Group two Dataframes with MultiIndex columns












0















I have two dataframes and I would like to create a new one, which will include all the unique columns of the two source dataframes and the aggregation of the common columns.



These are two samples:



enter image description here



enter image description here



And this is the result:



enter image description here



All the column indexes should match in order to be aggregated.



I have written the following code:



df_all = pd.DataFrame
for dfColumn in df_1:
if dfColumn in df_2.columns:
df_all[dfColumn] = df_1.loc[:, dfColumn].add(df_2.loc[:, dfColumn])
else:
df_all[dfColumn] = df_1[dfColumn]

for dfColumn in df_2:
if dfColumn not in df_all.columns:
df_all[dfColumn] = df_2[dfColumn]


However, I get an error on the following line:



df_all[dfColumn] = df_1.loc[:, dfColumn].add(df_2.loc[:, dfColumn])


when I am trying to assign the value to df_all[dfColumn]



It drives me crazy all the different possibilities that you have with Python.



But I cannot find one to make it work.



Thanks for your help and time.










share|improve this question



























    0















    I have two dataframes and I would like to create a new one, which will include all the unique columns of the two source dataframes and the aggregation of the common columns.



    These are two samples:



    enter image description here



    enter image description here



    And this is the result:



    enter image description here



    All the column indexes should match in order to be aggregated.



    I have written the following code:



    df_all = pd.DataFrame
    for dfColumn in df_1:
    if dfColumn in df_2.columns:
    df_all[dfColumn] = df_1.loc[:, dfColumn].add(df_2.loc[:, dfColumn])
    else:
    df_all[dfColumn] = df_1[dfColumn]

    for dfColumn in df_2:
    if dfColumn not in df_all.columns:
    df_all[dfColumn] = df_2[dfColumn]


    However, I get an error on the following line:



    df_all[dfColumn] = df_1.loc[:, dfColumn].add(df_2.loc[:, dfColumn])


    when I am trying to assign the value to df_all[dfColumn]



    It drives me crazy all the different possibilities that you have with Python.



    But I cannot find one to make it work.



    Thanks for your help and time.










    share|improve this question

























      0












      0








      0








      I have two dataframes and I would like to create a new one, which will include all the unique columns of the two source dataframes and the aggregation of the common columns.



      These are two samples:



      enter image description here



      enter image description here



      And this is the result:



      enter image description here



      All the column indexes should match in order to be aggregated.



      I have written the following code:



      df_all = pd.DataFrame
      for dfColumn in df_1:
      if dfColumn in df_2.columns:
      df_all[dfColumn] = df_1.loc[:, dfColumn].add(df_2.loc[:, dfColumn])
      else:
      df_all[dfColumn] = df_1[dfColumn]

      for dfColumn in df_2:
      if dfColumn not in df_all.columns:
      df_all[dfColumn] = df_2[dfColumn]


      However, I get an error on the following line:



      df_all[dfColumn] = df_1.loc[:, dfColumn].add(df_2.loc[:, dfColumn])


      when I am trying to assign the value to df_all[dfColumn]



      It drives me crazy all the different possibilities that you have with Python.



      But I cannot find one to make it work.



      Thanks for your help and time.










      share|improve this question














      I have two dataframes and I would like to create a new one, which will include all the unique columns of the two source dataframes and the aggregation of the common columns.



      These are two samples:



      enter image description here



      enter image description here



      And this is the result:



      enter image description here



      All the column indexes should match in order to be aggregated.



      I have written the following code:



      df_all = pd.DataFrame
      for dfColumn in df_1:
      if dfColumn in df_2.columns:
      df_all[dfColumn] = df_1.loc[:, dfColumn].add(df_2.loc[:, dfColumn])
      else:
      df_all[dfColumn] = df_1[dfColumn]

      for dfColumn in df_2:
      if dfColumn not in df_all.columns:
      df_all[dfColumn] = df_2[dfColumn]


      However, I get an error on the following line:



      df_all[dfColumn] = df_1.loc[:, dfColumn].add(df_2.loc[:, dfColumn])


      when I am trying to assign the value to df_all[dfColumn]



      It drives me crazy all the different possibilities that you have with Python.



      But I cannot find one to make it work.



      Thanks for your help and time.







      python-3.x pandas dataframe






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 16 '18 at 10:16









      ThanasisThanasis

      1821112




      1821112
























          1 Answer
          1






          active

          oldest

          votes


















          0














          Actually,



          I fixed it only with the following:



          df_all = pd.concat([df_1, df_2], axis=1)
          df_all = df_all.groupby(level=[0, 1, 2], axis=1).sum()


          Is there a way to replace level=[0, 1, 2] with something like level=df_all.columns.levels ?






          share|improve this answer
























          • Okay @Thanasis I edited my answer to do what you wanted. In this case I keep the unique columns and add the duplicated ones.

            – yatu
            Nov 16 '18 at 11:37












          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',
          autoActivateHeartbeat: false,
          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%2f53335718%2fgroup-two-dataframes-with-multiindex-columns%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0














          Actually,



          I fixed it only with the following:



          df_all = pd.concat([df_1, df_2], axis=1)
          df_all = df_all.groupby(level=[0, 1, 2], axis=1).sum()


          Is there a way to replace level=[0, 1, 2] with something like level=df_all.columns.levels ?






          share|improve this answer
























          • Okay @Thanasis I edited my answer to do what you wanted. In this case I keep the unique columns and add the duplicated ones.

            – yatu
            Nov 16 '18 at 11:37
















          0














          Actually,



          I fixed it only with the following:



          df_all = pd.concat([df_1, df_2], axis=1)
          df_all = df_all.groupby(level=[0, 1, 2], axis=1).sum()


          Is there a way to replace level=[0, 1, 2] with something like level=df_all.columns.levels ?






          share|improve this answer
























          • Okay @Thanasis I edited my answer to do what you wanted. In this case I keep the unique columns and add the duplicated ones.

            – yatu
            Nov 16 '18 at 11:37














          0












          0








          0







          Actually,



          I fixed it only with the following:



          df_all = pd.concat([df_1, df_2], axis=1)
          df_all = df_all.groupby(level=[0, 1, 2], axis=1).sum()


          Is there a way to replace level=[0, 1, 2] with something like level=df_all.columns.levels ?






          share|improve this answer













          Actually,



          I fixed it only with the following:



          df_all = pd.concat([df_1, df_2], axis=1)
          df_all = df_all.groupby(level=[0, 1, 2], axis=1).sum()


          Is there a way to replace level=[0, 1, 2] with something like level=df_all.columns.levels ?







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 16 '18 at 11:02









          ThanasisThanasis

          1821112




          1821112













          • Okay @Thanasis I edited my answer to do what you wanted. In this case I keep the unique columns and add the duplicated ones.

            – yatu
            Nov 16 '18 at 11:37



















          • Okay @Thanasis I edited my answer to do what you wanted. In this case I keep the unique columns and add the duplicated ones.

            – yatu
            Nov 16 '18 at 11:37

















          Okay @Thanasis I edited my answer to do what you wanted. In this case I keep the unique columns and add the duplicated ones.

          – yatu
          Nov 16 '18 at 11:37





          Okay @Thanasis I edited my answer to do what you wanted. In this case I keep the unique columns and add the duplicated ones.

          – yatu
          Nov 16 '18 at 11:37




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Stack Overflow!


          • 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.


          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%2fstackoverflow.com%2fquestions%2f53335718%2fgroup-two-dataframes-with-multiindex-columns%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

          List item for chat from Array inside array React Native

          Thiostrepton

          Caerphilly