function works on one element of list, doesn't work on full list, R
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I'm trying to make a function to add tags to a data frame.
The name of the data frame contains the information I need (date, selection, treatment, etc...). So I made a function that extracts the information I need. I have a large list containing all the data frames, and when I apply the function to the list, it does create the new columns for the tags but the values are NA-s. Every data frame has the same name structure, and if I extract a data frame from the list and run the function it works. Can you help me find out why doesn't it work when I apply it to the list?
here is my function :
library(stringr)
tagging <- function(H){
namey<-deparse(substitute(H)) #get the name of the data frame
namey<- str_sub(namey,1, -5) #drop the .csv
H$date<-substring(namey,1, 6) # get the first 6 characters containing the date
H$selection<- word(namey, -1) #get the last word
H$treatment<- word(namey, -2) # get the second last word
H$REP<- word(namey, -3) # get the third last word
return(H)
}
And I apply it like this
ListofData.tagged<-lapply(ListofData, tagging)
The name of the data frames looks like this:
180503 xyz1-6 R4_A6_xyz 5 yes.csv
r list function user-defined-functions lapply
edited Nov 16 '18 at 18:11
Rui Barradas
18.3k51833
asked Nov 16 '18 at 18:06
user163731user163731
61
|
show 1 more comment
I'm trying to make a function to add tags to a data frame.
The name of the data frame contains the information I need (date, selection, treatment, etc...). So I made a function that extracts the information I need. I have a large list containing all the data frames, and when I apply the function to the list, it does create the new columns for the tags but the values are NA-s. Every data frame has the same name structure, and if I extract a data frame from the list and run the function it works. Can you help me find out why doesn't it work when I apply it to the list?
here is my function :
library(stringr)
tagging <- function(H){
namey<-deparse(substitute(H)) #get the name of the data frame
namey<- str_sub(namey,1, -5) #drop the .csv
H$date<-substring(namey,1, 6) # get the first 6 characters containing the date
H$selection<- word(namey, -1) #get the last word
H$treatment<- word(namey, -2) # get the second last word
H$REP<- word(namey, -3) # get the third last word
return(H)
}
And I apply it like this
ListofData.tagged<-lapply(ListofData, tagging)
The name of the data frames looks like this:
180503 xyz1-6 R4_A6_xyz 5 yes.csv
r list function user-defined-functions lapply
edited Nov 16 '18 at 18:11
Rui Barradas
18.3k51833
asked Nov 16 '18 at 18:06
user163731user163731
61
4
Could you prove a minimal example of whatListofDatacontains?
– Anders Ellern Bilgrau
Nov 16 '18 at 18:08
Looks like it is the file name instead of the object name. I think you need to loop through the names of theListofData?
– akrun
Nov 16 '18 at 18:10
Trymap(names(ListofDatalst), ~ str_sub(.x, 1, -5))
– akrun
Nov 16 '18 at 18:15
The problem is alapplyproblem: it passesX[[1]], thenX[[2]], etc to the function. Putcat("namey:", namey, "n")right afterdeparse/substituteto see what is the df name in the function.
– Rui Barradas
Nov 16 '18 at 18:17
Anders Ellern Bilgrau : ListofData contains 300 data frames. Each dataframe has the same structure, same variables and the same "structure" for the name. akrun : the name of the file is the name of the object, because I import the .csv files as follows : temp = list.files(pattern="*.csv") ListofData = lapply(temp, read.csv)
– user163731
Nov 16 '18 at 18:19
|
show 1 more comment
I'm trying to make a function to add tags to a data frame.
The name of the data frame contains the information I need (date, selection, treatment, etc...). So I made a function that extracts the information I need. I have a large list containing all the data frames, and when I apply the function to the list, it does create the new columns for the tags but the values are NA-s. Every data frame has the same name structure, and if I extract a data frame from the list and run the function it works. Can you help me find out why doesn't it work when I apply it to the list?
here is my function :
library(stringr)
tagging <- function(H){
namey<-deparse(substitute(H)) #get the name of the data frame
namey<- str_sub(namey,1, -5) #drop the .csv
H$date<-substring(namey,1, 6) # get the first 6 characters containing the date
H$selection<- word(namey, -1) #get the last word
H$treatment<- word(namey, -2) # get the second last word
H$REP<- word(namey, -3) # get the third last word
return(H)
}
And I apply it like this
ListofData.tagged<-lapply(ListofData, tagging)
The name of the data frames looks like this:
180503 xyz1-6 R4_A6_xyz 5 yes.csv
r list function user-defined-functions lapply
edited Nov 16 '18 at 18:11
Rui Barradas
18.3k51833
asked Nov 16 '18 at 18:06
user163731user163731
61
I'm trying to make a function to add tags to a data frame.
