使用 R 的线性回归模型

Question

我是 R 的新手，但我正在努力学习。我在 excel 中有一个数据集，我将其导入到 R 中：

stockPrice<-read.csv("C:/Users/Desktop/prova.csv", sep=";", header=T, check.names = FALSE, stringsAsFactors=FALSE)

导入的结果是这样的。有 100 行和列。

        1       2       3      4       5       6       7      8       9     10      11      12
1  -1,8669 -1,2096  1,0358 0,0239 -1,0284 -0,0259  0,8801 0,4778  1,1449 0,4397 -0,1530 -0,3123
2  -2,1469 -0,4331 -0,0891 1,3842 -1,4148  0,1138 -0,8275 0,5115 -1,2898 1,8105  0,8521 -1,4327
3  -1,8919 -0,6469 -0,4098 2,8243 -1,3704 -1,6783 -0,6159 1,2910 -1,4260 2,4720  0,5230 -1,6965
4  -0,7912  0,4075  0,1092 3,8167 -0,9085 -1,0804  0,4104 0,9577 -0,2531 1,1191  1,5688 -0,8727
5  -0,2726  0,1827  0,7973 3,3848  1,0666  1,1254 -1,4111 1,2030 -0,9559 1,7813  1,8331 -1,0933
6   0,0539 -0,8640  2,0607 3,4989  2,1625  0,5226 -1,3890 2,6475 -0,6684 0,4587  0,7694  0,3462
7   0,6813 -1,9639  0,1362 1,9797  2,8645 -0,1524 -1,2367 4,6739 -1,7459 2,2648  1,8341 -0,4107
8  -0,4228 -0,3357  0,1201 2,1603  4,2053 -0,3679 -0,5577 3,7251 -1,6288 2,0168  1,1571 -0,8601
9   0,3020 -0,0523  1,4912 2,6993  5,2069 -0,0497 -0,3139 3,2010 -1,1773 1,8993  0,3357 -3,4239
10 -0,0832  0,2051  2,2387 2,9303  6,1984  1,9706 -0,3759 2,7283 -2,1752 2,0772  0,3298 -4,3092

我只是复制数据集的一部分。每列都称为一项资产。现在，我要做的是计算线性回归，例如，第一个资产 y 将是从 1 到 9 的第 1 行，x 将是从 2 到 10 的第 1 行。我必须为每个assets.i 只需要系数的值。

Answer 1

一种选择是循环使用lapply 的列，提取x 和'y'，并在删除, 后使用lm 创建一个模型并将其转换为numeric 类型

out <- lapply(stockPrice, function(vec) {
                      vec <- as.numeric(sub(",", "", vec))
                      y <- vec[1:(length(vec)-1)]
                     x <- vec[2:length(vec)]
                     coef(lm(y ~ x))
                  }
    )

out[[1]]
#  (Intercept)             x 
#-2798.4234922     0.8392437

如果我们想要斜率，那么rbind list 元素并提取第二列

do.call(rbind, out)[, 2]
#   1           2           3           4           5           6           7           8           9          10          11 
# 0.83924375  0.21597272  0.21761992  0.95551414  0.86204662  0.10036499  0.02051160  0.84014384  0.01129873 -0.31601104  0.18362571 
#        12 
# 0.46161256 

-在 excel 中检查第一列输出

-数据

-输出

---

除了上述之外，我们还可以利用lmList from nlme reshaping 为'long'格式

library(dplyr)
library(tidyr)
library(nlme)
stockPrice %>%
    mutate_all(readr::parse_number) %>% 
    pivot_longer(everything(), values_to = 'y') %>% 
    group_by(name  = factor(name, levels = unique(name))) 
    mutate(x = lead(y)) %>% 
    ungroup %>% 
    na.omit %>%
    lmList(y ~ x|name, data = .)
#Call:
#  Model: y ~ x | name 
#   Data: . 

