Count of values grouped per month, year - PandasMySQL Query GROUP BY day / month / yearHow do I get the row count of a pandas DataFrame?Select rows from a DataFrame based on values in a column in pandasGet statistics for each group (such as count, mean, etc) using pandas GroupBy?Count unique dates in pandas dataframeCounting values using pandas groupbysplitting of date column to day, month, year in python 2.7Count Unique Days Groupby Month/YearCounting events per contributor per dayHow to count the number of dropoffs per month for dataframe column

What if a quote contains an error

This fell out of my toilet when I unscrewed the supply line. What is it?

How did the Fried Liver Attack get its name?

Does UPDATE without WHERE clause lock a table in PostgreSQL?

True Polymorph, Dragon Change Shape, and True Sight interaction

Had there been instances of national states banning harmful imports before the Opium wars?

Does the Creighton Method of Natural Family Planning have a failure rate of 3.2% or less?

Concrete description of lift in Arens-Eells space

how do you value what your leisure time is worth?

Meaning/translation of title "The Light Fantastic" By Terry Pratchett

Over powered shield?

Why is there "Il" in "Il mio tesoro intanto"?

Can I bring alcohol to Dubai?

Who became a professor?

What kind of mission objective would make a parabolic escape trajectory desirable?

Generalized Assignment Problem as the sub-problem

Would Great Old Ones care about the Blood War?

Python Bingo game that stores card in a dictionary

Looking for PC graphics demo software from the early 90s called "Unreal"

"Es gefällt ihm." How to identify similar exceptions?

How are steel imports supposed to threaten US national security?

How do I know how many sub-shells deep I am?

Find the percentage

universal string conversion



Count of values grouped per month, year - Pandas


MySQL Query GROUP BY day / month / yearHow do I get the row count of a pandas DataFrame?Select rows from a DataFrame based on values in a column in pandasGet statistics for each group (such as count, mean, etc) using pandas GroupBy?Count unique dates in pandas dataframeCounting values using pandas groupbysplitting of date column to day, month, year in python 2.7Count Unique Days Groupby Month/YearCounting events per contributor per dayHow to count the number of dropoffs per month for dataframe column






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;









8















I am trying to groupby counts of dates per month and year in a specific output. I can do it per day but can't get the same output per month/year.



d = (
'Date' : ['1/1/18','1/1/18','2/1/18','3/1/18','1/2/18','1/3/18','2/1/19','3/1/19'],
'Val' : ['A','B','C','D','A','B','C','D'],
)

df = pd.DataFrame(data = d)

df['Date'] = pd.to_datetime(df['Date'], format= '%d/%m/%y')

df['Count_d'] = df.Date.map(df.groupby('Date').size())


This is the output I want:



 Date Val Count_d
0 2018-01-01 A 2
1 2018-01-01 B 2
2 2018-01-02 C 1
3 2018-01-03 D 1
4 2018-02-01 A 1
5 2018-03-01 B 1
6 2019-01-02 C 1
7 2019-01-03 D 1


When I attempt to do similar but per month and year I use the following:



df1 = df.groupby([df['Date'].dt.year.rename('year'), df['Date'].dt.month.rename('month')]).agg('count')
print(df)


But the output is:



 Date Val
count count
year month
2018 1 4 4
2 1 1
3 1 1
2019 1 2 2


Intended Output:



 Date Val Count_d Count_m Count_y
0 2018-01-01 A 2 4 6
1 2018-01-01 B 2 4 6
2 2018-01-02 C 1 4 6
3 2018-01-03 D 1 4 6
4 2018-02-01 A 1 1 6
5 2018-03-01 B 1 1 6
6 2019-01-02 C 1 2 2
7 2019-01-03 D 1 2 2









share|improve this question
































    8















    I am trying to groupby counts of dates per month and year in a specific output. I can do it per day but can't get the same output per month/year.



    d = (
    'Date' : ['1/1/18','1/1/18','2/1/18','3/1/18','1/2/18','1/3/18','2/1/19','3/1/19'],
    'Val' : ['A','B','C','D','A','B','C','D'],
    )

    df = pd.DataFrame(data = d)

    df['Date'] = pd.to_datetime(df['Date'], format= '%d/%m/%y')

    df['Count_d'] = df.Date.map(df.groupby('Date').size())


    This is the output I want:



     Date Val Count_d
    0 2018-01-01 A 2
    1 2018-01-01 B 2
    2 2018-01-02 C 1
    3 2018-01-03 D 1
    4 2018-02-01 A 1
    5 2018-03-01 B 1
    6 2019-01-02 C 1
    7 2019-01-03 D 1


