why can't subset data frame by some keys?











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0
down vote

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such data frame as follows:



df <- data.frame(yr = rep(2000:2017,each=46),

                 dayorder = rep(seq(1,365,8),time=18),

                 fid = rep(1:46,time=18),

                 Value = runif(46*18)

                 )


when I sieve by keys:



df[which(dayorder==9),]

      yr dayorder fid      Value

2   2000        9   2 0.3424053

48  2001        9   2 0.6639720

94  2002        9   2 0.5530076

140 2003        9   2 0.9757845

.....


When use 'yr' key ,report error:



df[which(yr==2001),]

Error in which(yr == 2001) : object 'yr' not found


same error by 'fid' key"



df[which(fid==2),]

Error in which(fid == 2) : object 'fid' not found


but df$yr is existed:



df$yr

  [1] 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000

 .....


What's the problem of data frame?






r







share|improve this question




























    up vote
    0
    down vote

    favorite












    such data frame as follows:



    df <- data.frame(yr = rep(2000:2017,each=46),
    
                 dayorder = rep(seq(1,365,8),time=18),
    
                 fid = rep(1:46,time=18),
    
                 Value = runif(46*18)
    
                 )


    when I sieve by keys:



    df[which(dayorder==9),]
    
      yr dayorder fid      Value
    
2   2000        9   2 0.3424053
    
48  2001        9   2 0.6639720
    
94  2002        9   2 0.5530076
    
140 2003        9   2 0.9757845
    
.....


    When use 'yr' key ,report error:



    df[which(yr==2001),]
    
Error in which(yr == 2001) : object 'yr' not found


    same error by 'fid' key"



    df[which(fid==2),]
    
Error in which(fid == 2) : object 'fid' not found


    but df$yr is existed:



    df$yr
    
  [1] 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000
    
 .....


    What's the problem of data frame?






    r







    share|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      such data frame as follows:



      df <- data.frame(yr = rep(2000:2017,each=46),
      
                 dayorder = rep(seq(1,365,8),time=18),
      
                 fid = rep(1:46,time=18),
      
                 Value = runif(46*18)
      
                 )


      when I sieve by keys:



      df[which(dayorder==9),]
      
      yr dayorder fid      Value
      
2   2000        9   2 0.3424053
      
48  2001        9   2 0.6639720
      
94  2002        9   2 0.5530076
      
140 2003        9   2 0.9757845
      
.....


      When use 'yr' key ,report error:



      df[which(yr==2001),]
      
Error in which(yr == 2001) : object 'yr' not found


      same error by 'fid' key"



      df[which(fid==2),]
      
Error in which(fid == 2) : object 'fid' not found


      but df$yr is existed:



      df$yr
      
  [1] 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000
      
 .....


      What's the problem of data frame?






      r







      share|improve this question















      such data frame as follows:



      df <- data.frame(yr = rep(2000:2017,each=46),
      
                 dayorder = rep(seq(1,365,8),time=18),
      
                 fid = rep(1:46,time=18),
      
                 Value = runif(46*18)
      
                 )


      when I sieve by keys:



      df[which(dayorder==9),]
      
      yr dayorder fid      Value
      
2   2000        9   2 0.3424053
      
48  2001        9   2 0.6639720
      
94  2002        9   2 0.5530076
      
140 2003        9   2 0.9757845
      
.....


      When use 'yr' key ,report error:



      df[which(yr==2001),]
      
Error in which(yr == 2001) : object 'yr' not found


      same error by 'fid' key"



      df[which(fid==2),]
      
Error in which(fid == 2) : object 'fid' not found


      but df$yr is existed:



      df$yr
      
  [1] 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000 2000
      
 .....


      What's the problem of data frame?






      r




      r






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 22 hours ago

























      asked 22 hours ago









      Cobin

      16211




      16211
























          2 Answers
          2






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted










          You need to add df$ before the column name, in the first case you don't get an error probably because dayorder is a variable defined in your environment.



          It should've thrown an error like the others.



          df[which(df$dayorder==9),] # or df[df$dayorder==9,]


          When I call like this df[which(df$dayorder==9),] I get:




          Error in which(dayorder == 9) : object 'dayorder' not found




          Another solution using with, avoids too many df$ (also note you can remove which():



          with(df, df[dayorder==9, ])





          share|improve this answer





















          • lol reason for the downvote? you people need to grow up.
            – RLave
            1 hour ago


















          up vote
          0
          down vote













          you can try a tidyverse as well



          library(tidyverse)
          
filter(df, yr == 2001)





          share|improve this answer





















          • or base R: subset(df, yr == 2001)
            – snoram
            22 hours ago










          • Yes of course...but IMO should be included in the other answer as a baseR solution ;)
            – Jimbou
            22 hours ago











          Your Answer






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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          You need to add df$ before the column name, in the first case you don't get an error probably because dayorder is a variable defined in your environment.