The name of the data frame contains the information I need (date, selection, treatment, etc...). So I made a function that extracts the information I need. I have a large list containing all the data frames, and when I apply the function to the list, it does create the new columns for the tags but the values are NA-s. Every data frame has the same name structure, and if I extract a data frame from the list and run the function it works. Can you help me find out why doesn't it work when I apply it to the list?
here is my function :
library(stringr)
tagging <- function(H){
namey<-deparse(substitute(H)) #get the name of the data frame
namey<- str_sub(namey,1, -5) #drop the .csv
H$date<-substring(namey,1, 6) # get the first 6 characters containing the date
H$selection<- word(namey, -1) #get the last word
H$treatment<- word(namey, -2) # get the second last word
H$REP<- word(namey, -3) # get the third last word
return(H)
}
And I apply it like this
ListofData.tagged<-lapply(ListofData, tagging)
The name of the data frames looks like this:
180503 xyz1-6 R4_A6_xyz 5 yes.csv
r list function user-defined-functions lapply
r list function user-defined-functions lapply
edited Nov 16 '18 at 18:11
Rui Barradas
18.3k51833
asked Nov 16 '18 at 18:06
user163731user163731
61
edited Nov 16 '18 at 18:11
Rui Barradas
18.3k51833
asked Nov 16 '18 at 18:06
user163731user163731
61
edited Nov 16 '18 at 18:11
Rui Barradas
18.3k51833
edited Nov 16 '18 at 18:11
Rui Barradas
18.3k51833
edited Nov 16 '18 at 18:11
Rui Barradas
18.3k51833
18.3k51833
asked Nov 16 '18 at 18:06
user163731user163731
61
asked Nov 16 '18 at 18:06
user163731user163731
61
asked Nov 16 '18 at 18:06
user163731user163731
61
61
4
Could you prove a minimal example of whatListofDatacontains?
– Anders Ellern Bilgrau
Nov 16 '18 at 18:08
Looks like it is the file name instead of the object name. I think you need to loop through the names of theListofData?
– akrun
Nov 16 '18 at 18:10
Trymap(names(ListofDatalst), ~ str_sub(.x, 1, -5))
– akrun
Nov 16 '18 at 18:15
The problem is alapplyproblem: it passesX[[1]], thenX[[2]], etc to the function. Putcat("namey:", namey, "n")right afterdeparse/substituteto see what is the df name in the function.
– Rui Barradas
Nov 16 '18 at 18:17
Anders Ellern Bilgrau : ListofData contains 300 data frames. Each dataframe has the same structure, same variables and the same "structure" for the name. akrun : the name of the file is the name of the object, because I import the .csv files as follows : temp = list.files(pattern="*.csv") ListofData = lapply(temp, read.csv)
– user163731
Nov 16 '18 at 18:19
|
show 1 more comment
4
Could you prove a minimal example of whatListofDatacontains?
– Anders Ellern Bilgrau
Nov 16 '18 at 18:08
Looks like it is the file name instead of the object name. I think you need to loop through the names of theListofData?
– akrun
Nov 16 '18 at 18:10
Trymap(names(ListofDatalst), ~ str_sub(.x, 1, -5))
– akrun
Nov 16 '18 at 18:15
The problem is alapplyproblem: it passesX[[1]], thenX[[2]], etc to the function. Putcat("namey:", namey, "n")right afterdeparse/substituteto see what is the df name in the function.
– Rui Barradas
Nov 16 '18 at 18:17
Anders Ellern Bilgrau : ListofData contains 300 data frames. Each dataframe has the same structure, same variables and the same "structure" for the name. akrun : the name of the file is the name of the object, because I import the .csv files as follows : temp = list.files(pattern="*.csv") ListofData = lapply(temp, read.csv)
– user163731
Nov 16 '18 at 18:19
4
4
Could you prove a minimal example of what
ListofData contains?– Anders Ellern Bilgrau
Nov 16 '18 at 18:08
Could you prove a minimal example of what
ListofData contains?– Anders Ellern Bilgrau
Nov 16 '18 at 18:08
Looks like it is the file name instead of the object name. I think you need to loop through the names of the
ListofData?– akrun
Nov 16 '18 at 18:10
Looks like it is the file name instead of the object name. I think you need to loop through the names of the
ListofData?– akrun
Nov 16 '18 at 18:10
Try
map(names(ListofDatalst), ~ str_sub(.x, 1, -5))– akrun
Nov 16 '18 at 18:15
Try
map(names(ListofDatalst), ~ str_sub(.x, 1, -5))– akrun
Nov 16 '18 at 18:15
The problem is a
lapply problem: it passes X[[1]], then X[[2]], etc to the function. Put cat("namey:", namey, "n") right after deparse/substitute to see what is the df name in the function.– Rui Barradas
Nov 16 '18 at 18:17
The problem is a
lapply problem: it passes X[[1]], then X[[2]], etc to the function. Put cat("namey:", namey, "n") right after deparse/substitute to see what is the df name in the function.– Rui Barradas
