Comparing Column names in R across various data frames
I am currently try to compare the column classes and names of various data frames in R prior to undertaking any transformations and calculations.
The code I have is noted below::
library(dplyr)
m1 <- mtcars
m2 <- mtcars %>% mutate(cyl = factor(cyl), xxxx1 = factor(cyl))
m3 <- mtcars %>% mutate(cyl = factor(cyl), xxxx2 = factor(cyl))
out <- cbind(sapply(m1, class), sapply(m2, class), sapply(m3, class))
If someone can solve this for dataframes stored in a list, that would be great. All my dataframes are currently stored in a list, for easier processing.
All.list <- list(m1,m2,m3)
I am expecting that the output is displayed in a matrix form as shown in the dataframe "out". The output in "out" is not desireable as it is incorrect. I am expecting the output to be more along the following::
r class dataframe lapply sapply
asked Nov 12 '18 at 15:10
Seb
1488
add a comment |
I am currently try to compare the column classes and names of various data frames in R prior to undertaking any transformations and calculations.
The code I have is noted below::
library(dplyr)
m1 <- mtcars
m2 <- mtcars %>% mutate(cyl = factor(cyl), xxxx1 = factor(cyl))
m3 <- mtcars %>% mutate(cyl = factor(cyl), xxxx2 = factor(cyl))
out <- cbind(sapply(m1, class), sapply(m2, class), sapply(m3, class))
If someone can solve this for dataframes stored in a list, that would be great. All my dataframes are currently stored in a list, for easier processing.
All.list <- list(m1,m2,m3)
I am expecting that the output is displayed in a matrix form as shown in the dataframe "out". The output in "out" is not desireable as it is incorrect. I am expecting the output to be more along the following::
r class dataframe lapply sapply
asked Nov 12 '18 at 15:10
Seb
1488
add a comment |
I am currently try to compare the column classes and names of various data frames in R prior to undertaking any transformations and calculations.
The code I have is noted below::
library(dplyr)
m1 <- mtcars
m2 <- mtcars %>% mutate(cyl = factor(cyl), xxxx1 = factor(cyl))
m3 <- mtcars %>% mutate(cyl = factor(cyl), xxxx2 = factor(cyl))
out <- cbind(sapply(m1, class), sapply(m2, class), sapply(m3, class))
If someone can solve this for dataframes stored in a list, that would be great. All my dataframes are currently stored in a list, for easier processing.
All.list <- list(m1,m2,m3)
I am expecting that the output is displayed in a matrix form as shown in the dataframe "out". The output in "out" is not desireable as it is incorrect. I am expecting the output to be more along the following::
r class dataframe lapply sapply
asked Nov 12 '18 at 15:10
Seb
1488
I am currently try to compare the column classes and names of various data frames in R prior to undertaking any transformations and calculations.
The code I have is noted below::
library(dplyr)
m1 <- mtcars
m2 <- mtcars %>% mutate(cyl = factor(cyl), xxxx1 = factor(cyl))
m3 <- mtcars %>% mutate(cyl = factor(cyl), xxxx2 = factor(cyl))
out <- cbind(sapply(m1, class), sapply(m2, class), sapply(m3, class))
If someone can solve this for dataframes stored in a list, that would be great. All my dataframes are currently stored in a list, for easier processing.
All.list <- list(m1,m2,m3)
I am expecting that the output is displayed in a matrix form as shown in the dataframe "out". The output in "out" is not desireable as it is incorrect. I am expecting the output to be more along the following::
r class dataframe lapply sapply
r class dataframe lapply sapply
asked Nov 12 '18 at 15:10
Seb
1488
asked Nov 12 '18 at 15:10
Seb
1488
asked Nov 12 '18 at 15:10
Seb
1488
asked Nov 12 '18 at 15:10
Seb
1488
asked Nov 12 '18 at 15:10
Seb
1488
1488
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I think the easiest way would be to define a function, and then use a combination of lapply and dplyr to obtain the result you want. Here is how I did it.
library(dplyr)
m1 <- mtcars
m2 <- mtcars %>% mutate(cyl = factor(cyl), xxxx1 = factor(cyl))
m3 <- mtcars %>% mutate(cyl = factor(cyl), xxxx2 = factor(cyl))
All.list <- list(m1,m2,m3)
##Define a function to get variable names and types
my_function <- function(data_frame){
require(dplyr)
x <- tibble(`var_name` = colnames(data_frame),
`var_type` = sapply(data_frame, class))
return(x)
}
target <- lapply(1:length(All.list),function(i)my_function(All.list[[i]]) %>%
mutate(element =i)) %>%
bind_rows() %>%
spread(element, var_type)
target
answered Nov 12 '18 at 15:23
Harro Cyranka
1,1601513
1
thank you - this works perfectly
– Seb
Nov 12 '18 at 18:36
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think the easiest way would be to define a function, and then use a combination of lapply and dplyr to obtain the result you want. Here is how I did it.
