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我编写了一个函数来获取（线性）模型的系数并将它们应用到原始变量上，以给出一个数据框，当它们相加时，将等效于 predict() 的结果。这种能力对我来说似乎很有用，以便更好地理解每个变量（或更复杂的交互项等）对模型的影响。

有没有更好的办法？我觉得自己像个黑客。我已经研究了模型的 str() ，但目前还没有看到更简单的解决方案。棘手的部分是捕捉和应用交互术语。

library(plyr)
nospredict = function(model, data = model$model, sorted = TRUE) {  # model is model (from lm, glm...), data is data.frame to be applied to                                                                    
    c = coef(model)  # model must support coef()                                                                                                                                                              
    my.names = names(c) =  gsub(':', '*', names(c) )  # ':' equals multiplication in formulas, coefs                                                                                                          
    data = data[ , colnames(data) %in% my.names] # don't do the attach() below with a zillion variables...                                                                                                    
    final.out = adply(data, 1, function(y) { # did I mention I like plyr?                                                                                                                                     
        attach(as.list(y), warn.conflicts = FALSE) # so you can do eval algebra blackRmagic                                                                                                                   
        out = ldply(my.names, function (x) { # did I mention...                                                                                                                                               
            Intercept = 1  # (Intercept) from model is a constant, multiply it by 1                                                                                                                           
            eval(  parse(  text = paste( c[x], "*", x )  )  )       }) # blackRmagic                                                                                                                          
        out = as.data.frame(t(out)) ; colnames(out) = my.names ; out
    })
    rownames(final.out) = rownames(data)
    final.out$Predict = predict(model, data) ## add predict() as column                                                                                                                                       
    if ( sorted ) {
        final.out[order(final.out$Predict), ]  ## return df sorted by predict()                                                                                                                               
    }
    final.out
}
set.seed(10538)
df = data.frame(a = 1:10, b = rnorm(10), c = 1:10 + rnorm(10) )
lmf = lm( c ~ a * b, data = df)

> df
a           b         c
1   1 -0.41184664 1.3739709
2   2  1.06464586 0.8975101
3   3 -0.07522363 3.4910425
4   4  1.21643049 2.8856876
5   5  0.34061917 4.3851439
6   6 -1.00020786 6.1836535
7   7 -0.36954963 6.4734150
8   8 -1.47754640 6.8150569
9   9 -0.19312147 9.6432687
10 10  2.32220098 9.4276813


> nospredict(lmf)
(Intercept)         a           b         a*b   Predict
1   0.09801818 0.9282185  0.48332671 -0.05438652 1.4551769
2   0.09801818 1.8564370 -1.24942570  0.28118420 0.9862137
3   0.09801818 2.7846555  0.08827944 -0.02980103 2.9411521
4   0.09801818 3.7128740 -1.42755405  0.64254425 3.0258824
5   0.09801818 4.6410925 -0.39973700  0.22490279 4.5642765
6   0.09801818 5.5693110  1.17380385 -0.79249635 6.0486367
7   0.09801818 6.4975295  0.43368863 -0.34160685 6.6876294
8   0.09801818 7.4257480  1.73398922 -1.56094237 7.6968130
9   0.09801818 8.3539665  0.22663962 -0.22952439 8.4490999
10  0.09801818 9.2821850 -2.72524198  3.06658890 9.7215501
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1 回答 1

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junk <- data.frame( x1 = rnorm(100), x2 = rnorm(100))

junk$YY <- 2 * junk$x1 + 4 * junk$x2 + 6 * junk$x1 * junk$x2 + 7 + rnorm(100)

out <- lm(YY ~ x1 * x2, data = junk, x = TRUE) # The x = TRUE part is key!

head(out$x)
   (Intercept)          x1         x2        x1:x2
 1           1 -0.34885894 -0.8127228  0.283525629
 2           1 -0.04482498 -0.1601600  0.007179167
 3           1 -1.11721391  0.3266213 -0.364905892
 4           1 -0.08530188  0.3482372 -0.029705287
 5           1  0.19138684 -0.1659683 -0.031764149
 6           1 -1.89493717  1.0261454 -1.944481020

coef(out)
(Intercept)          x1          x2       x1:x2 
   7.053434    1.804441    4.130249    5.970553

nomThings <- t( t(out$x[, names(coef(out))]) * coef(out) )

如果你有一些因素作为自变量，或者它在所有情况下都能正常工作，我不完全确定这会正常工作。但我怀疑是这样。

当然，您可以将 coef(out) 保存为临时对象，等等，以使代码更具可读性和效率。

鉴于您可以轻松做到这一点，我会认真考虑是否应该这样做。

于 2016-03-21T15:47:25.180 回答