How to create a new column based on two other columns in Pandas?












0












$begingroup$


I am searching for a way to create a new column in my data. I have tried using iterows() but found it extremely time consuming in my dataset containing 40 lakh rows. So here is what I want.
Consider I have 2 columns:
Event ID, TeamID ,I want to find the no. of unique TeamID under each EventID as a new column. In other words, I want to find the number of teams participating in each event as a new column.






machine-learning python pandas dataframe







share|improve this question









$endgroup$












  • $begingroup$
    Something like a groupby?
    $endgroup$
    – Matthieu Brucher
    Jan 3 at 16:59
















0












$begingroup$


I am searching for a way to create a new column in my data. I have tried using iterows() but found it extremely time consuming in my dataset containing 40 lakh rows. So here is what I want.
Consider I have 2 columns:
Event ID, TeamID ,I want to find the no. of unique TeamID under each EventID as a new column. In other words, I want to find the number of teams participating in each event as a new column.






machine-learning python pandas dataframe







share|improve this question









$endgroup$












  • $begingroup$
    Something like a groupby?
    $endgroup$
    – Matthieu Brucher
    Jan 3 at 16:59














0












0








0





$begingroup$


I am searching for a way to create a new column in my data. I have tried using iterows() but found it extremely time consuming in my dataset containing 40 lakh rows. So here is what I want.
Consider I have 2 columns:
Event ID, TeamID ,I want to find the no. of unique TeamID under each EventID as a new column. In other words, I want to find the number of teams participating in each event as a new column.






machine-learning python pandas dataframe







share|improve this question









$endgroup$




I am searching for a way to create a new column in my data. I have tried using iterows() but found it extremely time consuming in my dataset containing 40 lakh rows. So here is what I want.
Consider I have 2 columns:
Event ID, TeamID ,I want to find the no. of unique TeamID under each EventID as a new column. In other words, I want to find the number of teams participating in each event as a new column.






machine-learning python pandas dataframe




machine-learning python pandas dataframe






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jan 3 at 16:56









Arjun ChandraArjun Chandra

31




31












  • $begingroup$
    Something like a groupby?
    $endgroup$
    – Matthieu Brucher
    Jan 3 at 16:59


















  • $begingroup$
    Something like a groupby?
    $endgroup$
    – Matthieu Brucher
    Jan 3 at 16:59
















$begingroup$
Something like a groupby?
$endgroup$
– Matthieu Brucher
Jan 3 at 16:59




$begingroup$
Something like a groupby?
$endgroup$
– Matthieu Brucher
Jan 3 at 16:59










2 Answers
2






active

oldest

votes


















1












$begingroup$

You can try something like this to get a new dataframe that has pairs of (EventID, TeamCount):



event_id_team_count = data.groupby('EventID').agg({'TeamID': lambda x: x.nunique()})

event_id_team_count.rename(columns={"TeamID": "TeamCount"}, inplace=True)


If you want to have this new column in the original dataframe, all you need to do is to join the original dataframe with the one you have just created:



data = data.join(other=event_id_team_count, on="EventID")





share|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot... It worked!!!
    $endgroup$
    – Arjun Chandra
    Jan 4 at 6:25



















0












$begingroup$


  1. Create a dictionary with the unique count of TeamID with respective to EventID


uCountDict = dict(data.groupby("EventID").TeamID.count())

uCountDict


Sample output
{'A': 4,
'C': 3,
'D': 2,
'F': 1
}




  1. Now create a new column with unique count with respective to TeamID using apply function


data["TeamCount"] = data.EventID.apply(lambda x : uCountDict[x])





share|improve this answer








New contributor




Noor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "557"
    };
    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',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    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%2fdatascience.stackexchange.com%2fquestions%2f43439%2fhow-to-create-a-new-column-based-on-two-other-columns-in-pandas%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









    1












    $begingroup$

    You can try something like this to get a new dataframe that has pairs of (EventID, TeamCount):



    event_id_team_count = data.groupby('EventID').agg({'TeamID': lambda x: x.nunique()})
    
event_id_team_count.rename(columns={"TeamID": "TeamCount"}, inplace=True)


    If you want to have this new column in the original dataframe, all you need to do is to join the original dataframe with the one you have just created:



    data = data.join(other=event_id_team_count, on="EventID")





    share|improve this answer









    $endgroup$













    • $begingroup$
      Thanks a lot... It worked!!!
      $endgroup$
      – Arjun Chandra
      Jan 4 at 6:25
















    1












    $begingroup$

    You can try something like this to get a new dataframe that has pairs of (EventID, TeamCount):



    event_id_team_count = data.groupby('EventID').agg({'TeamID': lambda x: x.nunique()})
    
event_id_team_count.rename(columns={"TeamID": "TeamCount"}, inplace=True)


