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












1












$begingroup$


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











$endgroup$

















    1












    $begingroup$


    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











    $endgroup$















      1












      1








      1





      $begingroup$


      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 43 mins ago









      m_goldberg

      87k872197




      87k872197










      asked 3 hours ago









      zongxianzongxian

      1074




      1074






















          1 Answer
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          $begingroup$

          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 computer-less♦
            2 hours ago











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          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 computer-less♦
            2 hours ago
















          4












          $begingroup$

          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 computer-less♦
            2 hours ago














          4












          4








          4





          $begingroup$

          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 31 mins ago









          m_goldberg

          87k872197




          87k872197










          answered 3 hours ago









          NasserNasser

          58k489206




          58k489206








          • 2




            $begingroup$
            Counts should also work.
            $endgroup$
            – J. M. is computer-less♦
            2 hours ago














          • 2




            $begingroup$
            Counts should also work.
            $endgroup$
            – J. M. is computer-less♦
            2 hours ago








          2




          2




          $begingroup$
          Counts should also work.
          $endgroup$
          – J. M. is computer-less♦
          2 hours ago




          $begingroup$
          Counts should also work.
          $endgroup$
          – J. M. is computer-less♦
          2 hours ago


















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