R data.table 按因子循环子集并执行 lm() [英] R data.table loop subset by factor and do lm()

问题描述

我正在尝试创建一个函数，甚至只是想出如何使用 data.table 语法运行循环，我可以按因子对表进行子集化，在这种情况下是 id 变量，然后在每个子集上运行线性模型，然后出结果.示例数据如下.

I am trying to create a function or even just work out how to run a loop using data.table syntax where I can subset the table by factor, in this case the id variable, then run a linear model on each subset and out the results. Sample data below.

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?

预期结果类似于 fit[1] 模型、fit[2] 模型等.

Expected outcome is sopmething like fit[1] model, fit[2] model etc..

推荐答案

我知道你想用数据表来做这件事，如果你想要拟合的某些特定方面，比如系数，那么@MartinBel 的方法是一个不错的选择一.

I know you want to do this with data tables, and if you want some specific aspect of the fit, like the coefficients, then @MartinBel's approach is a good one.

另一方面，如果您想自己存储拟合，lapply(...) 可能是更好的选择:

On the other hand, if you want to store the fits themselves, lapply(...) might be a better option:

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" 在对 data.table(...) 的调用中的使用，以及 if dt[J(z)] 对数据表进行子集化.这真的没有必要，除非 dt 很大.

Note the use of key="id" in the call to data.table(...), and the use if dt[J(z)] to subset the data table. This really isn't necessary unless dt is enormous.

这篇关于R data.table 按因子循环子集并执行 lm()的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！