Stacking lapply results

I am using the following code to generate data, and i am estimating regression models across a list of variables (covar1 and covar2). I have also created confidence intervals for the coefficients and merged them together.

I have been examining all sorts of examples here and on other sites, but i can't seem to accomplish what i want. I want to stack the results for each covar into a single data frame, labeling each cluster of results by the covar it is attributable to (i.e., "covar1" and "covar2"). Here is the code for generating data and results using lapply:

##creating a fake dataset (N=1000, 500 at treated, 500 at control group)

#outcome variable

outcome <- c(rnorm(500, mean = 50, sd = 10),  rnorm(500, mean = 70, sd = 10))



#running variable

running.var <- seq(0, 1, by = .0001)

running.var <- sample(running.var, size = 1000, replace = T)



##Put negative values for the running variable in the control group

running.var[1:500] <- -running.var[1:500]



#treatment indicator (just a binary variable indicating treated and control groups)

treat.ind <- c(rep(0,500), rep(1,500))



#create covariates

set.seed(123)

covar1 <- c(rnorm(500, mean = 50, sd = 10), rnorm(500, mean = 50, sd = 20))

covar2 <- c(rnorm(500, mean = 10, sd = 20), rnorm(500, mean = 10, sd = 30))

data <- data.frame(cbind(outcome, running.var, treat.ind, covar1, covar2))

data$treat.ind <- as.factor(data$treat.ind)



#Bundle the covariates names together

covars <- c("covar1", "covar2")



#loop over them using a convenient feature of the "as.formula" function

models <- lapply(covars, function(x){

  regres <- lm(as.formula(paste(x," ~ running.var + treat.ind",sep = "")), data = d)

  ci <-confint(regres, level=0.95)

  regres_ci <- cbind(summary(regres)$coefficient, ci)

})

names(models) <- covars

print(models)

Any nudge in the right direction, or link to a post i just haven't come across, is greatly appreciated.

asked Nov 21 '18 at 19:30

Jason Schoeneberger

What is d in the code?

– m0nhawk
Nov 21 '18 at 19:32

In the lm() call within the lapply(), is d meant to be data? Also, it would help if you could outline the expected output (dimensions and colnames of the expected dataframe)

– 12b345b6b78
Nov 21 '18 at 19:33

good points above, I'm guessing something like models %>% purrr::map_df(broom::tidy, .id = "covar_id") will get close to what you want

– Nate
Nov 21 '18 at 19:35

add a comment |

##creating a fake dataset (N=1000, 500 at treated, 500 at control group)

#outcome variable

outcome <- c(rnorm(500, mean = 50, sd = 10),  rnorm(500, mean = 70, sd = 10))



#running variable

running.var <- seq(0, 1, by = .0001)

running.var <- sample(running.var, size = 1000, replace = T)



##Put negative values for the running variable in the control group

running.var[1:500] <- -running.var[1:500]



#treatment indicator (just a binary variable indicating treated and control groups)

treat.ind <- c(rep(0,500), rep(1,500))



#create covariates

set.seed(123)

covar1 <- c(rnorm(500, mean = 50, sd = 10), rnorm(500, mean = 50, sd = 20))

covar2 <- c(rnorm(500, mean = 10, sd = 20), rnorm(500, mean = 10, sd = 30))

data <- data.frame(cbind(outcome, running.var, treat.ind, covar1, covar2))

data$treat.ind <- as.factor(data$treat.ind)



#Bundle the covariates names together

covars <- c("covar1", "covar2")



#loop over them using a convenient feature of the "as.formula" function

models <- lapply(covars, function(x){

  regres <- lm(as.formula(paste(x," ~ running.var + treat.ind",sep = "")), data = d)

  ci <-confint(regres, level=0.95)

  regres_ci <- cbind(summary(regres)$coefficient, ci)

})

names(models) <- covars

print(models)

Any nudge in the right direction, or link to a post i just haven't come across, is greatly appreciated.

asked Nov 21 '18 at 19:30

Jason Schoeneberger

What is d in the code?

