如何将公式作为参数传递给lm in sapply？

Question

似乎lm在sapply中不会将公式作为参数。

只是lm

虽然lm单独接受公式参数FO，

summary(lm(y ~ x, df1, df1[["z"]] == 1, df1[["w"]]))$coef[1, ]
summary(lm(FO, data, data[[st]] == st1, data[[ws]]))$coef[1, ]

在lm的sapply

同样在sapply内

sapply(unique(df1$z), function(s) 
  summary(lm(y ~ x, df1, df1[["z"]] == s, df1[[ws]]))$coef[1, ])
sapply(unique(data[[st]]), function(s) 
  summary(lm(FO, data, data[[st]] == s, data[[ws]]))$coef[1, ])

导致错误：

 Error in eval(substitute(subset), data, env) : object 's' not found 

把所有东西作为参数，但公式FO它仍然有效：

sapply(unique(data[[st]]), function(s) 
  summary(lm(y ~ x, data, data[[st]] == s, data[[ws]]))$coef[1, ])

lm在for循环

所有参数都在for循环中工作：

m <- matrix(NA, 4, length(unique(data[[st]])))
for (s in unique(data[[st]])) {
  m[, s] <- summary(lm(FO, data, data[[st]] == s, data[[ws]]))$coef[1, ]
}
m
#           [,1]       [,2]         [,3]
# [1,] 1.6269038 -0.1404174 -0.010338774
# [2,] 0.9042738  0.4577001  1.858138516
# [3,] 1.7991275 -0.3067890 -0.005564049
# [4,] 0.3229600  0.8104951  0.996457853

---

数据：

df1 <- structure(list(x = c(1.37095844714667, -0.564698171396089, 0.363128411337339, 
0.63286260496104, 0.404268323140999, -0.106124516091484, 1.51152199743894, 
-0.0946590384130976, 2.01842371387704), y = c(1.30824434809425, 
0.740171482827397, 2.64977380403845, -0.755998096151299, 0.125479556323628, 
-0.239445852485142, 2.14747239550901, -0.37891195982917, -0.638031707027734
), z = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L), w = c(0.7, 0.8, 
1.2, 0.9, 1.3, 1.2, 0.8, 1, 1)), class = "data.frame", row.names = c(NA, 
-9L))

FO <- y ~ x; data <- df1; st <- "z"; ws <- "w"; st1 <- 1

---

sessionInfo()：

R version 3.6.0 (2019-04-26)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1

Matrix products: default

locale:
[1] LC_COLLATE=German_Switzerland.1252  LC_CTYPE=German_Switzerland.1252   
[3] LC_MONETARY=German_Switzerland.1252 LC_NUMERIC=C                       
[5] LC_TIME=German_Switzerland.1252    

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

loaded via a namespace (and not attached):
[1] compiler_3.6.0 tools_3.6.0    yaj_0.0.0.9044 packrat_0.5.0 

Answer 1

这在我尝试时起作用了。看起来你在公式中使用x会干扰你喜欢这个函数的行为方式。用num替换这个参数会产生听起来像你正在寻找的结果。这样，它可以确保公式中的x引用数据集而不是函数参数。

sapply(unique(dat$z), function(num) summary(lm(y ~ x, dat, z == num))$coef[1, ])

Answer 2

感谢来自R-help的@David提示尝试使用do.call，我可以搞清楚。解决方案是：

sapply(unique(data[[st]]), function(s)
  summary(do.call("lm", list(FO, data, data[[st]] == s, 
                             data[[ws]])))$coef[1, ])
#                 [,1]       [,2]         [,3]
# Estimate   1.6269038 -0.1404174 -0.010338774
# Std. Error 0.9042738  0.4577001  1.858138516
# t value    1.7991275 -0.3067890 -0.005564049
# Pr(>|t|)   0.3229600  0.8104951  0.996457853

说明:(来自R-help的@Duncan学分）sapply的调用者可能会忽略附加的> environment(FO) # <environment: R_GlobalEnv>，其中创建了公式。这可能是它与do.call和参数列表一起工作的原因。