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Compare 'n' number of lists in Tcl



2019 Community Moderator ElectionTCL How to read , extract and count the occurent in .txt file (Current Directory)Floating number comparing in tclSearching for a number in a sorted list in TclTCL Program that Compare StringSearching from list in tclWhy can't I use `w` for `while` in Tcl's interactive mode?How can I compare two lists (containing character and number in it) in tcl?Compare two paths in tclTCL Group list number by ruleGetting an element from a list in Tcl at a particular index for a given condition










1















I have the following lists:



set w 1 2 3 4 5 6 7
set x 1 2 3 4 5 8 9
set y 1 2 3 4 0 9 1
set z 1 2 3 4 5 6 7


I want to compare all the lists - w.r.t corresponding indices - and find out the common elements and append those common elements to a new list. If I compare the above lists, I see that 1 2 3 4 are common across all the lists and have the same index, so my output should be:



1 2 3 4


If there are no common elements(even at 0'th index) my new list will be empty.



I first start off by creating a list of lists:



set l1 [list $w $x $y $z]


Then I create a nested loop to compare the lists and extract my common elements, I will be using list 'x' as my reference list:



for set j 0 $j < [llength $x] incr j 
for set i 1 $i < [llength $l1] incr i

set a [lindex [lindex $l1 $i] $j]

if $a == [lindex $x $j] && [lindex $l2 $j] ==
lappend l2 $a
else
break






What I'm getting is:



1 2 3 4 5









share|improve this question


























    1















    I have the following lists:



    set w 1 2 3 4 5 6 7
    set x 1 2 3 4 5 8 9
    set y 1 2 3 4 0 9 1
    set z 1 2 3 4 5 6 7


    I want to compare all the lists - w.r.t corresponding indices - and find out the common elements and append those common elements to a new list. If I compare the above lists, I see that 1 2 3 4 are common across all the lists and have the same index, so my output should be:



    1 2 3 4


    If there are no common elements(even at 0'th index) my new list will be empty.



    I first start off by creating a list of lists:



    set l1 [list $w $x $y $z]


    Then I create a nested loop to compare the lists and extract my common elements, I will be using list 'x' as my reference list:



    for set j 0 $j < [llength $x] incr j 
    for set i 1 $i < [llength $l1] incr i

    set a [lindex [lindex $l1 $i] $j]

    if $a == [lindex $x $j] && [lindex $l2 $j] ==
    lappend l2 $a
    else
    break






    What I'm getting is:



    1 2 3 4 5









    share|improve this question
























      1












      1








      1








      I have the following lists:



      set w 1 2 3 4 5 6 7
      set x 1 2 3 4 5 8 9
      set y 1 2 3 4 0 9 1
      set z 1 2 3 4 5 6 7


      I want to compare all the lists - w.r.t corresponding indices - and find out the common elements and append those common elements to a new list. If I compare the above lists, I see that 1 2 3 4 are common across all the lists and have the same index, so my output should be:



      1 2 3 4


      If there are no common elements(even at 0'th index) my new list will be empty.



      I first start off by creating a list of lists:



      set l1 [list $w $x $y $z]


      Then I create a nested loop to compare the lists and extract my common elements, I will be using list 'x' as my reference list:



      for set j 0 $j < [llength $x] incr j 
      for set i 1 $i < [llength $l1] incr i

      set a [lindex [lindex $l1 $i] $j]

      if $a == [lindex $x $j] && [lindex $l2 $j] ==
      lappend l2 $a
      else
      break






      What I'm getting is:



      1 2 3 4 5









      share|improve this question














      I have the following lists:



      set w 1 2 3 4 5 6 7
      set x 1 2 3 4 5 8 9
      set y 1 2 3 4 0 9 1
      set z 1 2 3 4 5 6 7


      I want to compare all the lists - w.r.t corresponding indices - and find out the common elements and append those common elements to a new list. If I compare the above lists, I see that 1 2 3 4 are common across all the lists and have the same index, so my output should be:



      1 2 3 4


      If there are no common elements(even at 0'th index) my new list will be empty.