    When I attempt to do similar but per month and year I use the following:



    df1 = df.groupby([df['Date'].dt.year.rename('year'), df['Date'].dt.month.rename('month')]).agg('count')
    print(df)


    But the output is:



     Date Val
    count count
    year month
    2018 1 4 4
    2 1 1
    3 1 1
    2019 1 2 2


    Intended Output:



     Date Val Count_d Count_m Count_y
    0 2018-01-01 A 2 4 6
    1 2018-01-01 B 2 4 6
    2 2018-01-02 C 1 4 6
    3 2018-01-03 D 1 4 6
    4 2018-02-01 A 1 1 6
    5 2018-03-01 B 1 1 6
    6 2019-01-02 C 1 2 2
    7 2019-01-03 D 1 2 2









    share|improve this question




























      8












      8








      8








      I am trying to groupby counts of dates per month and year in a specific output. I can do it per day but can't get the same output per month/year.



      d = (
      'Date' : ['1/1/18','1/1/18','2/1/18','3/1/18','1/2/18','1/3/18','2/1/19','3/1/19'],
      'Val' : ['A','B','C','D','A','B','C','D'],
      )

      df = pd.DataFrame(data = d)

      df['Date'] = pd.to_datetime(df['Date'], format= '%d/%m/%y')

      df['Count_d'] = df.Date.map(df.groupby('Date').size())


      This is the output I want:



       Date Val Count_d
      0 2018-01-01 A 2
      1 2018-01-01 B 2
      2 2018-01-02 C 1
      3 2018-01-03 D 1
      4 2018-02-01 A 1
      5 2018-03-01 B 1
      6 2019-01-02 C 1
      7 2019-01-03 D 1


      When I attempt to do similar but per month and year I use the following:



      df1 = df.groupby([df['Date'].dt.year.rename('year'), df['Date'].dt.month.rename('month')]).agg('count')
      print(df)


      But the output is:



       Date Val
      count count
      year month
      2018 1 4 4
      2 1 1
      3 1 1
      2019 1 2 2


      Intended Output:



       Date Val Count_d Count_m Count_y
      0 2018-01-01 A 2 4 6
      1 2018-01-01 B 2 4 6
      2 2018-01-02 C 1 4 6
      3 2018-01-03 D 1 4 6
      4 2018-02-01 A 1 1 6
      5 2018-03-01 B 1 1 6
      6 2019-01-02 C 1 2 2
      7 2019-01-03 D 1 2 2









      share|improve this question
















      I am trying to groupby counts of dates per month and year in a specific output. I can do it per day but can't get the same output per month/year.



      d = (
      'Date' : ['1/1/18','1/1/18','2/1/18','3/1/18','1/2/18','1/3/18','2/1/19','3/1/19'],
      'Val' : ['A','B','C','D','A','B','C','D'],
      )

      df = pd.DataFrame(data = d)

      df['Date'] = pd.to_datetime(df['Date'], format= '%d/%m/%y')

      df['Count_d'] = df.Date.map(df.groupby('Date').size())


      This is the output I want:



       Date Val Count_d
      0 2018-01-01 A 2
      1 2018-01-01 B 2
      2 2018-01-02 C 1
      3 2018-01-03 D 1
      4 2018-02-01 A 1
      5 2018-03-01 B 1
      6 2019-01-02 C 1
      7 2019-01-03 D 1


      When I attempt to do similar but per month and year I use the following:



      df1 = df.groupby([df['Date'].dt.year.rename('year'), df['Date'].dt.month.rename('month')]).agg('count')
      print(df)


      But the output is:



       Date Val
      count count
      year month
      2018 1 4 4
      2 1 1
      3 1 1
      2019 1 2 2


      Intended Output:



       Date Val Count_d Count_m Count_y
      0 2018-01-01 A 2 4 6
      1 2018-01-01 B 2 4 6
      2 2018-01-02 C 1 4 6
      3 2018-01-03 D 1 4 6
      4 2018-02-01 A 1 1 6
      5 2018-03-01 B 1 1 6
      6 2019-01-02 C 1 2 2
      7 2019-01-03 D 1 2 2






      python pandas group-by count transform






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Apr 19 at 3:08







      jonboy

















      asked Apr 17 at 11:05









      jonboyjonboy

      492 silver badges15 bronze badges




      492 silver badges15 bronze badges

























          2 Answers
          2






          active

          oldest

          votes


















          6
















          Use GroupBy.transform for columns with same size like original DataFrame:



