R中动态组的线性回归
Linear regression on dynamic groups in R
我有一个 data.table data_dt
,我想对其进行 运行 线性回归,以便用户可以选择组 G1
和 [=14] 中的列数=] 使用变量 n_col
。以下代码运行完美,但由于创建矩阵需要花费额外的时间,所以速度很慢。为了提高下面代码的性能,是否有办法通过调整 lm
函数的公式来完全删除步骤 1、2 和 3 并仍然得到相同的结果?
library(timeSeries)
library(data.table)
data_dt = as.data.table(LPP2005REC[, -1])
n_col = 3 # Choose a number from 1 to 3
######### Step 1 ######### Create independent variable
xx <- as.matrix(data_dt[, "SPI"])
######### Step 2 ######### Create Group 1 of dependent variables
G1 <- as.matrix(data_dt[, .SD, .SDcols=c(1:n_col + 2)])
######### Step 3 ######### Create Group 2 of dependent variables
G2 <- as.matrix(data_dt[, .SD, .SDcols=c(1:n_col + 2 + n_col)])
lm(xx ~ G1 + G2)
结果 -
summary(lm(xx ~ G1 + G2))
Call:
lm(formula = xx ~ G1 + G2)
Residuals:
Min 1Q Median 3Q Max
-3.763e-07 -4.130e-09 3.000e-09 9.840e-09 4.401e-07
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -4.931e-09 3.038e-09 -1.623e+00 0.1054
G1LMI -5.000e-01 4.083e-06 -1.225e+05 <2e-16 ***
G1MPI -2.000e+00 4.014e-06 -4.982e+05 <2e-16 ***
G1ALT -1.500e+00 5.556e-06 -2.700e+05 <2e-16 ***
G2LPP25 3.071e-04 1.407e-04 2.184e+00 0.0296 *
G2LPP40 -5.001e+00 2.360e-04 -2.119e+04 <2e-16 ***
G2LPP60 1.000e+01 8.704e-05 1.149e+05 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 5.762e-08 on 370 degrees of freedom
Multiple R-squared: 1, Adjusted R-squared: 1
F-statistic: 1.104e+12 on 6 and 370 DF, p-value: < 2.2e-16
只需使用 reformulate
创建公式,这可能会更容易
out <- lm(reformulate(names(data_dt)[c(1:n_col + 2, 1:n_col + 2 + n_col)],
response = 'SPI'), data = data_dt)
-正在检查
> summary(out)
Call:
lm(formula = reformulate(names(data_dt)[c(1:n_col + 2, 1:n_col +
2 + n_col)], response = "SPI"), data = data_dt)
Residuals:
Min 1Q Median 3Q Max
-3.763e-07 -4.130e-09 3.000e-09 9.840e-09 4.401e-07
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -4.931e-09 3.038e-09 -1.623e+00 0.1054
LMI -5.000e-01 4.083e-06 -1.225e+05 <2e-16 ***
MPI -2.000e+00 4.014e-06 -4.982e+05 <2e-16 ***
ALT -1.500e+00 5.556e-06 -2.700e+05 <2e-16 ***
LPP25 3.071e-04 1.407e-04 2.184e+00 0.0296 *
LPP40 -5.001e+00 2.360e-04 -2.119e+04 <2e-16 ***
LPP60 1.000e+01 8.704e-05 1.149e+05 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 5.762e-08 on 370 degrees of freedom
Multiple R-squared: 1, Adjusted R-squared: 1
F-statistic: 1.104e+12 on 6 and 370 DF, p-value: < 2.2e-16
我有一个 data.table data_dt
,我想对其进行 运行 线性回归,以便用户可以选择组 G1
和 [=14] 中的列数=] 使用变量 n_col
。以下代码运行完美,但由于创建矩阵需要花费额外的时间,所以速度很慢。为了提高下面代码的性能,是否有办法通过调整 lm
函数的公式来完全删除步骤 1、2 和 3 并仍然得到相同的结果?
library(timeSeries)
library(data.table)
data_dt = as.data.table(LPP2005REC[, -1])
n_col = 3 # Choose a number from 1 to 3
######### Step 1 ######### Create independent variable
xx <- as.matrix(data_dt[, "SPI"])
######### Step 2 ######### Create Group 1 of dependent variables
G1 <- as.matrix(data_dt[, .SD, .SDcols=c(1:n_col + 2)])
######### Step 3 ######### Create Group 2 of dependent variables
G2 <- as.matrix(data_dt[, .SD, .SDcols=c(1:n_col + 2 + n_col)])
lm(xx ~ G1 + G2)
结果 -
summary(lm(xx ~ G1 + G2))
Call:
lm(formula = xx ~ G1 + G2)
Residuals:
Min 1Q Median 3Q Max
-3.763e-07 -4.130e-09 3.000e-09 9.840e-09 4.401e-07
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -4.931e-09 3.038e-09 -1.623e+00 0.1054
G1LMI -5.000e-01 4.083e-06 -1.225e+05 <2e-16 ***
G1MPI -2.000e+00 4.014e-06 -4.982e+05 <2e-16 ***
G1ALT -1.500e+00 5.556e-06 -2.700e+05 <2e-16 ***
G2LPP25 3.071e-04 1.407e-04 2.184e+00 0.0296 *
G2LPP40 -5.001e+00 2.360e-04 -2.119e+04 <2e-16 ***
G2LPP60 1.000e+01 8.704e-05 1.149e+05 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 5.762e-08 on 370 degrees of freedom
Multiple R-squared: 1, Adjusted R-squared: 1
F-statistic: 1.104e+12 on 6 and 370 DF, p-value: < 2.2e-16
只需使用 reformulate
out <- lm(reformulate(names(data_dt)[c(1:n_col + 2, 1:n_col + 2 + n_col)],
response = 'SPI'), data = data_dt)
-正在检查
> summary(out)
Call:
lm(formula = reformulate(names(data_dt)[c(1:n_col + 2, 1:n_col +
2 + n_col)], response = "SPI"), data = data_dt)
Residuals:
Min 1Q Median 3Q Max
-3.763e-07 -4.130e-09 3.000e-09 9.840e-09 4.401e-07
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -4.931e-09 3.038e-09 -1.623e+00 0.1054
LMI -5.000e-01 4.083e-06 -1.225e+05 <2e-16 ***
MPI -2.000e+00 4.014e-06 -4.982e+05 <2e-16 ***
ALT -1.500e+00 5.556e-06 -2.700e+05 <2e-16 ***
LPP25 3.071e-04 1.407e-04 2.184e+00 0.0296 *
LPP40 -5.001e+00 2.360e-04 -2.119e+04 <2e-16 ***
LPP60 1.000e+01 8.704e-05 1.149e+05 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 5.762e-08 on 370 degrees of freedom
Multiple R-squared: 1, Adjusted R-squared: 1
F-statistic: 1.104e+12 on 6 and 370 DF, p-value: < 2.2e-16