Nov 16 '18 at 18:17
Anders Ellern Bilgrau : ListofData contains 300 data frames. Each dataframe has the same structure, same variables and the same "structure" for the name. akrun : the name of the file is the name of the object, because I import the .csv files as follows : temp = list.files(pattern="*.csv") ListofData = lapply(temp, read.csv)
– user163731
Nov 16 '18 at 18:19
Anders Ellern Bilgrau : ListofData contains 300 data frames. Each dataframe has the same structure, same variables and the same "structure" for the name. akrun : the name of the file is the name of the object, because I import the .csv files as follows : temp = list.files(pattern="*.csv") ListofData = lapply(temp, read.csv)
– user163731
Nov 16 '18 at 18:19
|
show 1 more comment
1 Answer
1
active
oldest
votes
Import this way so you keep your names:
library(tidyverse)
ListofData <- map(set_names(temp), read_csv)
then we modify your function to add the name as a second parameter, which will be used through imap, so we get rid of the first line too :
tagging <- function(H, namey){
namey<- str_sub(namey,1, -5) #drop the .csv
H$date<-substring(namey,1, 6) # get the first 6 characters containing the date
H$selection<- word(namey, -1) #get the last word
H$treatment<- word(namey, -2) # get the second last word
H$REP<- word(namey, -3) # get the third last word
return(H)
}
Then imap passes the data to the H argument and the element names to the namey argument.
ListofData.tagged <- imap(ListofData, tagging)
base R translation
ListofData <- lapply(setNames(temp,temp), read.csv)
ListofData.tagged <- Map(tagging, ListofData, names(ListofData))
Or if you don't care about naming the elements of ListofData you can do directly Map(tagging, ListofData, temp) (still keeping the new definition of tagging).
edited Nov 16 '18 at 18:55
Rui Barradas
18.3k51833
answered Nov 16 '18 at 18:47
Moody_MudskipperMoody_Mudskipper
25k33673
thanks @rui-barradas
– Moody_Mudskipper
Nov 16 '18 at 18:57
It works!!! Thank you so so so much!
– user163731
Nov 17 '18 at 14:08
add a comment |
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Import this way so you keep your names:
library(tidyverse)
ListofData <- map(set_names(temp), read_csv)
then we modify your function to add the name as a second parameter, which will be used through imap, so we get rid of the first line too :
tagging <- function(H, namey){
namey<- str_sub(namey,1, -5) #drop the .csv
H$date<-substring(namey,1, 6) # get the first 6 characters containing the date
H$selection<- word(namey, -1) #get the last word
H$treatment<- word(namey, -2) # get the second last word
H$REP<- word(namey, -3) # get the third last word
return(H)
}
Then imap passes the data to the H argument and the element names to the namey argument.
ListofData.tagged <- imap(ListofData, tagging)
base R translation
ListofData <- lapply(setNames(temp,temp), read.csv)
ListofData.tagged <- Map(tagging, ListofData, names(ListofData))
Or if you don't care about naming the elements of ListofData you can do directly Map(tagging, ListofData, temp) (still keeping the new definition of tagging).
edited Nov 16 '18 at 18:55
Rui Barradas
18.3k51833
answered Nov 16 '18 at 18:47
Moody_MudskipperMoody_Mudskipper
25k33673
thanks @rui-barradas
– Moody_Mudskipper
Nov 16 '18 at 18:57
It works!!! Thank you so so so much!
– user163731
Nov 17 '18 at 14:08
add a comment |
Import this way so you keep your names:
library(tidyverse)
ListofData <- map(set_names(temp), read_csv)
then we modify your function to add the name as a second parameter, which will be used through imap, so we get rid of the first line too :
tagging <- function(H, namey){
namey<- str_sub(namey,1, -5) #drop the .csv
H$date<-substring(namey,1, 6) # get the first 6 characters containing the date
H$selection<- word(namey, -1) #get the last word
H$treatment<- word(namey, -2) # get the second last word
H$REP<- word(namey, -3) # get the third last word
return(H)
}
Then imap passes the data to the H argument and the element names to the namey argument.
ListofData.tagged <- imap(ListofData, tagging)
base R translation
ListofData <- lapply(setNames(temp,temp), read.csv)
ListofData.tagged <- Map(tagging, ListofData, names(ListofData))
Or if you don't care about naming the elements of ListofData you can do directly Map(tagging, ListofData, temp) (still keeping the new definition of tagging).
edited Nov 16 '18 at 18:55
Rui Barradas
18.3k51833
answered Nov 16 '18 at 18:47
Moody_MudskipperMoody_Mudskipper
25k33673
thanks @rui-barradas
– Moody_Mudskipper
Nov 16 '18 at 18:57
It works!!! Thank you so so so much!
