGVALUE2 Second Always MORE THAN 200 11.704404 11.611072 11.749586
# 5: AVGVALUE2 Second Amost always 11-25 8.086409 8.225030 8.028928
答案 1 :(得分：0)
<强>予。 lapply解决方案
all.weights <- c("WGT1", "WGT2", "WGT3")
avg.vars <- c("AVGVALUE1", "AVGVALUE2")
split.vars <- c("ITNUM", "SATS", "ASSETS")
myfun <- function(avg.vars){
tmp <-
mydata[ , Map(f = weighted.mean,
x = .(get(avg.vars)),
w = mget(all.weights),
na.rm = TRUE),
by = c(key(mydata)[1], split.vars)]
return(tmp) # totally optional, a habit from using C and Java
}
lapply(avg.vars, myfun)
向上侧：
向下侧：
[[1]] CLID ITNUM SATS ASSETS V1 V2 V3 1: CNK First Always 0-10 11.66824 11.66819 11.66829 2: CNK First Never 101-200 11.37378 12.21008 11.60182 3: CNK First Sometimes 26-100 12.43004 13.13450 12.01330 4: CNK Second Always MORE THAN 200 12.32265 11.81613 12.56786 5: CNK Second Amost always 11-25 10.76556 11.34669 10.52458 [[2]] CLID ITNUM SATS ASSETS V1 V2 V3 1: CNK First Always 0-10 9.132899 9.060045 9.197005 2: CNK First Never 101-200 12.896584 13.278680 13.000772 3: CNK First Sometimes 26-100 10.972260 11.215390 10.828431 4: CNK Second Always MORE THAN 200 11.704404 11.611072 11.749586 5: CNK Second Amost always 11-25 8.086409 8.225030 8.028928
<强> II。 for循环解决方案
使用简单的for循环，其中avg.vars有2个值的示例：
all.weights <- c("WGT1", "WGT2", "WGT3")
avg.vars <- c("AVGVALUE1", "AVGVALUE2")
split.vars <- c("ITNUM", "SATS", "ASSETS")
result <- data.frame(matrix(nrow=0,ncol=7))
for(i in avg.vars){
tmp <-
mydata[ , Map(f = weighted.mean,
x = .(get(i)),
w = mget(all.weights),
na.rm = TRUE),
by = c(key(mydata)[1], split.vars)]
result <- rbind(result,tmp,use.names=F)
}
colnames(result) <- c("CLID", "ITNUM", "SATS", "ASSETS", "V1", "V2", "V3")
result
CLID ITNUM SATS ASSETS V1 V2 V3 1: CNK First Always 0-10 11.668243 11.668192 11.668287 2: CNK First Never 101-200 11.373780 12.210083 11.601819 3: CNK First Sometimes 26-100 12.430039 13.134499 12.013299 4: CNK Second Always MORE THAN 200 12.322651 11.816135 12.567860 5: CNK Second Amost always 11-25 10.765557 11.346688 10.524583 6: CNK First Always 0-10 9.132899 9.060045 9.197005 7: CNK First Never 101-200 12.896584 13.278680 13.000772 8: CNK First Sometimes 26-100 10.972260 11.215390 10.828431 9: CNK Second Always MORE THAN 200 11.704404 11.611072 11.749586 10: CNK Second Amost always 11-25 8.086409 8.225030 8.028928
向上侧：
data.table return初始化为列表（return <- list()），创建计数器变量（n <- 1）然后替换rbind来获取该列表带有return[n] <- tmp的语句并在循环中递增计数器（n <- n + 1）向下侧：
avg.var），则使用循环编写的任何循环或函数的性能都会很差