    If you want to have this new column in the original dataframe, all you need to do is to join the original dataframe with the one you have just created:



    data = data.join(other=event_id_team_count, on="EventID")





    share|improve this answer









    $endgroup$













    • $begingroup$
      Thanks a lot... It worked!!!
      $endgroup$
      – Arjun Chandra
      Jan 4 at 6:25














    1












    1








    1





    $begingroup$

    You can try something like this to get a new dataframe that has pairs of (EventID, TeamCount):



    event_id_team_count = data.groupby('EventID').agg({'TeamID': lambda x: x.nunique()})
    
event_id_team_count.rename(columns={"TeamID": "TeamCount"}, inplace=True)


    If you want to have this new column in the original dataframe, all you need to do is to join the original dataframe with the one you have just created:



    data = data.join(other=event_id_team_count, on="EventID")





    share|improve this answer









    $endgroup$



    You can try something like this to get a new dataframe that has pairs of (EventID, TeamCount):



    event_id_team_count = data.groupby('EventID').agg({'TeamID': lambda x: x.nunique()})
    
event_id_team_count.rename(columns={"TeamID": "TeamCount"}, inplace=True)


    If you want to have this new column in the original dataframe, all you need to do is to join the original dataframe with the one you have just created:



    data = data.join(other=event_id_team_count, on="EventID")






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Jan 3 at 17:32









    msklmskl

    712




    712












    • $begingroup$
      Thanks a lot... It worked!!!
      $endgroup$
      – Arjun Chandra
      Jan 4 at 6:25


















    • $begingroup$
      Thanks a lot... It worked!!!
      $endgroup$
      – Arjun Chandra
      Jan 4 at 6:25
















    $begingroup$
    Thanks a lot... It worked!!!
    $endgroup$
    – Arjun Chandra
    Jan 4 at 6:25




    $begingroup$
    Thanks a lot... It worked!!!
    $endgroup$
    – Arjun Chandra
    Jan 4 at 6:25











    0












    $begingroup$


    1. Create a dictionary with the unique count of TeamID with respective to EventID


    uCountDict = dict(data.groupby("EventID").TeamID.count())
    
uCountDict


    Sample output
    {'A': 4,
    'C': 3,
    'D': 2,
    'F': 1
    }




    1. Now create a new column with unique count with respective to TeamID using apply function


    data["TeamCount"] = data.EventID.apply(lambda x : uCountDict[x])





    share|improve this answer








    New contributor




    Noor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$


















      0












      $begingroup$


      1. Create a dictionary with the unique count of TeamID with respective to EventID


      uCountDict = dict(data.groupby("EventID").TeamID.count())
      
uCountDict


      Sample output
      {'A': 4,
      'C': 3,
      'D': 2,
      'F': 1
      }




      1. Now create a new column with unique count with respective to TeamID using apply function


      data["TeamCount"] = data.EventID.apply(lambda x : uCountDict[x])





      share|improve this answer








      New contributor




      Noor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$
















        0












        0








        0





        $begingroup$


        1. Create a dictionary with the unique count of TeamID with respective to EventID


        uCountDict = dict(data.groupby("EventID").TeamID.count())
        
uCountDict


        Sample output
        {'A': 4,
        'C': 3,
        'D': 2,
        'F': 1
        }




        1. Now create a new column with unique count with respective to TeamID using apply function


        data["TeamCount"] = data.EventID.apply(lambda x : uCountDict[x])





        share|improve this answer








        New contributor




        Noor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$




        1. Create a dictionary with the unique count of TeamID with respective to EventID


        uCountDict = dict(data.groupby("EventID").TeamID.count())
        
uCountDict


        Sample output
        {'A': 4,
        'C': 3,
        'D': 2,
        'F': 1
        }




        1. Now create a new column with unique count with respective to TeamID using apply function


        data["TeamCount"] = data.EventID.apply(lambda x : uCountDict[x])






        share|improve this answer








        New contributor




        Noor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|improve this answer



        share|improve this answer






        New contributor




        Noor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered 5 hours ago









        NoorNoor

        12




        12




        New contributor




        Noor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.





        New contributor





        Noor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        Noor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Data Science Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdatascience.stackexchange.com%2fquestions%2f43439%2fhow-to-create-a-new-column-based-on-two-other-columns-in-pandas%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

            How to label and detect the document text images

            Image
            1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C...
            Read more

            Vallis Paradisi

            Image
            Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para...
            Read more

            Tabula Rosettana

            Image
            Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
            Read more