Question

我有这段代码完全可以实现我想要的功能，但是我不得不为每个组调用三次函数，这似乎效率很低。

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

有什么主意我可以重写它以便仅调用一次函数吗？

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

试图将所有内容包装在一个函数中，但实际上放慢了速度：

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)

Answer 1

如果使函数返回列表，则只需要调用

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

使用

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

您可以将x和y的值作为子集，然后与函数的结果联接

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

Answer 2

您可以修改函数以返回向量和dcast最终结果：

library(data.table)
myRegr = function(x, y) {
  regr <- lm.fit(cbind(1, x), y)
  c(
    regr$coef[[1]],
    regr$coef[[2]],
    1 - var(regr$residuals) / var(y)
  )
}
result <- df[, .(minX = min(x), minY = min(y), myRegr(x, y), c("m", "k", "r2")), a]
dcast(result, a + minX + minY ~ V4, value.var = "V3")

此解决方案并不完美，因为我必须创建V4（添加c("m", "k", "r2")向量）。应该有一个更好的方法来做到这一点（甚至不使用dcast）。也许更有经验的data.table用户可以对此提出建议？

数据：

df <- 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)
)

分组时从函数分配多个结果

2 个答案: