R data.table循环子集by factor and do lm()

Question

我正在尝试创建一个函数,或者甚至只是计算如何使用data.table语法运行循环,我可以通过因子对表进行子集化,在本例中为id变量,然后在每个子集上运行线性模型并输出结果.以下示例数据.

df <- data.frame(id = letters[1:3], 
                 cyl = sample(c("a","b","c"), 30, replace = TRUE),
                 factor = sample(c(TRUE, FALSE), 30, replace = TRUE),   
                 hp = sample(c(20:50), 30, replace = TRUE))

dt=as.data.table(df)

fit <- lm(hp ~ cyl + factor, data = df) #how do I get the [i] to work here to subset and iterate by each factor and also do it in data.table syntax?

预期的结果是适合[1]模型,拟合[2]模型等.

Answer 1

我知道你想用数据表做这个,如果你想要一些特定的拟合方面,比如系数,那么@ MartinBel的方法是一个很好的方法.

另一方面,如果您想自己存储拟合,lapply(...)可能是更好的选择:

set.seed(1)
df <- data.frame(id = letters[1:3], 
                 cyl = sample(c("a","b","c"), 30, replace = TRUE),
                 factor = sample(c(TRUE, FALSE), 30, replace = TRUE),   
                 hp = sample(c(20:50), 30, replace = TRUE))
dt <- data.table(df,key="id")

fits <- lapply(unique(df$id),
               function(z)lm(hp~cyl+factor, data=dt[J(z),], y=T))
# coefficients
sapply(fits,coef)
#                   [,1]      [,2]          [,3]
# (Intercept)  44.117647 35.000000  3.933333e+01
# cylb         -6.117647 -6.321429 -1.266667e+01
# cylc        -13.176471  3.821429 -7.833333e+00
# factorTRUE    1.176471  5.535714  2.325797e-15

# predicted values
sapply(fits,predict)
#        [,1]     [,2]     [,3]
# 1  45.29412 28.67857 26.66667
# 2  32.11765 35.00000 31.50000
# 3  30.94118 34.21429 26.66667
# ...

# residuals
sapply(fits,residuals)
#           [,1]        [,2]      [,3]
# 1    2.7058824   0.3214286  7.333333
# 2   -2.1176471   5.0000000 -4.500000
# 3    3.0588235   8.7857143 -4.666667
# ...

# se and r-sq
sapply(fits, function(x)c(se=summary(x)$sigma, rsq=summary(x)$r.squared))
#         [,1]      [,2]      [,3]
# se  7.923655 8.6358196 6.4592741
# rsq 0.463076 0.3069017 0.4957024

# Q-Q plots
par(mfrow=c(1,length(fits)))
lapply(fits,plot,2)

注意key="id"在调用中data.table(...)的使用,以及dt[J(z)]如何对数据表进行子集化.除非dt是巨大的,否则这确实没有必要.