Bulk rename files from csv files r











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I am struggling to find a clear answer to by problem.



I have a dataframe that has a file path and a filename.



data.frame': 6 obs. of 2 variables: $ Path : Factor w/ 97 levels "J:\GBD2017\China\Splitweekly\281_2010_10.pdf",..: 1 2 3 4 5 6 $ Filename: Factor w/ 97 levels "1000095452.pdf",..: 97 1 2 3 4 5



I want to rename each filepath J:\GBD2017\China\Splitweekly\281_2010_10.pdf so the last xx_xx_xx.pdf is replaced with the value in the filename column.



eg J:\GBD2017\China\Splitweekly\281_2010_10.pdf would become J:\GBD2017\China\Splitweekly\1000095452.pdf.



I am going to have hundreds of files so any method needs to be efficient.



Dput:



structure(list(Path = structure(1:6, .Label = c("J:\GBD2017\China    \Splitweekly\281_2010_10.pdf", 
`"J:\GBD2017\China\Splitweekly\282_2010_11.pdf", "J:\GBD2017\China \Splitweekly\283_2010_12.pdf"), class = "factor"), Filename = `structure(c(97L, 1L, 2L, 3L, L, 5L), .Label = c("1000095452.pdf","1000095453.pdf", "1000095454.pdf"), class = "factor")), .Names = c("Path", "Filename"), row.names = c(NA, 6L), class = "data.frame")









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  • ?dirname ; ?basename ; ?file.path
    – hrbrmstr
    Nov 14 at 14:08










  • Im not sure what you are saying
    – Bohnston
    Nov 14 at 14:12










  • I'm suggesting you read the documentation for those three functions as they're all you'll need to accomplish your goal.
    – hrbrmstr
    Nov 14 at 14:22















up vote
0
down vote

favorite












I am struggling to find a clear answer to by problem.



I have a dataframe that has a file path and a filename.



data.frame': 6 obs. of 2 variables: $ Path : Factor w/ 97 levels "J:\GBD2017\China\Splitweekly\281_2010_10.pdf",..: 1 2 3 4 5 6 $ Filename: Factor w/ 97 levels "1000095452.pdf",..: 97 1 2 3 4 5



I want to rename each filepath J:\GBD2017\China\Splitweekly\281_2010_10.pdf so the last xx_xx_xx.pdf is replaced with the value in the filename column.



eg J:\GBD2017\China\Splitweekly\281_2010_10.pdf would become J:\GBD2017\China\Splitweekly\1000095452.pdf.



I am going to have hundreds of files so any method needs to be efficient.



Dput:



structure(list(Path = structure(1:6, .Label = c("J:\GBD2017\China    \Splitweekly\281_2010_10.pdf", 
`"J:\GBD2017\China\Splitweekly\282_2010_11.pdf", "J:\GBD2017\China \Splitweekly\283_2010_12.pdf"), class = "factor"), Filename = `structure(c(97L, 1L, 2L, 3L, L, 5L), .Label = c("1000095452.pdf","1000095453.pdf", "1000095454.pdf"), class = "factor")), .Names = c("Path", "Filename"), row.names = c(NA, 6L), class = "data.frame")









share|improve this question






















  • ?dirname ; ?basename ; ?file.path
    – hrbrmstr
    Nov 14 at 14:08










  • Im not sure what you are saying
    – Bohnston
    Nov 14 at 14:12










  • I'm suggesting you read the documentation for those three functions as they're all you'll need to accomplish your goal.
    – hrbrmstr
    Nov 14 at 14:22













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am struggling to find a clear answer to by problem.



I have a dataframe that has a file path and a filename.



data.frame': 6 obs. of 2 variables: $ Path : Factor w/ 97 levels "J:\GBD2017\China\Splitweekly\281_2010_10.pdf",..: 1 2 3 4 5 6 $ Filename: Factor w/ 97 levels "1000095452.pdf",..: 97 1 2 3 4 5



I want to rename each filepath J:\GBD2017\China\Splitweekly\281_2010_10.pdf so the last xx_xx_xx.pdf is replaced with the value in the filename column.



eg J:\GBD2017\China\Splitweekly\281_2010_10.pdf would become J:\GBD2017\China\Splitweekly\1000095452.pdf.



I am going to have hundreds of files so any method needs to be efficient.



Dput:



structure(list(Path = structure(1:6, .Label = c("J:\GBD2017\China    \Splitweekly\281_2010_10.pdf", 
`"J:\GBD2017\China\Splitweekly\282_2010_11.pdf", "J:\GBD2017\China \Splitweekly\283_2010_12.pdf"), class = "factor"), Filename = `structure(c(97L, 1L, 2L, 3L, L, 5L), .Label = c("1000095452.pdf","1000095453.pdf", "1000095454.pdf"), class = "factor")), .Names = c("Path", "Filename"), row.names = c(NA, 6L), class = "data.frame")









share|improve this question













I am struggling to find a clear answer to by problem.



I have a dataframe that has a file path and a filename.



data.frame': 6 obs. of 2 variables: $ Path : Factor w/ 97 levels "J:\GBD2017\China\Splitweekly\281_2010_10.pdf",..: 1 2 3 4 5 6 $ Filename: Factor w/ 97 levels "1000095452.pdf",..: 97 1 2 3 4 5



I want to rename each filepath J:\GBD2017\China\Splitweekly\281_2010_10.pdf so the last xx_xx_xx.pdf is replaced with the value in the filename column.



eg J:\GBD2017\China\Splitweekly\281_2010_10.pdf would become J:\GBD2017\China\Splitweekly\1000095452.pdf.



I am going to have hundreds of files so any method needs to be efficient.



Dput:



structure(list(Path = structure(1:6, .Label = c("J:\GBD2017\China    \Splitweekly\281_2010_10.pdf", 
`"J:\GBD2017\China\Splitweekly\282_2010_11.pdf", "J:\GBD2017\China \Splitweekly\283_2010_12.pdf"), class = "factor"), Filename = `structure(c(97L, 1L, 2L, 3L, L, 5L), .Label = c("1000095452.pdf","1000095453.pdf", "1000095454.pdf"), class = "factor")), .Names = c("Path", "Filename"), row.names = c(NA, 6L), class = "data.frame")






r file bulk renaming






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share|improve this question










asked Nov 14 at 14:02









Bohnston

277




277












  • ?dirname ; ?basename ; ?file.path
    – hrbrmstr
    Nov 14 at 14:08










  • Im not sure what you are saying
    – Bohnston
    Nov 14 at 14:12










  • I'm suggesting you read the documentation for those three functions as they're all you'll need to accomplish your goal.
    – hrbrmstr
    Nov 14 at 14:22


















  • ?dirname ; ?basename ; ?file.path
    – hrbrmstr
    Nov 14 at 14:08










  • Im not sure what you are saying
    – Bohnston
    Nov 14 at 14:12










  • I'm suggesting you read the documentation for those three functions as they're all you'll need to accomplish your goal.
    – hrbrmstr
    Nov 14 at 14:22
















?dirname ; ?basename ; ?file.path
– hrbrmstr
Nov 14 at 14:08




?dirname ; ?basename ; ?file.path
– hrbrmstr
Nov 14 at 14:08












Im not sure what you are saying
– Bohnston
Nov 14 at 14:12




Im not sure what you are saying
– Bohnston
Nov 14 at 14:12












I'm suggesting you read the documentation for those three functions as they're all you'll need to accomplish your goal.
– hrbrmstr
Nov 14 at 14:22




I'm suggesting you read the documentation for those three functions as they're all you'll need to accomplish your goal.
– hrbrmstr
Nov 14 at 14:22

















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