Use of if/else in Dataframe.assign() results in ValueError: The truth value of a SeriesCheck if string is in a pandas dataframePandas data frame ValueError: The truth value of a Series is ambiguousDataFrame column comparison raises ValueError: The truth value of a Series is ambiguous.Python: If statement return ValueErrorValueError in pandas: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all()pandas add columns ,note The truth value of a Series is ambiguousValueError (while creating a function in python): The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all()Loop return error that the true value of the series is ambiguous

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Use of if/else in Dataframe.assign() results in ValueError: The truth value of a Series


Check if string is in a pandas dataframePandas data frame ValueError: The truth value of a Series is ambiguousDataFrame column comparison raises ValueError: The truth value of a Series is ambiguous.Python: If statement return ValueErrorValueError in pandas: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all()pandas add columns ,note The truth value of a Series is ambiguousValueError (while creating a function in python): The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all()Loop return error that the true value of the series is ambiguous






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









1

















I've got a ton of data transforms defined in a batch transform language that supports this structure: x = iif(condition, a, b). I want to rewrite these using dataframes.



I'm using Dataframe.assign() but get ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().



 import pandas as pd
df = pd.DataFrame(['apple', 'orange', 'granite'], columns=['name'])
df.assign(taste = lambda x: 'rocky' if (x.name=='granite') else 'yummy')

---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-39-e9ad71ccc45b> in <module>()
----> 1 df.assign(taste = lambda x: 'rocky' if (x.name=='granite') else 'yummy')

~Anaconda3libsite-packagespandascoreframe.py in assign(self, **kwargs)
3305 if PY36:
3306 for k, v in kwargs.items():
-> 3307 data[k] = com._apply_if_callable(v, data)
3308 else:
3309 # <= 3.5: do all calculations first...

~Anaconda3libsite-packagespandascorecommon.py in _apply_if_callable(maybe_callable, obj, **kwargs)
403
404 if callable(maybe_callable):
--> 405 return maybe_callable(obj, **kwargs)
406
407 return maybe_callable

<ipython-input-39-e9ad71ccc45b> in <lambda>(x)
----> 1 df.assign(taste = lambda x: 'rocky' if (x.name=='granite') else 'yummy')

~Anaconda3libsite-packagespandascoregeneric.py in __nonzero__(self)
1571 raise ValueError("The truth value of a 0 is ambiguous. "
1572 "Use a.empty, a.bool(), a.item(), a.any() or a.all()."
-> 1573 .format(self.__class__.__name__))
1574
1575 __bool__ = __nonzero__

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().


At first I thought this was due to a limitation of the keywords allowed in assign(), but a very similar construct works with apply():



df['name'].apply(lambda x: 'rocky' if (x=='granite') else 'yummy')

0 yummy
1 yummy
2 rocky
Name: name, dtype: object


However, this doesn't allow me use to an if-condition that uses multiple columns from the dataframe. Is there a way to get assign() to work?










share|improve this question
































    1

















    I've got a ton of data transforms defined in a batch transform language that supports this structure: x = iif(condition, a, b). I want to rewrite these using dataframes.



    I'm using Dataframe.assign() but get ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().



     import pandas as pd
    df = pd.DataFrame(['apple', 'orange', 'granite'], columns=['name'])
    df.assign(taste = lambda x: 'rocky' if (x.name=='granite') else 'yummy')

    ---------------------------------------------------------------------------
    ValueError Traceback (most recent call last)
    <ipython-input-39-e9ad71ccc45b> in <module>()
    ----> 1 df.assign(taste = lambda x: 'rocky' if (x.name=='granite') else 'yummy')

    ~Anaconda3libsite-packagespandascoreframe.py in assign(self, **kwargs)
    3305 if PY36:
    3306 for k, v in kwargs.items():
    -> 3307 data[k] = com._apply_if_callable(v, data)
    3308 else:
    3309 # <= 3.5: do all calculations first...