          df['Date'] = pd.to_datetime(df['Date'], format= '%d/%m/%y')
          y = df['Date'].dt.year
          m = df['Date'].dt.month

          df['Count_d'] = df.groupby('Date')['Date'].transform('size')
          df['Count_m'] = df.groupby([y, m])['Date'].transform('size')
          df['Count_y'] = df.groupby(y)['Date'].transform('size')

          print(df)
          Date Val Count_d Count_m Count_y
          0 2018-01-01 A 2 4 6
          1 2018-01-01 B 2 4 6
          2 2018-01-02 C 1 4 6
          3 2018-01-03 D 1 4 6
          4 2018-02-01 A 1 1 6
          5 2018-03-01 B 1 1 6
          6 2019-01-02 C 1 2 2
          7 2019-01-03 D 1 2 2





          share|improve this answer



























          • just found that they are removing agg with dict. any idea why?

            – anky_91
            Apr 17 at 11:14











          • @anky_91 - because same size columns like original df.

            – jezrael
            Apr 17 at 11:15











          • where did you see that @anky_91

            – Erfan
            Apr 17 at 11:15












          • @Erfan got a future warning. i was implementing wrongly i guess, jez made that clear

            – anky_91
            Apr 17 at 11:16



















          1
















          You can do this with pd.Grouper



          df['Count_d'] = df.groupby([pd.Grouper(key='Date', freq='D')])['Date'].transform('size').astype(int)
          df['Count_m'] = df.groupby([pd.Grouper(key='Date', freq='M')])['Date'].transform('size').astype(int)
          df['Count_y'] = df.groupby([pd.Grouper(key='Date', freq='Y')])['Date'].transform('size').astype(int)


          Which will give



           Date Val Count_d Count_m Count_y
          0 2018-01-01 A 2 4 6
          1 2018-01-01 B 2 4 6
          2 2018-01-02 C 1 4 6
          3 2018-01-03 D 1 4 6
          4 2018-02-01 A 1 1 6
          5 2018-03-01 B 1 1 6
          6 2019-01-02 C 1 2 2
          7 2019-01-03 D 1 2 2


          You can groupby various different frequencies with this, see the documentation on DateOffsets






          share|improve this answer


























            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/4.0/"u003ecc by-sa 4.0 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%2f55726107%2fcount-of-values-grouped-per-month-year-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









            6
















            Use GroupBy.transform for columns with same size like original DataFrame:



            df['Date'] = pd.to_datetime(df['Date'], format= '%d/%m/%y')
            y = df['Date'].dt.year
            m = df['Date'].dt.month

            df['Count_d'] = df.groupby('Date')['Date'].transform('size')
            df['Count_m'] = df.groupby([y, m])['Date'].transform('size')
            df['Count_y'] = df.groupby(y)['Date'].transform('size')

            print(df)
            Date Val Count_d Count_m Count_y
            0 2018-01-01 A 2 4 6
            1 2018-01-01 B 2 4 6
            2 2018-01-02 C 1 4 6
            3 2018-01-03 D 1 4 6
            4 2018-02-01 A 1 1 6
            5 2018-03-01 B 1 1 6
            6 2019-01-02 C 1 2 2
            7 2019-01-03 D 1 2 2





            share|improve this answer



























            • just found that they are removing agg with dict. any idea why?

              – anky_91
              Apr 17 at 11:14











            • @anky_91 - because same size columns like original df.

              – jezrael
              Apr 17 at 11:15











            • where did you see that @anky_91

              – Erfan
              Apr 17 at 11:15












            • @Erfan got a future warning. i was implementing wrongly i guess, jez made that clear

              – anky_91
              Apr 17 at 11:16
















            6
















            Use GroupBy.transform for columns with same size like original DataFrame:



            df['Date'] = pd.to_datetime(df['Date'], format= '%d/%m/%y')
            y = df['Date'].dt.year
            m = df['Date'].dt.month

            df['Count_d'] = df.groupby('Date')['Date'].transform('size')
            df['Count_m'] = df.groupby([y, m])['Date'].transform('size')
            df['Count_y'] = df.groupby(y)['Date'].transform('size')

            print(df)
            Date Val Count_d Count_m Count_y
            0 2018-01-01 A 2 4 6
            1 2018-01-01 B 2 4 6
            2 2018-01-02 C 1 4 6
            3 2018-01-03 D 1 4 6
            4 2018-02-01 A 1 1 6
            5 2018-03-01 B 1 1 6
            6 2019-01-02 C 1 2 2
            7 2019-01-03 D 1 2 2





            share|improve this answer



























            • just found that they are removing agg with dict. any idea why?

