r - 分组时从函数中分配多个结果
问题描述
我有这段代码,它完全符合我的要求,但我必须为每个组调用我的函数三次,这似乎非常低效。
library(data.table)
myRegr = function(x, y) {
regr = lm.fit(cbind(1, x), y)
coefs = regr$coef
k = coefs[[2]]
m = coefs[[1]]
r2 = 1 - var(regr$residuals) / var(y)
return (c(k = k, m = m, r2 = r2))
}
dt = data.table(a = c(0, 0, 0, 1, 1, 1),
x = c(12, 21, 15, 34, 32, 31),
y = c(3, 1, 6, 4, 2, 8))
result = dt[,list(minX = min(x),
minY = min(y),
k = myRegr(x, y)["k"],
m = myRegr(x, y)["m"],
r2 = myRegr(x, y)["r2"]
),
by = list(a)]
print(result)
输出:
a minX minY k m r2
0 12 1 -0.3095238 8.285714 0.3176692
1 31 2 -1.0000000 37.000000 0.2500000
知道如何将其重写为仅调用该函数一次吗?
更新:我的示例没有涵盖完整的问题,因为我选择了第四列,这是一个更好的示例:
library(data.table)
myRegr = function(x, y) {
regr = lm.fit(cbind(1, x), y)
coefs = regr$coef
k = coefs[[2]]
m = coefs[[1]]
r2 = 1 - var(regr$residuals) / var(y)
return (c(k = k, m = m, r2 = r2))
}
df = data.frame(a = c(0, 0, 0, 1, 1, 1),
x = c(12, 21, 15, 34, 32, 31),
y = c(3, 1, 6, 4, 2, 8),
time = as.POSIXct(c("2019-01-01 08:12:00", "2019-01-01 08:13:00", "2019-01-01 08:14:00", "2019-01-01 08:12:00", "2019-01-01 08:13:00", "2019-01-01 08:14:00")))
dt = data.table(df)
result = dt[, list(firstX = x[time == min(time)],
firstY = y[time == min(time)],
k = myRegr(x, y)["k"],
m = myRegr(x, y)["m"],
r2 = myRegr(x, y)["r2"]
),
by = a]
print(result)
输出:
a firstX firstY k m r2
0 12 3 -0.3095238 8.285714 0.3176692
1 34 4 -1.0000000 37.000000 0.2500000
尝试将它全部包装在一个函数中,但它实际上减慢了速度:
library(data.table)
myRegrList = function(group) {
firstX = group[,x[time == min(time)]]
firstY = group[,y[time == min(time)]]
regr = lm.fit(cbind(1, group$x), group$y)
coefs = regr$coef
k = coefs[[2]]
m = coefs[[1]]
r2 = 1 - var(regr$residuals) / var(group$y)
return (list(firstX = firstX, firstY = firstY, k = k, m = m, r2 = r2))
}
result = dt[, myRegrList(.SD), by = a]
print(result)
解决方案
如果你让你的函数返回一个列表,你只需要调用
dt[, myRegr(x, y), by = a]
# a minX minY k m r2
#1: 0 12 1 -0.3095238 8.285714 0.3176692
#2: 1 31 2 -1.0000000 37.000000 0.2500000
和
myRegr = function(x, y) {
regr = lm.fit(cbind(1, x), y)
coefs = regr$coef
k = coefs[[2]]
m = coefs[[1]]
r2 = 1 - var(regr$residuals) / var(y)
return (list(# minX = min(x),
# minY = min(y),
k = k,
m = m,
r2 = r2))
}
更新
您可以对x
和y
值进行子集化,然后加入函数的结果
result <- dt[dt[, .I[which.min(time)], by = a]$V1, .(a, x, y)]
result <- result[dt[, myRegr(x, y), by = a], on = .(a)]
result
# a x y k m r2
#1: 0 12 3 -0.3095238 8.285714 0.3176692
#2: 1 34 4 -1.0000000 37.000000 0.2500000