      I first start off by creating a list of lists:



      set l1 [list $w $x $y $z]


      Then I create a nested loop to compare the lists and extract my common elements, I will be using list 'x' as my reference list:



      for set j 0 $j < [llength $x] incr j 
      for set i 1 $i < [llength $l1] incr i

      set a [lindex [lindex $l1 $i] $j]

      if $a == [lindex $x $j] && [lindex $l2 $j] ==
      lappend l2 $a
      else
      break






      What I'm getting is:



      1 2 3 4 5






      tcl






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Mar 7 at 7:40









      LowerMoonLowerMoon

      315




      315






















          2 Answers
          2






          active

          oldest

          votes


















          1














          You are effectively only comparing list x against list x, and the actual output from your code above (assuming list l2 is initially empty) is actually:



          1 2 3 4 5 8 9


          You might ask:




          Why is it comparing list x against list x?




          Your inner loop starts at index 1 (set i 1), which is list x in l1.



          You might further ask:




          Why are the other lists not being compared?




          Once you have appended something to l2, lindex $l2 $j for the next list will never be empty, so the inner loop will break.




          So, how to do it?



          I would probably use something like this:



          set w 1 2 3 4 5 6 7
          set x 1 2 3 4 5 8 9
          set y 1 2 3 4 0 9 1
          set z 1 2 3 4 5 6 7

          set l1 [list $w $x $y $z]
          set l2 [list]

          set num [llength $x]

          for set i 0 $i < $num incr i
          # This variable will tell us how many matched. 0 indicating none.
          set same 0

          for set j 0 $j < [llength $l1] incr j
          # lindex accepts multiple indices, see the manual
          # using x as reference, if the current list's ith element is the same as x's ith element...
          if [lindex $l1 $j $i] == [lindex $x $i]
          incr same



          # if same reached 4, means 4 matched
          if $same == 4
          lappend l2 [lindex $x $i]




          Result:



          1 2 3 4



          You could make the inner loop break if the elements do not match, so it won't loop unnecessarily.



          Or instead of counting the number of matches, you could check if the inner loop broke with something like:



          for set i 0 $i < $num incr i 
          set broke 0

          for set j 0 $j < [llength $l1] incr j
          if [lindex $l1 $j $i] != [lindex $x $i]
          set broke 1
          break



          # if it did not break, then we know all matched
          if $broke == 0
          lappend l2 [lindex $x $i]







          share|improve this answer























          • Thanks, this works! I didn't notice that blunder of comparing against the same list, my initial goal was to avoid it and keep the first list in l1 (or) w as my reference and skip it during the comparison. Also, the reason I added && [lindex $l2 $j] == was because I was getting 1 1 1 2 2 2 3 3 3 4 4 4 5. I'm still a bit unsure on why my loop isn't terminating when a becomes 5.

            – LowerMoon
            Mar 7 at 8:36












          • Never mind, I understood. My loop just appends the element even if the element matches the element in the second list. I didn't add code to delete the already appended number if it finds an inequality in the other n-1 lists.

            – LowerMoon
            Mar 7 at 8:47


















          2














          Similar implementation, but using an array to store the unique elements of the slice



          set lists [list $w $x $y $z]
          set common [list]

          for set i 0 $i < [llength $w] incr i
          array unset elements
          foreach list $lists
          set elements([lindex $list $i]) dummyvalue

          set unique [array names elements]
          if [llength $unique] == 1
          lappend common $unique



          puts $common ;# => 1 2 3 4





          share|improve this answer






















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            You are effectively only comparing list x against list x, and the actual output from your code above (assuming list l2 is initially empty) is actually:



            1 2 3 4 5 8 9


            You might ask:




            Why is it comparing list x against list x?




            Your inner loop starts at index 1 (set i 1), which is list x in l1.



            You might further ask:




            Why are the other lists not being compared?




            Once you have appended something to l2, lindex $l2 $j for the next list will never be empty, so the inner loop will break.




            So, how to do it?