          It should've thrown an error like the others.



          df[which(df$dayorder==9),] # or df[df$dayorder==9,]


          When I call like this df[which(df$dayorder==9),] I get:




          Error in which(dayorder == 9) : object 'dayorder' not found




          Another solution using with, avoids too many df$ (also note you can remove which():



          with(df, df[dayorder==9, ])





          share|improve this answer





















          • lol reason for the downvote? you people need to grow up.
            – RLave
            1 hour ago















          up vote
          2
          down vote



          accepted










          You need to add df$ before the column name, in the first case you don't get an error probably because dayorder is a variable defined in your environment.



          It should've thrown an error like the others.



          df[which(df$dayorder==9),] # or df[df$dayorder==9,]


          When I call like this df[which(df$dayorder==9),] I get:




          Error in which(dayorder == 9) : object 'dayorder' not found




          Another solution using with, avoids too many df$ (also note you can remove which():



          with(df, df[dayorder==9, ])





          share|improve this answer





















          • lol reason for the downvote? you people need to grow up.
            – RLave
            1 hour ago













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          You need to add df$ before the column name, in the first case you don't get an error probably because dayorder is a variable defined in your environment.



          It should've thrown an error like the others.



          df[which(df$dayorder==9),] # or df[df$dayorder==9,]


          When I call like this df[which(df$dayorder==9),] I get:




          Error in which(dayorder == 9) : object 'dayorder' not found




          Another solution using with, avoids too many df$ (also note you can remove which():



          with(df, df[dayorder==9, ])





          share|improve this answer












          You need to add df$ before the column name, in the first case you don't get an error probably because dayorder is a variable defined in your environment.



          It should've thrown an error like the others.



          df[which(df$dayorder==9),] # or df[df$dayorder==9,]


          When I call like this df[which(df$dayorder==9),] I get:




          Error in which(dayorder == 9) : object 'dayorder' not found




          Another solution using with, avoids too many df$ (also note you can remove which():



          with(df, df[dayorder==9, ])






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 22 hours ago









          RLave

          2,8041820




          2,8041820












          • lol reason for the downvote? you people need to grow up.
            – RLave
            1 hour ago


















          • lol reason for the downvote? you people need to grow up.
            – RLave
            1 hour ago
















          lol reason for the downvote? you people need to grow up.
          – RLave
          1 hour ago




          lol reason for the downvote? you people need to grow up.
          – RLave
          1 hour ago












          up vote
          0
          down vote













          you can try a tidyverse as well



          library(tidyverse)
          
filter(df, yr == 2001)





          share|improve this answer





















          • or base R: subset(df, yr == 2001)
            – snoram
            22 hours ago










          • Yes of course...but IMO should be included in the other answer as a baseR solution ;)
            – Jimbou
            22 hours ago















          up vote
          0
          down vote













          you can try a tidyverse as well



          library(tidyverse)
          
filter(df, yr == 2001)





          share|improve this answer





















          • or base R: subset(df, yr == 2001)
            – snoram
            22 hours ago










          • Yes of course...but IMO should be included in the other answer as a baseR solution ;)
            – Jimbou
            22 hours ago













          up vote
          0
          down vote










          up vote
          0
          down vote









          you can try a tidyverse as well



          library(tidyverse)
          
filter(df, yr == 2001)





          share|improve this answer












          you can try a tidyverse as well



          library(tidyverse)
          
filter(df, yr == 2001)






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 22 hours ago









          Jimbou

          9,44111230




          9,44111230












          • or base R: subset(df, yr == 2001)
            – snoram
            22 hours ago










          • Yes of course...but IMO should be included in the other answer as a baseR solution ;)
            – Jimbou
            22 hours ago


















          • or base R: subset(df, yr == 2001)
            – snoram
            22 hours ago










          • Yes of course...but IMO should be included in the other answer as a baseR solution ;)
            – Jimbou
            22 hours ago
















          or base R: subset(df, yr == 2001)
          – snoram
          22 hours ago




          or base R: subset(df, yr == 2001)
          – snoram
          22 hours ago












          Yes of course...but IMO should be included in the other answer as a baseR solution ;)
          – Jimbou
          22 hours ago




          Yes of course...but IMO should be included in the other answer as a baseR solution ;)
          – Jimbou
          22 hours ago


















           

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