How can I get the count of how many times a string appears in my list? [closed]












2












$begingroup$


I have a list of strings, like so:



{"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss", "kk mm"}


I know WordCounts will return the numbers of every single word, but I want to the word group numbers, i.e., "aa bb" 3 times, and "cc dd" 2 times. How can I do this?






string-manipulation counting







share|improve this question











$endgroup$



closed as off-topic by MarcoB, Carl Lange, Henrik Schumacher, Öskå, march Mar 8 at 17:26


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – MarcoB, Carl Lange, Henrik Schumacher, Öskå, march

If this question can be reworded to fit the rules in the help center, please edit the question.





















    2












    $begingroup$


    I have a list of strings, like so:



    {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss", "kk mm"}


    I know WordCounts will return the numbers of every single word, but I want to the word group numbers, i.e., "aa bb" 3 times, and "cc dd" 2 times. How can I do this?






    string-manipulation counting







    share|improve this question











    $endgroup$



    closed as off-topic by MarcoB, Carl Lange, Henrik Schumacher, Öskå, march Mar 8 at 17:26


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – MarcoB, Carl Lange, Henrik Schumacher, Öskå, march

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      2












      2








      2





      $begingroup$


      I have a list of strings, like so:



      {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss", "kk mm"}


      I know WordCounts will return the numbers of every single word, but I want to the word group numbers, i.e., "aa bb" 3 times, and "cc dd" 2 times. How can I do this?






      string-manipulation counting







      share|improve this question











      $endgroup$




      I have a list of strings, like so:



      {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss", "kk mm"}


      I know WordCounts will return the numbers of every single word, but I want to the word group numbers, i.e., "aa bb" 3 times, and "cc dd" 2 times. How can I do this?






      string-manipulation counting




      string-manipulation counting






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 4 at 7:07









      m_goldberg

      88.2k872199




      88.2k872199










      asked Mar 4 at 4:22









      zongxianzongxian

      1144




      1144




      closed as off-topic by MarcoB, Carl Lange, Henrik Schumacher, Öskå, march Mar 8 at 17:26


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – MarcoB, Carl Lange, Henrik Schumacher, Öskå, march

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by MarcoB, Carl Lange, Henrik Schumacher, Öskå, march Mar 8 at 17:26


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – MarcoB, Carl Lange, Henrik Schumacher, Öskå, march

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
          1






          active

          oldest

          votes


















          8












          $begingroup$

          You can try Tally



          lst = {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss",   "kk mm"};
          
Tally[lst]


          Mathematica graphics



          Edit by m_goldberg



          As J.M. says in his comment below, Counts will give the same information as an association.



          Counts[data]



          <|"aa bb" -> 3, "cc dd" -> 2, "ww ss" -> 1, "ss ss" -> 1, "kk mm" -> 1|>




          This is equivalent to 



          Rule @@@ Tally[data] // Association





          share|improve this answer











          $endgroup$









          • 2




            $begingroup$
            Counts should also work.
            $endgroup$
            – J. M. is away♦
            Mar 4 at 5:48


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          8












          $begingroup$

          You can try Tally



          lst = {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss",   "kk mm"};
          
Tally[lst]


          Mathematica graphics



          Edit by m_goldberg



          As J.M. says in his comment below, Counts will give the same information as an association.



          Counts[data]



          <|"aa bb" -> 3, "cc dd" -> 2, "ww ss" -> 1, "ss ss" -> 1, "kk mm" -> 1|>




          This is equivalent to 



          Rule @@@ Tally[data] // Association





          share|improve this answer











          $endgroup$









          • 2




            $begingroup$
            Counts should also work.
            $endgroup$
            – J. M. is away♦
            Mar 4 at 5:48
















          8












          $begingroup$

          You can try Tally



          lst = {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss",   "kk mm"};
          
Tally[lst]


          Mathematica graphics



          Edit by m_goldberg



          As J.M. says in his comment below, Counts will give the same information as an association.



          Counts[data]



          <|"aa bb" -> 3, "cc dd" -> 2, "ww ss" -> 1, "ss ss" -> 1, "kk mm" -> 1|>




          This is equivalent to 



          Rule @@@ Tally[data] // Association





          share|improve this answer











          $endgroup$









          • 2




            $begingroup$
            Counts should also work.
            $endgroup$
            – J. M. is away♦
            Mar 4 at 5:48














          8












          8








          8





          $begingroup$

          You can try Tally



          lst = {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss",   "kk mm"};
          
Tally[lst]


          Mathematica graphics



          Edit by m_goldberg



          As J.M. says in his comment below, Counts will give the same information as an association.



          Counts[data]



          <|"aa bb" -> 3, "cc dd" -> 2, "ww ss" -> 1, "ss ss" -> 1, "kk mm" -> 1|>




          This is equivalent to 



          Rule @@@ Tally[data] // Association





          share|improve this answer











          $endgroup$



          You can try Tally



          lst = {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss",   "kk mm"};
          
Tally[lst]


          Mathematica graphics



          Edit by m_goldberg



          As J.M. says in his comment below, Counts will give the same information as an association.



          Counts[data]



          <|"aa bb" -> 3, "cc dd" -> 2, "ww ss" -> 1, "ss ss" -> 1, "kk mm" -> 1|>




          This is equivalent to 



          Rule @@@ Tally[data] // Association






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Mar 4 at 7:19









          m_goldberg

          88.2k872199




          88.2k872199










          answered Mar 4 at 4:36









          NasserNasser

          58.7k489206




          58.7k489206








          • 2




            $begingroup$
            Counts should also work.
            $endgroup$
            – J. M. is away♦
            Mar 4 at 5:48














          • 2




            $begingroup$
            Counts should also work.
            $endgroup$
            – J. M. is away♦
            Mar 4 at 5:48








          2




          2




          $begingroup$
          Counts should also work.
          $endgroup$
          – J. M. is away♦
          Mar 4 at 5:48




          $begingroup$
          Counts should also work.
          $endgroup$
          – J. M. is away♦
          Mar 4 at 5:48



          -

          Popular posts from this blog

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19
          Read more

          “minimization” problem in Euclidean space related to orthonormal basis

          Image
          3 $begingroup$ I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows: Let $(E, langle, rangle)$ be a euclidean space of dimension $n geq 2$ . 1) For any $x in E$ , show that $||x|| = sqrt{n}$ if and only if there exists $(e_1, dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ ldots + e_n$ . 2) For any $(x,y) in E$ , show that $||x|| = sqrt{n}$ , $||y|| = sqrt{frac{n(n+1)(2n+1)}{6}}$ and $langle x,y rangle = frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + ldots + e_n$ and $y = e_1 + 2e_2 + ldots + ne_n$ The first q
          Read more