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

edited Nov 16 '18 at 18:11

Rui Barradas

18.3k51833

asked Nov 16 '18 at 18:06

user163731

4

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

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

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

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

edited Nov 16 '18 at 18:11

Rui Barradas

18.3k51833

asked Nov 16 '18 at 18:06

user163731

4

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

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

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

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

edited Nov 16 '18 at 18:11

Rui Barradas

18.3k51833

asked Nov 16 '18 at 18:06

user163731

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

user163731

edited Nov 16 '18 at 18:11

Rui Barradas

18.3k51833

asked Nov 16 '18 at 18:06

user163731

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

asked Nov 16 '18 at 18:06

user163731

asked Nov 16 '18 at 18:06

user163731

asked Nov 16 '18 at 18:06

user163731

4

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

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

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 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

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

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

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

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

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_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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1 Answer
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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_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_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_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_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

answered Nov 16 '18 at 18:47

Moody_Mudskipper

25k33673

answered Nov 16 '18 at 18:47

Moody_Mudskipper

25k33673

answered Nov 16 '18 at 18:47

Moody_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 |

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

It works!!! Thank you so so so much!

– user163731
Nov 17 '18 at 14:08

add a comment |

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