– user163731
Nov 17 '18 at 14:08
add a comment |
Import this way so you keep your names:
library(tidyverse)
ListofData <- map(set_names(temp), read_csv)
then we modify your function to add the name as a second parameter, which will be used through imap, so we get rid of the first line too :
tagging <- function(H, namey){
namey<- str_sub(namey,1, -5) #drop the .csv
H$date<-substring(namey,1, 6) # get the first 6 characters containing the date
H$selection<- word(namey, -1) #get the last word
H$treatment<- word(namey, -2) # get the second last word
H$REP<- word(namey, -3) # get the third last word
return(H)
}
Then imap passes the data to the H argument and the element names to the namey argument.
ListofData.tagged <- imap(ListofData, tagging)
base R translation
ListofData <- lapply(setNames(temp,temp), read.csv)
ListofData.tagged <- Map(tagging, ListofData, names(ListofData))
Or if you don't care about naming the elements of ListofData you can do directly Map(tagging, ListofData, temp) (still keeping the new definition of tagging).
edited Nov 16 '18 at 18:55
Rui Barradas
18.3k51833
answered Nov 16 '18 at 18:47
Moody_MudskipperMoody_Mudskipper
25k33673
Import this way so you keep your names:
library(tidyverse)
ListofData <- map(set_names(temp), read_csv)
then we modify your function to add the name as a second parameter, which will be used through imap, so we get rid of the first line too :
tagging <- function(H, namey){
namey<- str_sub(namey,1, -5) #drop the .csv
H$date<-substring(namey,1, 6) # get the first 6 characters containing the date
H$selection<- word(namey, -1) #get the last word
H$treatment<- word(namey, -2) # get the second last word
H$REP<- word(namey, -3) # get the third last word
return(H)
}
Then imap passes the data to the H argument and the element names to the namey argument.
ListofData.tagged <- imap(ListofData, tagging)
base R translation
ListofData <- lapply(setNames(temp,temp), read.csv)
ListofData.tagged <- Map(tagging, ListofData, names(ListofData))
Or if you don't care about naming the elements of ListofData you can do directly Map(tagging, ListofData, temp) (still keeping the new definition of tagging).
edited Nov 16 '18 at 18:55
Rui Barradas
18.3k51833
answered Nov 16 '18 at 18:47
Moody_MudskipperMoody_Mudskipper
25k33673
edited Nov 16 '18 at 18:55
Rui Barradas
18.3k51833
edited Nov 16 '18 at 18:55
Rui Barradas
18.3k51833
edited Nov 16 '18 at 18:55
Rui Barradas
18.3k51833
18.3k51833
answered Nov 16 '18 at 18:47
Moody_MudskipperMoody_Mudskipper
25k33673
answered Nov 16 '18 at 18:47
Moody_MudskipperMoody_Mudskipper
25k33673
answered Nov 16 '18 at 18:47
Moody_MudskipperMoody_Mudskipper
25k33673
25k33673
thanks @rui-barradas
– Moody_Mudskipper
Nov 16 '18 at 18:57
It works!!! Thank you so so so much!
– user163731
Nov 17 '18 at 14:08
add a comment |
thanks @rui-barradas
– Moody_Mudskipper
Nov 16 '18 at 18:57
It works!!! Thank you so so so much!
– user163731
Nov 17 '18 at 14:08
thanks @rui-barradas
– Moody_Mudskipper
Nov 16 '18 at 18:57
thanks @rui-barradas
– Moody_Mudskipper
Nov 16 '18 at 18:57
It works!!! Thank you so so so much!
– user163731
Nov 17 '18 at 14:08
It works!!! Thank you so so so much!
– user163731
Nov 17 '18 at 14:08
add a comment |
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4
Could you prove a minimal example of what
ListofDatacontains?– Anders Ellern Bilgrau
Nov 16 '18 at 18:08
Looks like it is the file name instead of the object name. I think you need to loop through the names of the
ListofData?– akrun
Nov 16 '18 at 18:10
Try
map(names(ListofDatalst), ~ str_sub(.x, 1, -5))– akrun
Nov 16 '18 at 18:15
The problem is a
lapplyproblem: it passesX[[1]], thenX[[2]], etc to the function. Putcat("namey:", namey, "n")right afterdeparse/substituteto see what is the df name in the function.– Rui Barradas
Nov 16 '18 at 18:17
Anders Ellern Bilgrau : ListofData contains 300 data frames. Each dataframe has the same structure, same variables and the same "structure" for the name. akrun : the name of the file is the name of the object, because I import the .csv files as follows : temp = list.files(pattern="*.csv") ListofData = lapply(temp, read.csv)
– user163731
Nov 16 '18 at 18:19