#Coefficients:
#   (Intercept)           x
#1    -2798.423  0.83924375
#2    -4621.407  0.21597272
#3     4274.414  0.21761992
#4    -2009.506  0.95551414
#5    -5269.101  0.86204662
#6    -1814.797  0.10036499
#7    -5479.691  0.02051160
#8     1218.587  0.84014384
#9    -8747.104  0.01129873
#10   21429.645 -0.31601104
#11    7811.527  0.18362571
#12   -3786.098  0.46161256

#Degrees of freedom: 108 total; 84 residual
#Residual standard error: 8419.53

数据

stockPrice <- structure(list(`1` = c("-1,8669", "-2,1469", "-1,8919", "-0,7912", 
"-0,2726", "0,0539", "0,6813", "-0,4228", "0,3020", "-0,0832"
), `2` = c("-1,2096", "-0,4331", "-0,6469", "0,4075", "0,1827", 
"-0,8640", "-1,9639", "-0,3357", "-0,0523", "0,2051"), `3` = c("1,0358", 
"-0,0891", "-0,4098", "0,1092", "0,7973", "2,0607", "0,1362", 
"0,1201", "1,4912", "2,2387"), `4` = c("0,0239", "1,3842", "2,8243", 
"3,8167", "3,3848", "3,4989", "1,9797", "2,1603", "2,6993", "2,9303"
), `5` = c("-1,0284", "-1,4148", "-1,3704", "-0,9085", "1,0666", 
"2,1625", "2,8645", "4,2053", "5,2069", "6,1984"), `6` = c("-0,0259", 
"0,1138", "-1,6783", "-1,0804", "1,1254", "0,5226", "-0,1524", 
"-0,3679", "-0,0497", "1,9706"), `7` = c("0,8801", "-0,8275", 
"-0,6159", "0,4104", "-1,4111", "-1,3890", "-1,2367", "-0,5577", 
"-0,3139", "-0,3759"), `8` = c("0,4778", "0,5115", "1,2910", 
"0,9577", "1,2030", "2,6475", "4,6739", "3,7251", "3,2010", "2,7283"
), `9` = c("1,1449", "-1,2898", "-1,4260", "-0,2531", "-0,9559", 
"-0,6684", "-1,7459", "-1,6288", "-1,1773", "-2,1752"), `10` = c("0,4397", 
"1,8105", "2,4720", "1,1191", "1,7813", "0,4587", "2,2648", "2,0168", 
"1,8993", "2,0772"), `11` = c("-0,1530", "0,8521", "0,5230", 
"1,5688", "1,8331", "0,7694", "1,8341", "1,1571", "0,3357", "0,3298"
), `12` = c("-0,3123", "-1,4327", "-1,6965", "-0,8727", "-1,0933", 
"0,3462", "-0,4107", "-0,8601", "-3,4239", "-4,3092")),
class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10"))

Answer 2

作为@akrun 答案的替代方案，您可以使用apply 代替lapply：

使用假例子m

m <- data.frame(matrix(rnorm(100), ncol = 10, nrow = 10))

> head(m[1:3,1:3])
           X1        X2          X3
1 -0.81150290 0.3196615 -0.70848803
2  1.39105642 0.8232761  0.02241253
3 -0.01187938 0.9158422 -0.21934718

你可以这样做：

coeff = apply(m, 2, function(x) lm(x[1:9] ~ x[2:10])$coefficients[2])

并获得一个向量，其中包含从每个资产计算的所有系数：

> coeff
         X1          X2          X3          X4          X5          X6 
-0.19160847 -0.52686830  0.36973049  0.29217668 -0.70102686  0.22142335 
         X7          X8          X9         X10 
-0.13817910 -0.14292086  0.05105796 -0.22829763 

顺便说一句，当您使用 read.table 打开数据集时，您应该添加参数 dec = "," 以便不必处理非数值。所以像：

stockPrice<-read.csv("C:/Users/Desktop/prova.csv", sep=";", header=T, check.names = FALSE, stringsAsFactors=FALSE, dec = ",")