    ~Anaconda3libsite-packagespandascorecommon.py in _apply_if_callable(maybe_callable, obj, **kwargs)
    403
    404 if callable(maybe_callable):
    --> 405 return maybe_callable(obj, **kwargs)
    406
    407 return maybe_callable

    <ipython-input-39-e9ad71ccc45b> in <lambda>(x)
    ----> 1 df.assign(taste = lambda x: 'rocky' if (x.name=='granite') else 'yummy')

    ~Anaconda3libsite-packagespandascoregeneric.py in __nonzero__(self)
    1571 raise ValueError("The truth value of a 0 is ambiguous. "
    1572 "Use a.empty, a.bool(), a.item(), a.any() or a.all()."
    -> 1573 .format(self.__class__.__name__))
    1574
    1575 __bool__ = __nonzero__

    ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().


    At first I thought this was due to a limitation of the keywords allowed in assign(), but a very similar construct works with apply():



    df['name'].apply(lambda x: 'rocky' if (x=='granite') else 'yummy')

    0 yummy
    1 yummy
    2 rocky
    Name: name, dtype: object


    However, this doesn't allow me use to an if-condition that uses multiple columns from the dataframe. Is there a way to get assign() to work?










    share|improve this question




























      1












      1








      1


      1






      I've got a ton of data transforms defined in a batch transform language that supports this structure: x = iif(condition, a, b). I want to rewrite these using dataframes.



      I'm using Dataframe.assign() but get ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().



       import pandas as pd
      df = pd.DataFrame(['apple', 'orange', 'granite'], columns=['name'])
      df.assign(taste = lambda x: 'rocky' if (x.name=='granite') else 'yummy')

      ---------------------------------------------------------------------------
      ValueError Traceback (most recent call last)
      <ipython-input-39-e9ad71ccc45b> in <module>()
      ----> 1 df.assign(taste = lambda x: 'rocky' if (x.name=='granite') else 'yummy')

      ~Anaconda3libsite-packagespandascoreframe.py in assign(self, **kwargs)
      3305 if PY36:
      3306 for k, v in kwargs.items():
      -> 3307 data[k] = com._apply_if_callable(v, data)
      3308 else:
      3309 # <= 3.5: do all calculations first...

      ~Anaconda3libsite-packagespandascorecommon.py in _apply_if_callable(maybe_callable, obj, **kwargs)
      403
      404 if callable(maybe_callable):
      --> 405 return maybe_callable(obj, **kwargs)
      406
      407 return maybe_callable

      <ipython-input-39-e9ad71ccc45b> in <lambda>(x)
      ----> 1 df.assign(taste = lambda x: 'rocky' if (x.name=='granite') else 'yummy')

      ~Anaconda3libsite-packagespandascoregeneric.py in __nonzero__(self)
      1571 raise ValueError("The truth value of a 0 is ambiguous. "
      1572 "Use a.empty, a.bool(), a.item(), a.any() or a.all()."
      -> 1573 .format(self.__class__.__name__))
      1574
      1575 __bool__ = __nonzero__

      ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().


      At first I thought this was due to a limitation of the keywords allowed in assign(), but a very similar construct works with apply():



      df['name'].apply(lambda x: 'rocky' if (x=='granite') else 'yummy')

      0 yummy
      1 yummy
      2 rocky
      Name: name, dtype: object


      However, this doesn't allow me use to an if-condition that uses multiple columns from the dataframe. Is there a way to get assign() to work?










      share|improve this question















      I've got a ton of data transforms defined in a batch transform language that supports this structure: x = iif(condition, a, b). I want to rewrite these using dataframes.



      I'm using Dataframe.assign() but get ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().



       import pandas as pd
      df = pd.DataFrame(['apple', 'orange', 'granite'], columns=['name'])
      df.assign(taste = lambda x: 'rocky' if (x.name=='granite') else 'yummy')

      ---------------------------------------------------------------------------
      ValueError Traceback (most recent call last)
      <ipython-input-39-e9ad71ccc45b> in <module>()
      ----> 1 df.assign(taste = lambda x: 'rocky' if (x.name=='granite') else 'yummy')

      ~Anaconda3libsite-packagespandascoreframe.py in assign(self, **kwargs)
      3305 if PY36:
      3306 for k, v in kwargs.items():
      -> 3307 data[k] = com._apply_if_callable(v, data)
      3308 else:
      3309 # <= 3.5: do all calculations first...

