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Why can't I access the 'name' of an object when looping through the scene's objects?


How to get 3d view grid properties?How to create properties for new object typesWhy is Pie empty when no object in layerHow to emulate tracking to objects like the 'Nokia snake game'?TypeError: 'Scene' object is not iterable“'NoneType' object has no attribute 'action'” error when looping over objects with PythonAttributeError while changing the display units using python scriptPython 2.8 - why does linking the image texture node throw an error?How to remove certain part of the object name for hundreds of objects?






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








1












$begingroup$


I am trying to delete all the objects that start with 'LTR' but am getting the following error: 'tuple' object has no attribute 'name'



Any suggestions are much appreciated!



for obj in enumerate(bpy.context.scene.objects):
namestr=obj.name #this throws the error
if len(namestr)>=3:
name2=namestr[3:]
if name2=='LTR':
delete_object(namestr) #this is a function that deletes the object









share|improve this question











$endgroup$




















    1












    $begingroup$


    I am trying to delete all the objects that start with 'LTR' but am getting the following error: 'tuple' object has no attribute 'name'



    Any suggestions are much appreciated!



    for obj in enumerate(bpy.context.scene.objects):
    namestr=obj.name #this throws the error
    if len(namestr)>=3:
    name2=namestr[3:]
    if name2=='LTR':
    delete_object(namestr) #this is a function that deletes the object









    share|improve this question











    $endgroup$
















      1












      1








      1





      $begingroup$


      I am trying to delete all the objects that start with 'LTR' but am getting the following error: 'tuple' object has no attribute 'name'



      Any suggestions are much appreciated!



      for obj in enumerate(bpy.context.scene.objects):
      namestr=obj.name #this throws the error
      if len(namestr)>=3:
      name2=namestr[3:]
      if name2=='LTR':
      delete_object(namestr) #this is a function that deletes the object









      share|improve this question











      $endgroup$




      I am trying to delete all the objects that start with 'LTR' but am getting the following error: 'tuple' object has no attribute 'name'



      Any suggestions are much appreciated!



      for obj in enumerate(bpy.context.scene.objects):
      namestr=obj.name #this throws the error
      if len(namestr)>=3:
      name2=namestr[3:]
      if name2=='LTR':
      delete_object(namestr) #this is a function that deletes the object






      python scripting objects






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 12 hours ago









      Ray Mairlot

      23.3k6 gold badges72 silver badges102 bronze badges




      23.3k6 gold badges72 silver badges102 bronze badges










      asked 13 hours ago









      vndepvndep

      867 bronze badges




      867 bronze badges























          2 Answers
          2






          active

          oldest

          votes


















          3













          $begingroup$

          enumerate(iterable, start=0) returns a tuple for each element from any given iterable and its position based on the actual iteration. Proof using the Console:



          >>> for i in enumerate(["Hello","World"]):
          >>> print (i)
          (0, "Hello")
          (1, "World")


          A tuple is a composite data type, composed out of two elements (item1, item2). If you want to access its individual components (left, right) you can either use the index operator [] or create 2 variables for the actual return value (tuple) on the fly: left, right = ("Hello", "World"):



          >>> my_tuple = ("Hello", "World")
          >>> my_tuple[0]
          "Hello"

          left, right = ("Hello", "World")
          >>> left
          "Hello"


          Means in your case that you can't access the object and its name this way because the returned tuple by the enumerate() function in your loop has no name property. Proof using the Console again:



          >>> for ob in enumerate(C.scene.objects):
          ... print (ob)
          (0, bpy.data.objects['Cube'])
          (1, bpy.data.objects['Light'])
          (2, bpy.data.objects['Camera'])


          In summary: you have to find a way splitting the components of the tuple if you want to access each element separately which basically leads into the following pattern:



          >>> for left, right in enumerate(C.scene.objects):
          ... print (left, right, "Name:", right.name)
          0 <bpy_struct, Object("Cube")>, Name: Cube
          1 <bpy_struct, Object("Light")> Name: Light
          2 <bpy_struct, Object("Camera")> Name: Camera



          Also notice that python has a lot of awesome tools dealing with strings. In order to determine whether a string is part of a another string (substring), you could use the in operator to test its "membership":



          >>> "ell" in "HelloWorld"
          True


          To remove the objects, you might want to create some kind of black list before removing it. The following example removes all objects in the scene that have "Cam" in their name (substring):



          import bpy

          objects_to_remove = []

          for ob in bpy.context.scene.objects:
          if "Cam" in ob.name:
          objects_to_remove.append(ob)

          bpy.ops.object.delete("selected_objects": objects_to_remove)


          There is also str.startswith(prefix) returning True if the string begins with the prefix passed:



          >>> "Hello World".startswith("Hell")
          True


          The following example removes all found objects whose names begin with "Cam":



          import bpy

          objects_to_remove = []

          for ob in bpy.context.scene.objects:
          if ob.name.startswith("Cam"):
          objects_to_remove.append(ob)

          bpy.ops.object.delete("selected_objects": objects_to_remove)





          share|improve this answer











          $endgroup$






















            3













            $begingroup$

            When you use enumerate, an item in the list and an index/counter of where you are in the loop is returned to the variable (obj) instead of just the item. This is why you get the error, because in this case obj is a tuple ((index, listItem)) which does not have a name property itself.



            The common loop syntax when using enumerate is:



            for index, obj in enumerate(bpy.context.scene.objects):


            obj will now contain the actual object as expected.



            However, if you don't need an index of where you are in the loop you do not need to use enumerate. Remove it and it will work fine, e.g.:



            for obj in bpy.context.scene.objects:


            You could also access the second item in the tuple when trying to access the name:



            namestr=obj[1].name





            share|improve this answer









            $endgroup$

















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              2 Answers
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              2 Answers
              2






              active

              oldest

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              active

              oldest

              votes






              active

              oldest

              votes









              3













              $begingroup$

              enumerate(iterable, start=0) returns a tuple for each element from any given iterable and its position based on the actual iteration. Proof using the Console:



              >>> for i in enumerate(["Hello","World"]):
              >>> print (i)
              (0, "Hello")
              (1, "World")


              A tuple is a composite data type, composed out of two elements (item1, item2). If you want to access its individual components (left, right) you can either use the index operator [] or create 2 variables for the actual return value (tuple) on the fly: left, right = ("Hello", "World"):



              >>> my_tuple = ("Hello", "World")
              >>> my_tuple[0]
              "Hello"

              left, right = ("Hello", "World")
              >>> left
              "Hello"


              Means in your case that you can't access the object and its name this way because the returned tuple by the enumerate() function in your loop has no name property. Proof using the Console again:



              >>> for ob in enumerate(C.scene.objects):
              ... print (ob)
              (0, bpy.data.objects['Cube'])
              (1, bpy.data.objects['Light'])
              (2, bpy.data.objects['Camera'])


              In summary: you have to find a way splitting the components of the tuple if you want to access each element separately which basically leads into the following pattern:



              >>> for left, right in enumerate(C.scene.objects):
              ... print (left, right, "Name:", right.name)
              0 <bpy_struct, Object("Cube")>, Name: Cube
              1 <bpy_struct, Object("Light")> Name: Light
              2 <bpy_struct, Object("Camera")> Name: Camera



              Also notice that python has a lot of awesome tools dealing with strings. In order to determine whether a string is part of a another string (substring), you could use the in operator to test its "membership":



              >>> "ell" in "HelloWorld"
              True


              To remove the objects, you might want to create some kind of black list before removing it. The following example removes all objects in the scene that have "Cam" in their name (substring):



              import bpy

              objects_to_remove = []

              for ob in bpy.context.scene.objects:
              if "Cam" in ob.name:
              objects_to_remove.append(ob)

              bpy.ops.object.delete("selected_objects": objects_to_remove)


              There is also str.startswith(prefix) returning True if the string begins with the prefix passed:



              >>> "Hello World".startswith("Hell")
              True


              The following example removes all found objects whose names begin with "Cam":



              import bpy

              objects_to_remove = []

              for ob in bpy.context.scene.objects:
              if ob.name.startswith("Cam"):
              objects_to_remove.append(ob)

              bpy.ops.object.delete("selected_objects": objects_to_remove)





              share|improve this answer











              $endgroup$



















                3













                $begingroup$

                enumerate(iterable, start=0) returns a tuple for each element from any given iterable and its position based on the actual iteration. Proof using the Console:



                >>> for i in enumerate(["Hello","World"]):
                >>> print (i)
                (0, "Hello")
                (1, "World")


                A tuple is a composite data type, composed out of two elements (item1, item2). If you want to access its individual components (left, right) you can either use the index operator [] or create 2 variables for the actual return value (tuple) on the fly: left, right = ("Hello", "World"):



                >>> my_tuple = ("Hello", "World")
                >>> my_tuple[0]
                "Hello"

                left, right = ("Hello", "World")
                >>> left
                "Hello"


                Means in your case that you can't access the object and its name this way because the returned tuple by the enumerate() function in your loop has no name property. Proof using the Console again:



                >>> for ob in enumerate(C.scene.objects):
                ... print (ob)
                (0, bpy.data.objects['Cube'])
                (1, bpy.data.objects['Light'])
                (2, bpy.data.objects['Camera'])


                In summary: you have to find a way splitting the components of the tuple if you want to access each element separately which basically leads into the following pattern:



                >>> for left, right in enumerate(C.scene.objects):
                ... print (left, right, "Name:", right.name)
                0 <bpy_struct, Object("Cube")>, Name: Cube
                1 <bpy_struct, Object("Light")> Name: Light
                2 <bpy_struct, Object("Camera")> Name: Camera



                Also notice that python has a lot of awesome tools dealing with strings. In order to determine whether a string is part of a another string (substring), you could use the in operator to test its "membership":



                >>> "ell" in "HelloWorld"
                True


                To remove the objects, you might want to create some kind of black list before removing it. The following example removes all objects in the scene that have "Cam" in their name (substring):



                import bpy

                objects_to_remove = []

                for ob in bpy.context.scene.objects:
                if "Cam" in ob.name:
                objects_to_remove.append(ob)

                bpy.ops.object.delete("selected_objects": objects_to_remove)


                There is also str.startswith(prefix) returning True if the string begins with the prefix passed:



                >>> "Hello World".startswith("Hell")
                True


                The following example removes all found objects whose names begin with "Cam":



                import bpy

                objects_to_remove = []

                for ob in bpy.context.scene.objects:
                if ob.name.startswith("Cam"):
                objects_to_remove.append(ob)

                bpy.ops.object.delete("selected_objects": objects_to_remove)





                share|improve this answer











                $endgroup$

















                  3














                  3










                  3







                  $begingroup$

                  enumerate(iterable, start=0) returns a tuple for each element from any given iterable and its position based on the actual iteration. Proof using the Console:



                  >>> for i in enumerate(["Hello","World"]):
                  >>> print (i)
                  (0, "Hello")
                  (1, "World")


                  A tuple is a composite data type, composed out of two elements (item1, item2). If you want to access its individual components (left, right) you can either use the index operator [] or create 2 variables for the actual return value (tuple) on the fly: left, right = ("Hello", "World"):



                  >>> my_tuple = ("Hello", "World")
                  >>> my_tuple[0]
                  "Hello"

                  left, right = ("Hello", "World")
                  >>> left
                  "Hello"


                  Means in your case that you can't access the object and its name this way because the returned tuple by the enumerate() function in your loop has no name property. Proof using the Console again:



                  >>> for ob in enumerate(C.scene.objects):
                  ... print (ob)
                  (0, bpy.data.objects['Cube'])
                  (1, bpy.data.objects['Light'])
                  (2, bpy.data.objects['Camera'])


                  In summary: you have to find a way splitting the components of the tuple if you want to access each element separately which basically leads into the following pattern:



                  >>> for left, right in enumerate(C.scene.objects):
                  ... print (left, right, "Name:", right.name)
                  0 <bpy_struct, Object("Cube")>, Name: Cube
                  1 <bpy_struct, Object("Light")> Name: Light
                  2 <bpy_struct, Object("Camera")> Name: Camera



                  Also notice that python has a lot of awesome tools dealing with strings. In order to determine whether a string is part of a another string (substring), you could use the in operator to test its "membership":



                  >>> "ell" in "HelloWorld"
                  True


                  To remove the objects, you might want to create some kind of black list before removing it. The following example removes all objects in the scene that have "Cam" in their name (substring):



                  import bpy

                  objects_to_remove = []

                  for ob in bpy.context.scene.objects:
                  if "Cam" in ob.name:
                  objects_to_remove.append(ob)

                  bpy.ops.object.delete("selected_objects": objects_to_remove)


                  There is also str.startswith(prefix) returning True if the string begins with the prefix passed:



                  >>> "Hello World".startswith("Hell")
                  True


                  The following example removes all found objects whose names begin with "Cam":



                  import bpy

                  objects_to_remove = []

                  for ob in bpy.context.scene.objects:
                  if ob.name.startswith("Cam"):
                  objects_to_remove.append(ob)

                  bpy.ops.object.delete("selected_objects": objects_to_remove)





                  share|improve this answer











                  $endgroup$



                  enumerate(iterable, start=0) returns a tuple for each element from any given iterable and its position based on the actual iteration. Proof using the Console:



                  >>> for i in enumerate(["Hello","World"]):
                  >>> print (i)
                  (0, "Hello")
                  (1, "World")


                  A tuple is a composite data type, composed out of two elements (item1, item2). If you want to access its individual components (left, right) you can either use the index operator [] or create 2 variables for the actual return value (tuple) on the fly: left, right = ("Hello", "World"):



                  >>> my_tuple = ("Hello", "World")
                  >>> my_tuple[0]
                  "Hello"

                  left, right = ("Hello", "World")
                  >>> left
                  "Hello"


                  Means in your case that you can't access the object and its name this way because the returned tuple by the enumerate() function in your loop has no name property. Proof using the Console again:



                  >>> for ob in enumerate(C.scene.objects):
                  ... print (ob)
                  (0, bpy.data.objects['Cube'])
                  (1, bpy.data.objects['Light'])
                  (2, bpy.data.objects['Camera'])


                  In summary: you have to find a way splitting the components of the tuple if you want to access each element separately which basically leads into the following pattern:



                  >>> for left, right in enumerate(C.scene.objects):
                  ... print (left, right, "Name:", right.name)
                  0 <bpy_struct, Object("Cube")>, Name: Cube
                  1 <bpy_struct, Object("Light")> Name: Light
                  2 <bpy_struct, Object("Camera")> Name: Camera



                  Also notice that python has a lot of awesome tools dealing with strings. In order to determine whether a string is part of a another string (substring), you could use the in operator to test its "membership":



                  >>> "ell" in "HelloWorld"
                  True


                  To remove the objects, you might want to create some kind of black list before removing it. The following example removes all objects in the scene that have "Cam" in their name (substring):



                  import bpy

                  objects_to_remove = []

                  for ob in bpy.context.scene.objects:
                  if "Cam" in ob.name:
                  objects_to_remove.append(ob)

                  bpy.ops.object.delete("selected_objects": objects_to_remove)


                  There is also str.startswith(prefix) returning True if the string begins with the prefix passed:



                  >>> "Hello World".startswith("Hell")
                  True


                  The following example removes all found objects whose names begin with "Cam":



                  import bpy

                  objects_to_remove = []

                  for ob in bpy.context.scene.objects:
                  if ob.name.startswith("Cam"):
                  objects_to_remove.append(ob)

                  bpy.ops.object.delete("selected_objects": objects_to_remove)






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 2 hours ago

























                  answered 5 hours ago









                  brockmannbrockmann

                  1,5226 silver badges31 bronze badges




                  1,5226 silver badges31 bronze badges


























                      3













                      $begingroup$

                      When you use enumerate, an item in the list and an index/counter of where you are in the loop is returned to the variable (obj) instead of just the item. This is why you get the error, because in this case obj is a tuple ((index, listItem)) which does not have a name property itself.



                      The common loop syntax when using enumerate is:



                      for index, obj in enumerate(bpy.context.scene.objects):


                      obj will now contain the actual object as expected.



                      However, if you don't need an index of where you are in the loop you do not need to use enumerate. Remove it and it will work fine, e.g.:



                      for obj in bpy.context.scene.objects:


                      You could also access the second item in the tuple when trying to access the name:



                      namestr=obj[1].name





                      share|improve this answer









                      $endgroup$



















                        3













                        $begingroup$

                        When you use enumerate, an item in the list and an index/counter of where you are in the loop is returned to the variable (obj) instead of just the item. This is why you get the error, because in this case obj is a tuple ((index, listItem)) which does not have a name property itself.



                        The common loop syntax when using enumerate is:



                        for index, obj in enumerate(bpy.context.scene.objects):


                        obj will now contain the actual object as expected.



                        However, if you don't need an index of where you are in the loop you do not need to use enumerate. Remove it and it will work fine, e.g.:



                        for obj in bpy.context.scene.objects:


                        You could also access the second item in the tuple when trying to access the name:



                        namestr=obj[1].name





                        share|improve this answer









                        $endgroup$

















                          3














                          3










                          3







                          $begingroup$

                          When you use enumerate, an item in the list and an index/counter of where you are in the loop is returned to the variable (obj) instead of just the item. This is why you get the error, because in this case obj is a tuple ((index, listItem)) which does not have a name property itself.



                          The common loop syntax when using enumerate is:



                          for index, obj in enumerate(bpy.context.scene.objects):


                          obj will now contain the actual object as expected.



                          However, if you don't need an index of where you are in the loop you do not need to use enumerate. Remove it and it will work fine, e.g.:



                          for obj in bpy.context.scene.objects:


                          You could also access the second item in the tuple when trying to access the name:



                          namestr=obj[1].name





                          share|improve this answer









                          $endgroup$



                          When you use enumerate, an item in the list and an index/counter of where you are in the loop is returned to the variable (obj) instead of just the item. This is why you get the error, because in this case obj is a tuple ((index, listItem)) which does not have a name property itself.



                          The common loop syntax when using enumerate is:



                          for index, obj in enumerate(bpy.context.scene.objects):


                          obj will now contain the actual object as expected.



                          However, if you don't need an index of where you are in the loop you do not need to use enumerate. Remove it and it will work fine, e.g.:



                          for obj in bpy.context.scene.objects:


                          You could also access the second item in the tuple when trying to access the name:



                          namestr=obj[1].name






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 12 hours ago









                          Ray MairlotRay Mairlot

                          23.3k6 gold badges72 silver badges102 bronze badges




                          23.3k6 gold badges72 silver badges102 bronze badges






























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                              Tom Holland Mục lục Đầu đời và giáo dục | Sự nghiệp | Cuộc sống cá nhân | Phim tham gia | Giải thưởng và đề cử | Chú thích | Liên kết ngoài | Trình đơn chuyển hướngProfile“Person Details for Thomas Stanley Holland, "England and Wales Birth Registration Index, 1837-2008" — FamilySearch.org”"Meet Tom Holland... the 16-year-old star of The Impossible""Schoolboy actor Tom Holland finds himself in Oscar contention for role in tsunami drama"“Naomi Watts on the Prince William and Harry's reaction to her film about the late Princess Diana”lưu trữ"Holland and Pflueger Are West End's Two New 'Billy Elliots'""I'm so envious of my son, the movie star! British writer Dominic Holland's spent 20 years trying to crack Hollywood - but he's been beaten to it by a very unlikely rival"“Richard and Margaret Povey of Jersey, Channel Islands, UK: Information about Thomas Stanley Holland”"Tom Holland to play Billy Elliot""New Billy Elliot leaving the garage"Billy Elliot the Musical - Tom Holland - Billy"A Tale of four Billys: Tom Holland""The Feel Good Factor""Thames Christian College schoolboys join Myleene Klass for The Feelgood Factor""Government launches £600,000 arts bursaries pilot""BILLY's Chapman, Holland, Gardner & Jackson-Keen Visit Prime Minister""Elton John 'blown away' by Billy Elliot fifth birthday" (video with John's interview and fragments of Holland's performance)"First News interviews Arrietty's Tom Holland"“33rd Critics' Circle Film Awards winners”“National Board of Review Current Awards”Bản gốc"Ron Howard Whaling Tale 'In The Heart Of The Sea' Casts Tom Holland"“'Spider-Man' Finds Tom Holland to Star as New Web-Slinger”lưu trữ“Captain America: Civil War (2016)”“Film Review: ‘Captain America: Civil War’”lưu trữ“‘Captain America: Civil War’ review: Choose your own avenger”lưu trữ“The Lost City of Z reviews”“Sony Pictures and Marvel Studios Find Their 'Spider-Man' Star and Director”“‘Mary Magdalene’, ‘Current War’ & ‘Wind River’ Get 2017 Release Dates From Weinstein”“Lionsgate Unleashing Daisy Ridley & Tom Holland Starrer ‘Chaos Walking’ In Cannes”“PTA's 'Master' Leads Chicago Film Critics Nominations, UPDATED: Houston and Indiana Critics Nominations”“Nominaciones Goya 2013 Telecinco Cinema – ENG”“Jameson Empire Film Awards: Martin Freeman wins best actor for performance in The Hobbit”“34th Annual Young Artist Awards”Bản gốc“Teen Choice Awards 2016—Captain America: Civil War Leads Second Wave of Nominations”“BAFTA Film Award Nominations: ‘La La Land’ Leads Race”“Saturn Awards Nominations 2017: 'Rogue One,' 'Walking Dead' Lead”Tom HollandTom HollandTom HollandTom Hollandmedia.gettyimages.comWorldCat Identities300279794no20130442900000 0004 0355 42791085670554170004732cb16706349t(data)XX5557367