              – anky_91
              Apr 17 at 11:14











            • @anky_91 - because same size columns like original df.

              – jezrael
              Apr 17 at 11:15











            • where did you see that @anky_91

              – Erfan
              Apr 17 at 11:15












            • @Erfan got a future warning. i was implementing wrongly i guess, jez made that clear

              – anky_91
              Apr 17 at 11:16














            6














            6










            6









            Use GroupBy.transform for columns with same size like original DataFrame:



            df['Date'] = pd.to_datetime(df['Date'], format= '%d/%m/%y')
            y = df['Date'].dt.year
            m = df['Date'].dt.month

            df['Count_d'] = df.groupby('Date')['Date'].transform('size')
            df['Count_m'] = df.groupby([y, m])['Date'].transform('size')
            df['Count_y'] = df.groupby(y)['Date'].transform('size')

            print(df)
            Date Val Count_d Count_m Count_y
            0 2018-01-01 A 2 4 6
            1 2018-01-01 B 2 4 6
            2 2018-01-02 C 1 4 6
            3 2018-01-03 D 1 4 6
            4 2018-02-01 A 1 1 6
            5 2018-03-01 B 1 1 6
            6 2019-01-02 C 1 2 2
            7 2019-01-03 D 1 2 2





            share|improve this answer















            Use GroupBy.transform for columns with same size like original DataFrame:



            df['Date'] = pd.to_datetime(df['Date'], format= '%d/%m/%y')
            y = df['Date'].dt.year
            m = df['Date'].dt.month

            df['Count_d'] = df.groupby('Date')['Date'].transform('size')
            df['Count_m'] = df.groupby([y, m])['Date'].transform('size')
            df['Count_y'] = df.groupby(y)['Date'].transform('size')

            print(df)
            Date Val Count_d Count_m Count_y
            0 2018-01-01 A 2 4 6
            1 2018-01-01 B 2 4 6
            2 2018-01-02 C 1 4 6
            3 2018-01-03 D 1 4 6
            4 2018-02-01 A 1 1 6
            5 2018-03-01 B 1 1 6
            6 2019-01-02 C 1 2 2
            7 2019-01-03 D 1 2 2






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Apr 17 at 11:16

























            answered Apr 17 at 11:11









            jezraeljezrael

            416k33 gold badges437 silver badges499 bronze badges




            416k33 gold badges437 silver badges499 bronze badges















            • just found that they are removing agg with dict. any idea why?

              – anky_91
              Apr 17 at 11:14











            • @anky_91 - because same size columns like original df.

              – jezrael
              Apr 17 at 11:15











            • where did you see that @anky_91

              – Erfan
              Apr 17 at 11:15












            • @Erfan got a future warning. i was implementing wrongly i guess, jez made that clear

              – anky_91
              Apr 17 at 11:16


















            • just found that they are removing agg with dict. any idea why?

              – anky_91
              Apr 17 at 11:14











            • @anky_91 - because same size columns like original df.

              – jezrael
              Apr 17 at 11:15











            • where did you see that @anky_91

              – Erfan
              Apr 17 at 11:15












            • @Erfan got a future warning. i was implementing wrongly i guess, jez made that clear

              – anky_91
              Apr 17 at 11:16

















            just found that they are removing agg with dict. any idea why?

            – anky_91
            Apr 17 at 11:14





            just found that they are removing agg with dict. any idea why?

            – anky_91
            Apr 17 at 11:14













            @anky_91 - because same size columns like original df.

            – jezrael
            Apr 17 at 11:15





            @anky_91 - because same size columns like original df.

            – jezrael
            Apr 17 at 11:15













            where did you see that @anky_91

            – Erfan
            Apr 17 at 11:15






            where did you see that @anky_91

            – Erfan
            Apr 17 at 11:15














            @Erfan got a future warning. i was implementing wrongly i guess, jez made that clear

            – anky_91
            Apr 17 at 11:16






            @Erfan got a future warning. i was implementing wrongly i guess, jez made that clear

            – anky_91
            Apr 17 at 11:16














            1
















            You can do this with pd.Grouper



            df['Count_d'] = df.groupby([pd.Grouper(key='Date', freq='D')])['Date'].transform('size').astype(int)
            df['Count_m'] = df.groupby([pd.Grouper(key='Date', freq='M')])['Date'].transform('size').astype(int)
            df['Count_y'] = df.groupby([pd.Grouper(key='Date', freq='Y')])['Date'].transform('size').astype(int)