– m0nhawk
Nov 21 '18 at 19:32

In the lm() call within the lapply(), is d meant to be data? Also, it would help if you could outline the expected output (dimensions and colnames of the expected dataframe)

– 12b345b6b78
Nov 21 '18 at 19:33

good points above, I'm guessing something like models %>% purrr::map_df(broom::tidy, .id = "covar_id") will get close to what you want

– Nate
Nov 21 '18 at 19:35

add a comment |

##creating a fake dataset (N=1000, 500 at treated, 500 at control group)

#outcome variable

outcome <- c(rnorm(500, mean = 50, sd = 10),  rnorm(500, mean = 70, sd = 10))



#running variable

running.var <- seq(0, 1, by = .0001)

running.var <- sample(running.var, size = 1000, replace = T)



##Put negative values for the running variable in the control group

running.var[1:500] <- -running.var[1:500]



#treatment indicator (just a binary variable indicating treated and control groups)

treat.ind <- c(rep(0,500), rep(1,500))



#create covariates

set.seed(123)

covar1 <- c(rnorm(500, mean = 50, sd = 10), rnorm(500, mean = 50, sd = 20))

covar2 <- c(rnorm(500, mean = 10, sd = 20), rnorm(500, mean = 10, sd = 30))

data <- data.frame(cbind(outcome, running.var, treat.ind, covar1, covar2))

data$treat.ind <- as.factor(data$treat.ind)



#Bundle the covariates names together

covars <- c("covar1", "covar2")



#loop over them using a convenient feature of the "as.formula" function

models <- lapply(covars, function(x){

  regres <- lm(as.formula(paste(x," ~ running.var + treat.ind",sep = "")), data = d)

  ci <-confint(regres, level=0.95)

  regres_ci <- cbind(summary(regres)$coefficient, ci)

})

names(models) <- covars

print(models)

Any nudge in the right direction, or link to a post i just haven't come across, is greatly appreciated.

asked Nov 21 '18 at 19:30

Jason Schoeneberger

##creating a fake dataset (N=1000, 500 at treated, 500 at control group)

#outcome variable

outcome <- c(rnorm(500, mean = 50, sd = 10),  rnorm(500, mean = 70, sd = 10))



#running variable

running.var <- seq(0, 1, by = .0001)

running.var <- sample(running.var, size = 1000, replace = T)



##Put negative values for the running variable in the control group

running.var[1:500] <- -running.var[1:500]



#treatment indicator (just a binary variable indicating treated and control groups)

treat.ind <- c(rep(0,500), rep(1,500))



#create covariates

set.seed(123)

covar1 <- c(rnorm(500, mean = 50, sd = 10), rnorm(500, mean = 50, sd = 20))

covar2 <- c(rnorm(500, mean = 10, sd = 20), rnorm(500, mean = 10, sd = 30))

data <- data.frame(cbind(outcome, running.var, treat.ind, covar1, covar2))

data$treat.ind <- as.factor(data$treat.ind)



#Bundle the covariates names together

covars <- c("covar1", "covar2")



#loop over them using a convenient feature of the "as.formula" function

models <- lapply(covars, function(x){

  regres <- lm(as.formula(paste(x," ~ running.var + treat.ind",sep = "")), data = d)

  ci <-confint(regres, level=0.95)

  regres_ci <- cbind(summary(regres)$coefficient, ci)

})

names(models) <- covars

print(models)

Any nudge in the right direction, or link to a post i just haven't come across, is greatly appreciated.

asked Nov 21 '18 at 19:30

Jason Schoeneberger

asked Nov 21 '18 at 19:30

Jason Schoeneberger

asked Nov 21 '18 at 19:30

Jason Schoeneberger

asked Nov 21 '18 at 19:30

Jason Schoeneberger

asked Nov 21 '18 at 19:30

Jason Schoeneberger

What is d in the code?

– m0nhawk
Nov 21 '18 at 19:32

In the lm() call within the lapply(), is d meant to be data? Also, it would help if you could outline the expected output (dimensions and colnames of the expected dataframe)

– 12b345b6b78
Nov 21 '18 at 19:33

good points above, I'm guessing something like models %>% purrr::map_df(broom::tidy, .id = "covar_id") will get close to what you want

– Nate
Nov 21 '18 at 19:35

add a comment |

What is d in the code?

– m0nhawk
Nov 21 '18 at 19:32

In the lm() call within the lapply(), is d meant to be data? Also, it would help if you could outline the expected output (dimensions and colnames of the expected dataframe)

– 12b345b6b78
Nov 21 '18 at 19:33

good points above, I'm guessing something like models %>% purrr::map_df(broom::tidy, .id = "covar_id") will get close to what you want

– Nate
Nov 21 '18 at 19:35

What is d in the code?

