R: Splitting dataframe columnwise
I have a dataframe with the this structure
x = data.frame(let = letters, LET = LETTERS, num1 = 1:26, num2 = 21:46, num3 = 71:96, num4 = 68:93 )
I want to split it into list of 3 columns dataframes.The first two columns let and LET remains common, the third column varies. The first dataframe would be (let, LET, num1) and the second one would be (let, LET, num2) and so on so forth.
My current strategy is to convert the dataframe into long format and split it based on the num using plyr and dplyr packages. Is there an easier way to accomplish this task.
r split dataframe subset
edited Nov 16 '18 at 4:37
Cœur
19.1k9114155
asked Mar 12 '16 at 7:21
ArihantArihant
321618
add a comment |
I have a dataframe with the this structure
x = data.frame(let = letters, LET = LETTERS, num1 = 1:26, num2 = 21:46, num3 = 71:96, num4 = 68:93 )
I want to split it into list of 3 columns dataframes.The first two columns let and LET remains common, the third column varies. The first dataframe would be (let, LET, num1) and the second one would be (let, LET, num2) and so on so forth.
My current strategy is to convert the dataframe into long format and split it based on the num using plyr and dplyr packages. Is there an easier way to accomplish this task.
r split dataframe subset
edited Nov 16 '18 at 4:37
Cœur
19.1k9114155
asked Mar 12 '16 at 7:21
ArihantArihant
321618
add a comment |
I have a dataframe with the this structure
x = data.frame(let = letters, LET = LETTERS, num1 = 1:26, num2 = 21:46, num3 = 71:96, num4 = 68:93 )
I want to split it into list of 3 columns dataframes.The first two columns let and LET remains common, the third column varies. The first dataframe would be (let, LET, num1) and the second one would be (let, LET, num2) and so on so forth.
My current strategy is to convert the dataframe into long format and split it based on the num using plyr and dplyr packages. Is there an easier way to accomplish this task.
r split dataframe subset
edited Nov 16 '18 at 4:37
Cœur
19.1k9114155
asked Mar 12 '16 at 7:21
ArihantArihant
321618
I have a dataframe with the this structure
x = data.frame(let = letters, LET = LETTERS, num1 = 1:26, num2 = 21:46, num3 = 71:96, num4 = 68:93 )
I want to split it into list of 3 columns dataframes.The first two columns let and LET remains common, the third column varies. The first dataframe would be (let, LET, num1) and the second one would be (let, LET, num2) and so on so forth.
My current strategy is to convert the dataframe into long format and split it based on the num using plyr and dplyr packages. Is there an easier way to accomplish this task.
r split dataframe subset
r split dataframe subset
edited Nov 16 '18 at 4:37
Cœur
19.1k9114155
asked Mar 12 '16 at 7:21
ArihantArihant
321618
edited Nov 16 '18 at 4:37
Cœur
19.1k9114155
asked Mar 12 '16 at 7:21
ArihantArihant
321618
edited Nov 16 '18 at 4:37
Cœur
19.1k9114155
edited Nov 16 '18 at 4:37
Cœur
19.1k9114155
edited Nov 16 '18 at 4:37
Cœur
19.1k9114155
19.1k9114155
asked Mar 12 '16 at 7:21
ArihantArihant
321618
asked Mar 12 '16 at 7:21
ArihantArihant
321618
asked Mar 12 '16 at 7:21
ArihantArihant
321618
321618
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You can use lapply like so for your example
lapply(1:4, function(D) x[ ,c("let", "LET", paste0("num", D))])
If you don't know the column names of the num* columns you could use
nonLetNames <- names(x)[!(names(x) %in% c("let", "LET"))]
lapply(nonLetNames, function(nom) x[ ,c("let", "LET", nom)])
answered Mar 12 '16 at 7:25
HughHugh
7,87543669
add a comment |
Here is a Map based approach
Map(function(x,y,z) setNames(cbind(x,y), c(names(x), z)),
list(x[1:2]), x[-(1:2)], names(x)[-(1:2)])
answered Mar 12 '16 at 7:39
akrunakrun
416k13205278
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use lapply like so for your example
lapply(1:4, function(D) x[ ,c("let", "LET", paste0("num", D))])
If you don't know the column names of the num* columns you could use
nonLetNames <- names(x)[!(names(x) %in% c("let", "LET"))]
lapply(nonLetNames, function(nom) x[ ,c("let", "LET", nom)])
answered Mar 12 '16 at 7:25
HughHugh
7,87543669
add a comment |
You can use lapply like so for your example
lapply(1:4, function(D) x[ ,c("let", "LET", paste0("num", D))])
If you don't know the column names of the num* columns you could use
nonLetNames <- names(x)[!(names(x) %in% c("let", "LET"))]
lapply(nonLetNames, function(nom) x[ ,c("let", "LET", nom)])
answered Mar 12 '16 at 7:25
HughHugh
7,87543669
add a comment |
You can use lapply like so for your example
lapply(1:4, function(D) x[ ,c("let", "LET", paste0("num", D))])
If you don't know the column names of the num* columns you could use
nonLetNames <- names(x)[!(names(x) %in% c("let", "LET"))]
lapply(nonLetNames, function(nom) x[ ,c("let", "LET", nom)])
answered Mar 12 '16 at 7:25
HughHugh
7,87543669
You can use lapply like so for your example
lapply(1:4, function(D) x[ ,c("let", "LET", paste0("num", D))])
If you don't know the column names of the num* columns you could use
nonLetNames <- names(x)[!(names(x) %in% c("let", "LET"))]
lapply(nonLetNames, function(nom) x[ ,c("let", "LET", nom)])
answered Mar 12 '16 at 7:25
HughHugh
7,87543669
answered Mar 12 '16 at 7:25
HughHugh
7,87543669
answered Mar 12 '16 at 7:25
HughHugh
7,87543669
answered Mar 12 '16 at 7:25
HughHugh
7,87543669
7,87543669
add a comment |
add a comment |
Here is a Map based approach
Map(function(x,y,z) setNames(cbind(x,y), c(names(x), z)),
list(x[1:2]), x[-(1:2)], names(x)[-(1:2)])
answered Mar 12 '16 at 7:39
akrunakrun
416k13205278
add a comment |
Here is a Map based approach
Map(function(x,y,z) setNames(cbind(x,y), c(names(x), z)),
list(x[1:2]), x[-(1:2)], names(x)[-(1:2)])
answered Mar 12 '16 at 7:39
akrunakrun
416k13205278
add a comment |
Here is a Map based approach
Map(function(x,y,z) setNames(cbind(x,y), c(names(x), z)),
list(x[1:2]), x[-(1:2)], names(x)[-(1:2)])
answered Mar 12 '16 at 7:39
akrunakrun
416k13205278
Here is a Map based approach
Map(function(x,y,z) setNames(cbind(x,y), c(names(x), z)),
list(x[1:2]), x[-(1:2)], names(x)[-(1:2)])
answered Mar 12 '16 at 7:39
akrunakrun
416k13205278
answered Mar 12 '16 at 7:39
akrunakrun
416k13205278
answered Mar 12 '16 at 7:39
akrunakrun
416k13205278
answered Mar 12 '16 at 7:39
akrunakrun
416k13205278
416k13205278
add a comment |
add a comment |
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