library(dplyr)
m1 <- mtcars
m2 <- mtcars %>% mutate(cyl = factor(cyl), xxxx1 = factor(cyl))
m3 <- mtcars %>% mutate(cyl = factor(cyl), xxxx2 = factor(cyl))
All.list <- list(m1,m2,m3)
##Define a function to get variable names and types
my_function <- function(data_frame){
require(dplyr)
x <- tibble(`var_name` = colnames(data_frame),
`var_type` = sapply(data_frame, class))
return(x)
}
target <- lapply(1:length(All.list),function(i)my_function(All.list[[i]]) %>%
mutate(element =i)) %>%
bind_rows() %>%
spread(element, var_type)
target
answered Nov 12 '18 at 15:23
Harro Cyranka
1,1601513
1
thank you - this works perfectly
– Seb
Nov 12 '18 at 18:36
add a comment |
I think the easiest way would be to define a function, and then use a combination of lapply and dplyr to obtain the result you want. Here is how I did it.
library(dplyr)
m1 <- mtcars
m2 <- mtcars %>% mutate(cyl = factor(cyl), xxxx1 = factor(cyl))
m3 <- mtcars %>% mutate(cyl = factor(cyl), xxxx2 = factor(cyl))
All.list <- list(m1,m2,m3)
##Define a function to get variable names and types
my_function <- function(data_frame){
require(dplyr)
x <- tibble(`var_name` = colnames(data_frame),
`var_type` = sapply(data_frame, class))
return(x)
}
target <- lapply(1:length(All.list),function(i)my_function(All.list[[i]]) %>%
mutate(element =i)) %>%
bind_rows() %>%
spread(element, var_type)
target
answered Nov 12 '18 at 15:23
Harro Cyranka
1,1601513
1
thank you - this works perfectly
– Seb
Nov 12 '18 at 18:36
add a comment |
I think the easiest way would be to define a function, and then use a combination of lapply and dplyr to obtain the result you want. Here is how I did it.
library(dplyr)
m1 <- mtcars
m2 <- mtcars %>% mutate(cyl = factor(cyl), xxxx1 = factor(cyl))
m3 <- mtcars %>% mutate(cyl = factor(cyl), xxxx2 = factor(cyl))
All.list <- list(m1,m2,m3)
##Define a function to get variable names and types
my_function <- function(data_frame){
require(dplyr)
x <- tibble(`var_name` = colnames(data_frame),
`var_type` = sapply(data_frame, class))
return(x)
}
target <- lapply(1:length(All.list),function(i)my_function(All.list[[i]]) %>%
mutate(element =i)) %>%
bind_rows() %>%
spread(element, var_type)
target
answered Nov 12 '18 at 15:23
Harro Cyranka
1,1601513
I think the easiest way would be to define a function, and then use a combination of lapply and dplyr to obtain the result you want. Here is how I did it.
library(dplyr)
m1 <- mtcars
m2 <- mtcars %>% mutate(cyl = factor(cyl), xxxx1 = factor(cyl))
m3 <- mtcars %>% mutate(cyl = factor(cyl), xxxx2 = factor(cyl))
All.list <- list(m1,m2,m3)
##Define a function to get variable names and types
my_function <- function(data_frame){
require(dplyr)
x <- tibble(`var_name` = colnames(data_frame),
`var_type` = sapply(data_frame, class))
return(x)
}
target <- lapply(1:length(All.list),function(i)my_function(All.list[[i]]) %>%
mutate(element =i)) %>%
bind_rows() %>%
spread(element, var_type)
target
answered Nov 12 '18 at 15:23
Harro Cyranka
1,1601513
answered Nov 12 '18 at 15:23
Harro Cyranka
1,1601513
answered Nov 12 '18 at 15:23
Harro Cyranka
1,1601513
answered Nov 12 '18 at 15:23
Harro Cyranka
1,1601513
1,1601513
1
thank you - this works perfectly
– Seb
Nov 12 '18 at 18:36
add a comment |
1
thank you - this works perfectly
– Seb
Nov 12 '18 at 18:36
1
1
thank you - this works perfectly
– Seb
Nov 12 '18 at 18:36
thank you - this works perfectly
– Seb
Nov 12 '18 at 18:36
add a comment |
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