– m0nhawk
Nov 21 '18 at 19:32

In the lm() call within the lapply(), is d meant to be data? Also, it would help if you could outline the expected output (dimensions and colnames of the expected dataframe)

– 12b345b6b78
Nov 21 '18 at 19:33

good points above, I'm guessing something like models %>% purrr::map_df(broom::tidy, .id = "covar_id") will get close to what you want

– Nate
Nov 21 '18 at 19:35

add a comment |

1 Answer
1

active

oldest

votes

You can use do.call were de second argument is a list (like in here):

do.call(rbind, models)

I made a (possible) improve to your lapply function. This way you can save the estimated parameters and the variables in a data.frame:

models <- lapply(covars, function(x){

  regres <- lm(as.formula(paste(x," ~ running.var + treat.ind",sep = "")), data = data)

  ci <-confint(regres, level=0.95)

  regres_ci <- data.frame(covar=x,param=rownames(summary(regres)$coefficient),

                          summary(regres)$coefficient, ci)

})



do.call(rbind,models)

answered Nov 21 '18 at 19:46

P. Paccioretti

16616

Thanks! This is concise and gives me what i need. Much appreciated!

– Jason Schoeneberger
Nov 21 '18 at 22:38

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

You can use do.call were de second argument is a list (like in here):

do.call(rbind, models)

I made a (possible) improve to your lapply function. This way you can save the estimated parameters and the variables in a data.frame:

models <- lapply(covars, function(x){

  regres <- lm(as.formula(paste(x," ~ running.var + treat.ind",sep = "")), data = data)

  ci <-confint(regres, level=0.95)

  regres_ci <- data.frame(covar=x,param=rownames(summary(regres)$coefficient),

                          summary(regres)$coefficient, ci)

})



do.call(rbind,models)

answered Nov 21 '18 at 19:46

P. Paccioretti

16616

Thanks! This is concise and gives me what i need. Much appreciated!

– Jason Schoeneberger
Nov 21 '18 at 22:38

add a comment |

You can use do.call were de second argument is a list (like in here):

do.call(rbind, models)

I made a (possible) improve to your lapply function. This way you can save the estimated parameters and the variables in a data.frame:

models <- lapply(covars, function(x){

  regres <- lm(as.formula(paste(x," ~ running.var + treat.ind",sep = "")), data = data)

  ci <-confint(regres, level=0.95)

  regres_ci <- data.frame(covar=x,param=rownames(summary(regres)$coefficient),

                          summary(regres)$coefficient, ci)

})



do.call(rbind,models)

answered Nov 21 '18 at 19:46

P. Paccioretti

16616

Thanks! This is concise and gives me what i need. Much appreciated!

– Jason Schoeneberger
Nov 21 '18 at 22:38

add a comment |

You can use do.call were de second argument is a list (like in here):

do.call(rbind, models)

I made a (possible) improve to your lapply function. This way you can save the estimated parameters and the variables in a data.frame:

models <- lapply(covars, function(x){

  regres <- lm(as.formula(paste(x," ~ running.var + treat.ind",sep = "")), data = data)

  ci <-confint(regres, level=0.95)

  regres_ci <- data.frame(covar=x,param=rownames(summary(regres)$coefficient),

                          summary(regres)$coefficient, ci)

})



do.call(rbind,models)

answered Nov 21 '18 at 19:46

P. Paccioretti

16616

You can use do.call were de second argument is a list (like in here):

do.call(rbind, models)

I made a (possible) improve to your lapply function. This way you can save the estimated parameters and the variables in a data.frame:

models <- lapply(covars, function(x){

  regres <- lm(as.formula(paste(x," ~ running.var + treat.ind",sep = "")), data = data)

  ci <-confint(regres, level=0.95)

  regres_ci <- data.frame(covar=x,param=rownames(summary(regres)$coefficient),

                          summary(regres)$coefficient, ci)

})



do.call(rbind,models)

answered Nov 21 '18 at 19:46

P. Paccioretti

16616

answered Nov 21 '18 at 19:46

P. Paccioretti

16616

answered Nov 21 '18 at 19:46

P. Paccioretti

16616

answered Nov 21 '18 at 19:46

P. Paccioretti

16616

Thanks! This is concise and gives me what i need. Much appreciated!

– Jason Schoeneberger
Nov 21 '18 at 22:38

add a comment |

Thanks! This is concise and gives me what i need. Much appreciated!

– Jason Schoeneberger
Nov 21 '18 at 22:38

Thanks! This is concise and gives me what i need. Much appreciated!

– Jason Schoeneberger
Nov 21 '18 at 22:38

add a comment |

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