            I would probably use something like this:



            set w 1 2 3 4 5 6 7
            set x 1 2 3 4 5 8 9
            set y 1 2 3 4 0 9 1
            set z 1 2 3 4 5 6 7

            set l1 [list $w $x $y $z]
            set l2 [list]

            set num [llength $x]

            for set i 0 $i < $num incr i
            # This variable will tell us how many matched. 0 indicating none.
            set same 0

            for set j 0 $j < [llength $l1] incr j
            # lindex accepts multiple indices, see the manual
            # using x as reference, if the current list's ith element is the same as x's ith element...
            if [lindex $l1 $j $i] == [lindex $x $i]
            incr same



            # if same reached 4, means 4 matched
            if $same == 4
            lappend l2 [lindex $x $i]




            Result:



            1 2 3 4



            You could make the inner loop break if the elements do not match, so it won't loop unnecessarily.



            Or instead of counting the number of matches, you could check if the inner loop broke with something like:



            for set i 0 $i < $num incr i 
            set broke 0

            for set j 0 $j < [llength $l1] incr j
            if [lindex $l1 $j $i] != [lindex $x $i]
            set broke 1
            break



            # if it did not break, then we know all matched
            if $broke == 0
            lappend l2 [lindex $x $i]







            share|improve this answer























            • Thanks, this works! I didn't notice that blunder of comparing against the same list, my initial goal was to avoid it and keep the first list in l1 (or) w as my reference and skip it during the comparison. Also, the reason I added && [lindex $l2 $j] == was because I was getting 1 1 1 2 2 2 3 3 3 4 4 4 5. I'm still a bit unsure on why my loop isn't terminating when a becomes 5.

              – LowerMoon
              Mar 7 at 8:36












            • Never mind, I understood. My loop just appends the element even if the element matches the element in the second list. I didn't add code to delete the already appended number if it finds an inequality in the other n-1 lists.

              – LowerMoon
              Mar 7 at 8:47















            1














            You are effectively only comparing list x against list x, and the actual output from your code above (assuming list l2 is initially empty) is actually:



            1 2 3 4 5 8 9


            You might ask:




            Why is it comparing list x against list x?




            Your inner loop starts at index 1 (set i 1), which is list x in l1.



            You might further ask:




            Why are the other lists not being compared?




            Once you have appended something to l2, lindex $l2 $j for the next list will never be empty, so the inner loop will break.




            So, how to do it?



            I would probably use something like this:



            set w 1 2 3 4 5 6 7
            set x 1 2 3 4 5 8 9
            set y 1 2 3 4 0 9 1
            set z 1 2 3 4 5 6 7

            set l1 [list $w $x $y $z]
            set l2 [list]

            set num [llength $x]

            for set i 0 $i < $num incr i
            # This variable will tell us how many matched. 0 indicating none.
            set same 0

            for set j 0 $j < [llength $l1] incr j
            # lindex accepts multiple indices, see the manual
            # using x as reference, if the current list's ith element is the same as x's ith element...
            if [lindex $l1 $j $i] == [lindex $x $i]
            incr same



            # if same reached 4, means 4 matched
            if $same == 4
            lappend l2 [lindex $x $i]




            Result:



            1 2 3 4



            You could make the inner loop break if the elements do not match, so it won't loop unnecessarily.



            Or instead of counting the number of matches, you could check if the inner loop broke with something like:



            for set i 0 $i < $num incr i 
            set broke 0

            for set j 0 $j < [llength $l1] incr j
            if [lindex $l1 $j $i] != [lindex $x $i]
            set broke 1
            break



            # if it did not break, then we know all matched
            if $broke == 0
            lappend l2 [lindex $x $i]







            share|improve this answer























            • Thanks, this works! I didn't notice that blunder of comparing against the same list, my initial goal was to avoid it and keep the first list in l1 (or) w as my reference and skip it during the comparison. Also, the reason I added && [lindex $l2 $j] == was because I was getting 1 1 1 2 2 2 3 3 3 4 4 4 5. I'm still a bit unsure on why my loop isn't terminating when a becomes 5.

              – LowerMoon
              Mar 7 at 8:36












            • Never mind, I understood. My loop just appends the element even if the element matches the element in the second list. I didn't add code to delete the already appended number if it finds an inequality in the other n-1 lists.

              – LowerMoon
              Mar 7 at 8:47













            1












            1








            1







            You are effectively only comparing list x against list x, and the actual output from your code above (assuming list l2 is initially empty) is actually:



            1 2 3 4 5 8 9


            You might ask:




            Why is it comparing list x against list x?




            Your inner loop starts at index 1 (set i 1), which is list x in l1.



            You might further ask:




            Why are the other lists not being compared?




            Once you have appended something to l2, lindex $l2 $j for the next list will never be empty, so the inner loop will break.




            So, how to do it?



            I would probably use something like this:



            set w 1 2 3 4 5 6 7
            set x 1 2 3 4 5 8 9
            set y 1 2 3 4 0 9 1
            set z 1 2 3 4 5 6 7

            set l1 [list $w $x $y $z]
            set l2 [list]

            set num [llength $x]

            for set i 0 $i < $num incr i
            # This variable will tell us how many matched. 0 indicating none.
            set same 0

            for set j 0 $j < [llength $l1] incr j
            # lindex accepts multiple indices, see the manual
            # using x as reference, if the current list's ith element is the same as x's ith element...
            if [lindex $l1 $j $i] == [lindex $x $i]
            incr same



            # if same reached 4, means 4 matched
            if $same == 4
            lappend l2 [lindex $x $i]




            Result:



            1 2 3 4



            You could make the inner loop break if the elements do not match, so it won't loop unnecessarily.



            Or instead of counting the number of matches, you could check if the inner loop broke with something like:



            for set i 0 $i < $num incr i 
            set broke 0

            for set j 0 $j < [llength $l1] incr j
            if [lindex $l1 $j $i] != [lindex $x $i]
            set broke 1
            break



            # if it did not break, then we know all matched
            if $broke == 0
            lappend l2 [lindex $x $i]







            share|improve this answer













            You are effectively only comparing list x against list x, and the actual output from your code above (assuming list l2 is initially empty) is actually:



            1 2 3 4 5 8 9


            You might ask:




            Why is it comparing list x against list x?




            Your inner loop starts at index 1 (set i 1), which is list x in l1.



            You might further ask:




            Why are the other lists not being compared?




            Once you have appended something to l2, lindex $l2 $j for the next list will never be empty, so the inner loop will break.




            So, how to do it?



            I would probably use something like this:



            set w 1 2 3 4 5 6 7
            set x 1 2 3 4 5 8 9
            set y 1 2 3 4 0 9 1
            set z 1 2 3 4 5 6 7

            set l1 [list $w $x $y $z]
            set l2 [list]

            set num [llength $x]

            for set i 0 $i < $num incr i
            # This variable will tell us how many matched. 0 indicating none.
            set same 0

            for set j 0 $j < [llength $l1] incr j
            # lindex accepts multiple indices, see the manual
            # using x as reference, if the current list's ith element is the same as x's ith element...
            if [lindex $l1 $j $i] == [lindex $x $i]
            incr same



            # if same reached 4, means 4 matched
            if $same == 4
            lappend l2 [lindex $x $i]




            Result:



            1 2 3 4



            You could make the inner loop break if the elements do not match, so it won't loop unnecessarily.



            Or instead of counting the number of matches, you could check if the inner loop broke with something like:



            for set i 0 $i < $num incr i 
            set broke 0

            for set j 0 $j < [llength $l1] incr j
            if [lindex $l1 $j $i] != [lindex $x $i]
            set broke 1
            break



            # if it did not break, then we know all matched
            if $broke == 0
            lappend l2 [lindex $x $i]








            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Mar 7 at 8:02









            JerryJerry

            59.6k1070106




            59.6k1070106












            • Thanks, this works! I didn't notice that blunder of comparing against the same list, my initial goal was to avoid it and keep the first list in l1 (or) w as my reference and skip it during the comparison. Also, the reason I added && [lindex $l2 $j] == was because I was getting 1 1 1 2 2 2 3 3 3 4 4 4 5. I'm still a bit unsure on why my loop isn't terminating when a becomes 5.

              – LowerMoon
              Mar 7 at 8:36












            • Never mind, I understood. My loop just appends the element even if the element matches the element in the second list. I didn't add code to delete the already appended number if it finds an inequality in the other n-1 lists.

              – LowerMoon
              Mar 7 at 8:47

















            • Thanks, this works! I didn't notice that blunder of comparing against the same list, my initial goal was to avoid it and keep the first list in l1 (or) w as my reference and skip it during the comparison. Also, the reason I added && [lindex $l2 $j] == was because I was getting 1 1 1 2 2 2 3 3 3 4 4 4 5. I'm still a bit unsure on why my loop isn't terminating when a becomes 5.

              – LowerMoon
              Mar 7 at 8:36












            • Never mind, I understood. My loop just appends the element even if the element matches the element in the second list. I didn't add code to delete the already appended number if it finds an inequality in the other n-1 lists.

              – LowerMoon
              Mar 7 at 8:47
















            Thanks, this works! I didn't notice that blunder of comparing against the same list, my initial goal was to avoid it and keep the first list in l1 (or) w as my reference and skip it during the comparison. Also, the reason I added && [lindex $l2 $j] == was because I was getting 1 1 1 2 2 2 3 3 3 4 4 4 5. I'm still a bit unsure on why my loop isn't terminating when a becomes 5.

            – LowerMoon
            Mar 7 at 8:36






            Thanks, this works! I didn't notice that blunder of comparing against the same list, my initial goal was to avoid it and keep the first list in l1 (or) w as my reference and skip it during the comparison. Also, the reason I added && [lindex $l2 $j] == was because I was getting 1 1 1 2 2 2 3 3 3 4 4 4 5. I'm still a bit unsure on why my loop isn't terminating when a becomes 5.

            – LowerMoon
            Mar 7 at 8:36














            Never mind, I understood. My loop just appends the element even if the element matches the element in the second list. I didn't add code to delete the already appended number if it finds an inequality in the other n-1 lists.

            – LowerMoon
            Mar 7 at 8:47





            Never mind, I understood. My loop just appends the element even if the element matches the element in the second list. I didn't add code to delete the already appended number if it finds an inequality in the other n-1 lists.

            – LowerMoon
            Mar 7 at 8:47













            2














            Similar implementation, but using an array to store the unique elements of the slice



            set lists [list $w $x $y $z]
            set common [list]

            for set i 0 $i < [llength $w] incr i
            array unset elements
            foreach list $lists
            set elements([lindex $list $i]) dummyvalue

            set unique [array names elements]
            if [llength $unique] == 1
            lappend common $unique



            puts $common ;# => 1 2 3 4





            share|improve this answer



























              2














              Similar implementation, but using an array to store the unique elements of the slice



              set lists [list $w $x $y $z]
              set common [list]

              for set i 0 $i < [llength $w] incr i
              array unset elements
              foreach list $lists
              set elements([lindex $list $i]) dummyvalue

              set unique [array names elements]
              if [llength $unique] == 1
              lappend common $unique



              puts $common ;# => 1 2 3 4





              share|improve this answer

























                2












                2








                2







                Similar implementation, but using an array to store the unique elements of the slice



                set lists [list $w $x $y $z]
                set common [list]

                for set i 0 $i < [llength $w] incr i
                array unset elements
                foreach list $lists
                set elements([lindex $list $i]) dummyvalue

                set unique [array names elements]
                if [llength $unique] == 1
                lappend common $unique



                puts $common ;# => 1 2 3 4





                share|improve this answer













                Similar implementation, but using an array to store the unique elements of the slice



                set lists [list $w $x $y $z]
                set common [list]

                for set i 0 $i < [llength $w] incr i
                array unset elements
                foreach list $lists
                set elements([lindex $list $i]) dummyvalue

                set unique [array names elements]
                if [llength $unique] == 1
                lappend common $unique



                puts $common ;# => 1 2 3 4






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Mar 7 at 14:04









                glenn jackmanglenn jackman

                170k26147240




                170k26147240



























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