why can't subset data frame by some keys?











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












    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






          StackExchange.ifUsing("editor", function () {
          StackExchange.using("externalEditor", function () {
          StackExchange.using("snippets", function () {
          StackExchange.snippets.init();
          });
          });
          }, "code-snippets");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "1"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














           

          draft saved


          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53372098%2fwhy-cant-subset-data-frame-by-some-keys%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          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


















           

          draft saved


          draft discarded



















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53372098%2fwhy-cant-subset-data-frame-by-some-keys%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          'app-layout' is not a known element: how to share Component with different Modules

          Image
          0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
          Read more

          android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
          Read more

          WPF add header to Image with URL pettitions [duplicate]

          Image
          up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
          Read more