Question

我有一个数据框，其系数来自glm（下面betas）。数据框包含协变量标签，协变量形式和估计值。形式是线性（Li），平方/二次（Sq）和log（Ps）。

betas <- structure(list(CovGen = c("A", "B", "C", "D", "E", "F", "G", 
                                   "G", "H"), Form = c("Li", "Li", "Li", "Li", "Li", "Li", "Li", 
                                                       "Sq", "Ps"), Estimate = c(0.0294573176934061, 0.0100315121169383, 
                                                                                 -0.0155864186367343, -0.00871344935814372, 0.0362538988332902, 
                                                                                 -0.0263072916746069, 0.0865742118052235, 0.0614689145750204, 
                                                                                 0.00229745713752781)), .Names = c("CovGen", "Form", "Estimate"
                                                                                 ), row.names = c(NA, 9L), class = "data.frame")

betas
  CovGen Form     Estimate
1      A   Li  0.029457318
2      B   Li  0.010031512
3      C   Li -0.015586419
4      D   Li -0.008713449
5      E   Li  0.036253899
6      F   Li -0.026307292
7      G   Li  0.086574212
8      G   Sq  0.061468915
9      H   Ps  0.002297457

我尝试应用系数估算来手动预测新数据框的值（此处包含dat dput）。

dat <- structure(list(B = c(-1.47218074669544, -1.46929972689195, -1.46641870708846, 
                            -1.46353768728497, -1.46065666748148, -1.45777564767799), C = c(-1.09847692593512, 
                                                                                            -1.09375316152745, -1.08902939711978, -1.08430563271211, -1.07958186830444, 
                                                                                            -1.07485810389677), D = c(-1.0109875688763, -1.00407851818141, 
                                                                                                                      -0.997169467486518, -0.990260416791627, -0.983351366096736, -0.976442315401845
                                                                                            ), E = c(-3.19632050296668, -3.19041566990116, -3.18451083683563, 
                                                                                                     -3.17860600377011, -3.17270117070458, -3.16679633763906), F = c(-2.81211918021003, 
                                                                                                                                                                     -2.80673925496675, -2.80135932972346, -2.79597940448018, -2.7905994792369, 
                                                                                                                                                                     -2.78521955399362), G = c(-2.32916817000267, -2.32368219245727, 
                                                                                                                                                                                               -2.31819621491187, -2.31271023736647, -2.30722425982107, -2.30173828227567
                                                                                                                                                                     ), H = c(0.442067970883549, 0.417909464459238, 0.393750958034926, 
                                                                                                                                                                              0.369592451610615, 0.345433945186303, 0.321275438761992)), .Names = c("B", 
                                                                                                                                                                                                                                                    "C", "D", "E", "F", "G", "H"), row.names = c(NA, 6L), class = "data.frame")                                                                                                                                                                                                                                   "C", "D", "E", "F", "G", "H"), row.names = c(NA, 6L), class = "data.frame")



> dat
          B         C          D         E         F         G         H
1 -1.472181 -1.098477 -1.0109876 -3.196321 -2.812119 -2.329168 0.4420680
2 -1.469300 -1.093753 -1.0040785 -3.190416 -2.806739 -2.323682 0.4179095
3 -1.466419 -1.089029 -0.9971695 -3.184511 -2.801359 -2.318196 0.3937510
4 -1.463538 -1.084306 -0.9902604 -3.178606 -2.795979 -2.312710 0.3695925
5 -1.460657 -1.079582 -0.9833514 -3.172701 -2.790599 -2.307224 0.3454339
6 -1.457776 -1.074858 -0.9764423 -3.166796 -2.785220 -2.301738 0.3212754

我正在尝试将dat df中的新数据值乘以各自的beta，并考虑功能形式。更具体地说，在此处包含的示例中，我想将G beta的Sq形式应用于dat$G^2，将Ps H beta应用于log(dat$H)。所有其他测试版和值可以简单地直接相乘而不考虑函数形式。请注意，A beta未应用于dat df中的新值。

我可能需要通过奖励ifelse声明，但我想知道是否有其他想法和/或建议。

我在一个更大的循环中工作，并且每个协变量都没有一致的形式。

期望的结果是矩阵或df，其中列包含每个β形式组合的预测值。例如，将有一个列包含所有测试数据的预测值，除了G，它将具有G和G ^ 2的预测值。

提前致谢。

Answer 1

您可以尝试这样的解决方案

trans <- list(
  Li=identity,
  Sq=function(x) x^2,
  Ps=function(x) log(x)
)

cpredict<-function(betas, datas) {
  Map(function(var, fun, coef) {
    trans[[fun]](datas[[var]])*coef
  }, betas$CovGen, betas$Form, betas$Estimate)
}

cpredict(betas, dat)

但这不适用于您当前的数据，因为没有dat$A，您无法记录负数。

Answer 2

我尝试构建公式，然后使用model.matrix和矩阵乘法，如下所示：

betas$term = with(betas, ifelse(
    Form == "Li", CovGen,
    ifelse(Form == "Sq", sprintf("I(%s^2)", CovGen),
           ifelse(Form == "Ps", sprintf("log(%s)", CovGen), NA)
)))
betas
#   CovGen Form     Estimate   term
# 1      A   Li  0.029457318      A
# 2      B   Li  0.010031512      B
# 3      C   Li -0.015586419      C
# 4      D   Li -0.008713449      D
# 5      E   Li  0.036253899      E
# 6      F   Li -0.026307292      F
# 7      G   Li  0.086574212      G
# 8      G   Sq  0.061468915 I(G^2)
# 9      H   Ps  0.002297457 log(H)
(my_formula = as.formula(paste("~", paste(betas$term, collapse = " + "))))
#~A + B + C + D + E + F + G + I(G^2) + log(H)

X = model.matrix(my_formula, data = dat)
prediction = X %*% betas$Estimate

正如MrFlick所说，这不能用你当前的样本数据。

用不同的函数形式手动预测

2 个答案: