对于每个带有正则表达式的循环

Question

我有一个这样的数据集：

y_post = c(0,1,2,3) 
y_pre = c(0,1,5,3) 
x_post = c(0,1,4,3) 
x_pre = c(0,1,4,3) 
x_time_pre = c(0,4,2,3) 
x_time_post = c(0,4,2,3) 
y_time_pre = c(0,1,5,3) 
y_time_post = c(0,1,4,3)
strata = c(1,2,2,1)
df = data.frame(strata, y_post, y_pre, x_post, x_pre,x_time_pre,x_time_post, y_time_pre, y_time_post) 

我想要做的是创建一个循环来为每个包含“时间”一词的变量运行回归，这样

fit <- lm(var_contains_"time"_post ~ same_var_contains_"time"_pre +strata) 

我不确定如何指定？ 例如我想运行

lm (y_time_post ~y_time_pre+strata) 

并对所有包含时间一词的变量（我有 20 个）执行此操作

Answer 1

我们可以创建显示'time_pre'、'time_post'的列名向量并使用Map循环相应的元素，使用reformulate创建公式并应用lm

nm1 <- grep('time_pre', names(df), value = TRUE)
nm2 <- grep('time_post', names(df), value = TRUE)

Map(function(x, y) lm(reformulate(c(x, 'strata'), 
          response = y), data = df), nm1, nm2)

-输出

#$x_time_pre

#Call:
#lm(formula = reformulate(c(x, "strata"), response = y), data = df)

#Coefficients:
#(Intercept)   x_time_pre       strata  
#  4.441e-16    1.000e+00    0.000e+00  


#$y_time_pre

#Call:
#lm(formula = reformulate(c(x, "strata"), response = y), data = df)

#Coefficients:
#(Intercept)   y_time_pre       strata  
#       0.50         0.84        -0.26  

---

提取系数

library(broom)
do.call(rbind, Map(function(x, y) tidy(lm(reformulate(c(x, 'strata'), 
          response = y)), data = df), nm1, nm2))

-输出

# A tibble: 6 x 5
#  term         estimate std.error statistic  p.value
#* <chr>           <dbl>     <dbl>     <dbl>    <dbl>
#1 (Intercept)  4.44e-16     0       Inf       0     
#2 x_time_pre   1.00e+ 0     0       Inf       0     
#3 strata       0.           0       NaN     NaN     
#4 (Intercept)  5.00e- 1     0.671     0.745   0.592 
#5 y_time_pre   8.40e- 1     0.12      7.00    0.0903
#6 strata      -2.60e- 1     0.461    -0.564   0.673 

---

如果列名不同，使用bind_rows或map2_dfr

library(dplyr)
library(purrr)
map2_dfr(nm1, nm2, ~ tidy(lm(reformulate(c(.x, 'strata'), 
       response = .y), data = df)))