拆分因数，sapply和lm

Question

我想将lm（）应用于按主题分组的观察结果，但无法计算出sapply语法。最后，我想要一个数据帧，每个主题有1行，并具有截距和斜率（即，以下行：subj，lm $ coefficients [1] lm $ coefficients [2]）

set.seed(1)
subj <- rep(c("a","b","c"), 4) # 4 observations each on 3 experimental subjects
ind <- rnorm(12) #12 random numbers, the independent variable, the x axis
dep <- rnorm(12) + .5 #12 random numbers, the dependent variable, the y axis
df <- data.frame(subj=subj, ind=ind, dep=dep)
s <- (split(df,subj)) # create a list of observations by subject

我可以从s中获取一组观察值，制作一个数据框，然后得到我想要的：

df2 <- as.data.frame(s[1])
df2
lm1 <- lm(df2$a.dep ~ df2$a.ind)

lm1$coefficients[1]
lm1$coefficients[2]

我在遍历s的所有元素并将数据转换成我想要的最终形式时遇到麻烦：

lm.list <- sapply(s, FUN= function(x)
  (lm(x[ ,"dep"] ~ x[,"ind"])))
a <-as.data.frame(lm.list)

我觉得我需要以下结构的某种转置；列（a，b，c）是我想要的行，但是t（a）不起作用。

head(a)
                                                            a
coefficients                             0.1233519, 0.4610505
residuals        0.4471916, -0.3060402, 0.4460895, -0.5872409
effects          -0.6325478, 0.6332422, 0.5343949, -0.7429069
rank                                                        2
fitted.values 0.74977179, 0.09854505, -0.05843569, 0.47521446
assign                                                   0, 1
                                                             b
coefficients                              1.1220840, 0.2024222
residuals     -0.04461432, 0.02124541, 0.27103003, -0.24766112
effects           -2.0717363, 0.2228309, 0.2902311, -0.2302195
rank                                                         2
fitted.values       1.1012775, 0.8433366, 1.1100777, 1.0887808
assign                                                    0, 1
                                                         c
coefficients                          0.2982019, 0.1900459
residuals     -0.5606330, 1.0491990, 0.3908486, -0.8794147
effects       -0.6742600, 0.2271767, 1.1273566, -1.0345665
rank                                                     2
fitted.values   0.3718773, 0.2193339, 0.5072572, 0.2500516
assign                                                0, 1

Answer 1

听起来，这可能就是您要尝试执行的操作：

sapply(s, FUN= function(x)
  lm(x[ ,"dep"] ~ x[,"ind"])$coefficients[c(1, 2)])
#                      a          b          c
# (Intercept) 0.71379430 -0.6817331  0.5717372
# x[, "ind"]  0.07125591  1.1452096 -1.0303726

其他选择，如果这是您要寻找的

我已经看到它指出，一般来说，如果先拆分然后使用s/lapply，通常可以直接跳至by并跳过以下split步骤：

do.call(rbind, 
        by(data = df, INDICES=df$subj, FUN=function(x) 
          lm(x[, "dep"] ~ x[, "ind"])$coefficients[c(1, 2)]))
#   (Intercept)  x[, "ind"]
# a   0.7137943  0.07125591
# b  -0.6817331  1.14520962
# c   0.5717372 -1.03037257

或者，您可以使用一个包来使您更方便地进行此类计算，例如“ data.table”：

library(data.table)
DT <- data.table(df)
DT[, list(Int = lm(dep ~ ind)$coefficients[1],
          Slo = lm(dep ~ ind)$coefficients[2]), by = subj]
#    subj        Int         Slo
# 1:    a  0.7137943  0.07125591
# 2:    b -0.6817331  1.14520962
# 3:    c  0.5717372 -1.03037257