Issue with executing a loop in a dynamic linear regression with lagged time time series data in R

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In continuation to the earlier query as mentioned below

Regression with lagged time time series data in R

My objective is to extract the 4th coefficient of the regression model which are significant i.e p-value < 0.05

The general form of the model is

A ~ Lags(A, 1:2) + Lags(B, 1:2)

I have a data frame Deposits_2015Q2 having 10 columns and 24 rows. Stating below the first 6 rows of the data frame Deposits_2015Q2

Deposits_2015Q2

    B1       B2       B3       B4       B5       B6       B7       B8       B9      B10
234174.8 807526.7 93198.03 83191.27 79443.31 79593.98 52686.17 34536.29 70645.93 132601.4
236352.8 804852.0 92187.96 81834.52 77953.43 78687.54 51993.71 36098.29 69301.58 132588.6
227331.2 751653.9 83174.36 80386.75 74743.49 73838.38 50211.90 32020.65 67010.77 126405.7
219533.7 735303.5 79304.15 76730.38 73645.53 73450.54 48976.98 28927.36 64938.26 122845.9
210172.2 704675.6 79827.54 75258.59 69907.54 71725.17 47578.75 28821.78 63474.15 118728.5
198614.2 677182.0 76796.17 72260.30 67320.08 68386.68 45167.51 29215.77 59916.35 114406.4

I am able to run the below codes without a loop to extract the 4th coefficient of the model

# Loading the library
library(dyn)

# Performing Dynamic Regression and storing the results in fm
lag <- stats::lag

# Running the regression without loop
fm <- dyn$lm(B1 ~ lag(B1, -(1:2)) + lag(B10, -(1:2)), zoo(Deposits_15Q2_21Q1))

model_summary <- summary(fm)
model_summary

# Extracting the 4th coefficient
Beta <- model_summary$coefficients[5,1]
Beta

# Extracting the p-value of the 4th coefficient
Beta_p_value <- model_summary$coefficients[5,4]
Beta_p_value

My objective is to create a 10 x 10 matrix where I will store the value of the 4th coefficient of the model which is significant. The model will run w.r.t each column. I am unable to execute the below code

# Matrix to store the p-values
p=matrix(rep(0,10*10),nrow=10,ncol=10)

# Matrix to store the 4th coefficient values of the model 
K=matrix(rep(0,10*10),nrow=10,ncol=10)

# Binding the data in the data set 
  Deposits=rbind(Deposits_2015Q2$B1,Deposits_2015Q2$B2,Deposits_2015Q2$B3,Deposits_2015Q2$B4,Deposits_2015Q2$B5,Deposits_2015Q2$B6,Deposits_2015Q2$B7,Deposits_2015Q2$B8,Deposits_2015Q2$B9,Deposits_2015Q2$B10)

Deposits

# Converting dataframe into time series

B = zoo(Deposits)


for (i in 1:10){
for (j in 1:10){
if (j!=i){
  fm=dyn$lm(B[i] ~ lag(B[i], -(1:2)) + lag(B[j], -(1:2)))
  p[i,j]=summary(fm)$coefficients[5,4]
}
else{
  p[i,j]=0
  }
 }
}

 for (i in 1:10){
   for (j in 1:10){
   if (p[i,j]>0.05){
     K[i,j]=0
   }
   else {
     K[i,j]=summary(fm)$coefficients[5,1]
   }
 }
}

# Converting the Matrix K into data frame
 dataf <- as.data.frame(K)

The error message that I am getting is

Error in model.frame.default(formula = dyn(B[i] ~ lag(B[i], -(1:2)) +  : 
  variable lengths differ (found for 'lag(B[i], -(1:2))')
time-series linear-regression nested-loops r
1 Answers

There are several problems.

  1. The construction of B is incorrect. Also series should be stored in columns, not rows, and the ith column of a matrix or zoo object is B[, i].
  2. summary(fm)$coefficients[5, 1] and summary(fm)$coefficients[5, 4] work but would be better written as coef(fm)[[5]] and coef(summary(fm))[5, 4]
  3. K would be best left as a matrix, not converted to a data frame.
  4. The number of rows of data shown in the question is insufficient to run the problem. I have doubled the number of rows in the sample data in the Note at the end.
  5. Since p and K are set to all 0 we don't need to set them again.

Along with some other minor improvements this gives

library(dyn)
lag <- stats::lag  # only needed if dplyr is loaded

nc <- ncol(Deposits_2015Q2)
K <- p <- matrix(0, nc, nc)  # coefs and p values
B <- zoo(Deposits_2015Q2)

for (i in 1:nc) {
  for (j in 1:nc) {
    if (i != j) {
      fm <- dyn$lm(B[, i] ~ lag(B[, i], -(1:2)) + lag(B[, j], -(1:2)))
      p[i,j] <- coef(summary(fm))[5, 4]
      if (p[i, j] < 0.5) K[i,j] <- coef(fm)[[5]]
    }
  }
}

Note

Lines <- "    B1       B2       B3       B4       B5       B6       B7       B8       B9      B10
234174.8 807526.7 93198.03 83191.27 79443.31 79593.98 52686.17 34536.29 70645.93 132601.4
236352.8 804852.0 92187.96 81834.52 77953.43 78687.54 51993.71 36098.29 69301.58 132588.6
227331.2 751653.9 83174.36 80386.75 74743.49 73838.38 50211.90 32020.65 67010.77 126405.7
219533.7 735303.5 79304.15 76730.38 73645.53 73450.54 48976.98 28927.36 64938.26 122845.9
210172.2 704675.6 79827.54 75258.59 69907.54 71725.17 47578.75 28821.78 63474.15 118728.5
198614.2 677182.0 76796.17 72260.30 67320.08 68386.68 45167.51 29215.77 59916.35 114406.4"
Deposits_2015Q2 <- read.table(text = Lines, header = TRUE)[c(1:6, 1:6), ]
user_5165480
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