            Which will give



             Date Val Count_d Count_m Count_y
            0 2018-01-01 A 2 4 6
            1 2018-01-01 B 2 4 6
            2 2018-01-02 C 1 4 6
            3 2018-01-03 D 1 4 6
            4 2018-02-01 A 1 1 6
            5 2018-03-01 B 1 1 6
            6 2019-01-02 C 1 2 2
            7 2019-01-03 D 1 2 2


            You can groupby various different frequencies with this, see the documentation on DateOffsets






            share|improve this answer





























              1
















              You can do this with pd.Grouper



              df['Count_d'] = df.groupby([pd.Grouper(key='Date', freq='D')])['Date'].transform('size').astype(int)
              df['Count_m'] = df.groupby([pd.Grouper(key='Date', freq='M')])['Date'].transform('size').astype(int)
              df['Count_y'] = df.groupby([pd.Grouper(key='Date', freq='Y')])['Date'].transform('size').astype(int)


              Which will give



               Date Val Count_d Count_m Count_y
              0 2018-01-01 A 2 4 6
              1 2018-01-01 B 2 4 6
              2 2018-01-02 C 1 4 6
              3 2018-01-03 D 1 4 6
              4 2018-02-01 A 1 1 6
              5 2018-03-01 B 1 1 6
              6 2019-01-02 C 1 2 2
              7 2019-01-03 D 1 2 2


              You can groupby various different frequencies with this, see the documentation on DateOffsets






              share|improve this answer



























                1














                1










                1









                You can do this with pd.Grouper



                df['Count_d'] = df.groupby([pd.Grouper(key='Date', freq='D')])['Date'].transform('size').astype(int)
                df['Count_m'] = df.groupby([pd.Grouper(key='Date', freq='M')])['Date'].transform('size').astype(int)
                df['Count_y'] = df.groupby([pd.Grouper(key='Date', freq='Y')])['Date'].transform('size').astype(int)


                Which will give



                 Date Val Count_d Count_m Count_y
                0 2018-01-01 A 2 4 6
                1 2018-01-01 B 2 4 6
                2 2018-01-02 C 1 4 6
                3 2018-01-03 D 1 4 6
                4 2018-02-01 A 1 1 6
                5 2018-03-01 B 1 1 6
                6 2019-01-02 C 1 2 2
                7 2019-01-03 D 1 2 2


                You can groupby various different frequencies with this, see the documentation on DateOffsets






                share|improve this answer













                You can do this with pd.Grouper



                df['Count_d'] = df.groupby([pd.Grouper(key='Date', freq='D')])['Date'].transform('size').astype(int)
                df['Count_m'] = df.groupby([pd.Grouper(key='Date', freq='M')])['Date'].transform('size').astype(int)
                df['Count_y'] = df.groupby([pd.Grouper(key='Date', freq='Y')])['Date'].transform('size').astype(int)


                Which will give



                 Date Val Count_d Count_m Count_y
                0 2018-01-01 A 2 4 6
                1 2018-01-01 B 2 4 6
                2 2018-01-02 C 1 4 6
                3 2018-01-03 D 1 4 6
                4 2018-02-01 A 1 1 6
                5 2018-03-01 B 1 1 6
                6 2019-01-02 C 1 2 2
                7 2019-01-03 D 1 2 2


                You can groupby various different frequencies with this, see the documentation on DateOffsets







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Apr 17 at 11:11









                Ken SymeKen Syme

                2,5061 gold badge12 silver badges17 bronze badges




                2,5061 gold badge12 silver badges17 bronze badges































                    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%2f55726107%2fcount-of-values-grouped-per-month-year-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

                    Distance measures on a map of a game The 2019 Stack Overflow Developer Survey Results Are Inmin distance in a graphShortest distance path on contour plotHow to plot a tilted map?Finding points outside of a diskDelaunay link distanceAnnulus from GeoDisks: drawing a ring on a mapNegative Correlation DistanceFind distance along a path (GPS coordinates)Finding position at given distance in a GeoPathMathematics behind distance estimation using camera

                    How to get a smooth, uniform ParametricPlot of a 2D Region?How to plot a complicated Region?How to exclude a region from ParametricPlotHow discretize a region placing vertices on a specific non-uniform gridHow to transform a Plot or a ParametricPlot into a RegionHow can I get a smooth plot of a bounded region?Smooth ParametricPlot3D with RegionFunction?Smooth border of a region ParametricPlotSmooth region boundarySmooth region plot from list of pointsGet minimum y of a certain x in a region

                    Genealogie vun de Merowenger Vum Merowech bis zum Chilperich I. | Navigatiounsmenü