      ~Anaconda3libsite-packagespandascorecommon.py in _apply_if_callable(maybe_callable, obj, **kwargs)
      403
      404 if callable(maybe_callable):
      --> 405 return maybe_callable(obj, **kwargs)
      406
      407 return maybe_callable

      <ipython-input-39-e9ad71ccc45b> in <lambda>(x)
      ----> 1 df.assign(taste = lambda x: 'rocky' if (x.name=='granite') else 'yummy')

      ~Anaconda3libsite-packagespandascoregeneric.py in __nonzero__(self)
      1571 raise ValueError("The truth value of a 0 is ambiguous. "
      1572 "Use a.empty, a.bool(), a.item(), a.any() or a.all()."
      -> 1573 .format(self.__class__.__name__))
      1574
      1575 __bool__ = __nonzero__

      ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().


      At first I thought this was due to a limitation of the keywords allowed in assign(), but a very similar construct works with apply():



      df['name'].apply(lambda x: 'rocky' if (x=='granite') else 'yummy')

      0 yummy
      1 yummy
      2 rocky
      Name: name, dtype: object


      However, this doesn't allow me use to an if-condition that uses multiple columns from the dataframe. Is there a way to get assign() to work?







      python pandas dataframe






      share|improve this question














      share|improve this question











      share|improve this question




      share|improve this question










      asked Mar 28 at 21:58









      Scott WilsonScott Wilson

      84 bronze badges




      84 bronze badges

























          2 Answers
          2






          active

          oldest

          votes


















          2


















          When calling Series.apply, the lambda receives each row value (i.e., a scalar value). With assign, the lambda receives the entire DataFrame. Understanding this means you can now do something such as



          df.assign(taste=lambda x: np.where(x['name'] == 'granite', 'r', 'y'))

          name taste
          0 apple y
          1 orange y
          2 granite r


          Or,



          df.assign(taste=np.where(df['name'] == 'granite', 'r', 'y'))
          name taste
          0 apple y
          1 orange y
          2 granite r


          Or, more simply, for in-place assignment,



          df['taste'] = np.where(df['name'] == 'granite', 'r', 'y')
          df

          name taste
          0 apple y
          1 orange y
          2 granite r





          share|improve this answer


























          • Perfect! Thanks for the answer and the explanation of why apply() worked in the initial formulation.

            – Scott Wilson
            Mar 29 at 17:38











          • @ScottWilson awesome ... Please accept one of our answers if your question was solved. Click the grey check to the left of the answer to toggle it green. Thanks.

            – cs95
            Mar 29 at 17:55


















          2


















          assign is not the function you should using with condition assignment



          df['taste']=np.where(df['name'].eq('granite'),'rocky','yummy')
          df
          Out[513]:
          name taste
          0 apple yummy
          1 orange yummy
          2 granite rocky





          share|improve this answer


























          • Thanks! The important point is to use np.where() in place of if/else. The answer below shows that assign() can be used too.

            – Scott Wilson
            Mar 29 at 17:40












          Your Answer






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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2


















          When calling Series.apply, the lambda receives each row value (i.e., a scalar value). With assign, the lambda receives the entire DataFrame. Understanding this means you can now do something such as



          df.assign(taste=lambda x: np.where(x['name'] == 'granite', 'r', 'y'))

          name taste
          0 apple y
          1 orange y
          2 granite r


          Or,



          df.assign(taste=np.where(df['name'] == 'granite', 'r', 'y'))
          name taste
          0 apple y
          1 orange y
          2 granite r


          Or, more simply, for in-place assignment,



          df['taste'] = np.where(df['name'] == 'granite', 'r', 'y')
          df

          name taste
          0 apple y
          1 orange y
          2 granite r





          share|improve this answer


























          • Perfect! Thanks for the answer and the explanation of why apply() worked in the initial formulation.

            – Scott Wilson
            Mar 29 at 17:38











          • @ScottWilson awesome ... Please accept one of our answers if your question was solved. Click the grey check to the left of the answer to toggle it green. Thanks.

            – cs95
            Mar 29 at 17:55















          2


















          When calling Series.apply, the lambda receives each row value (i.e., a scalar value). With assign, the lambda receives the entire DataFrame. Understanding this means you can now do something such as



          df.assign(taste=lambda x: np.where(x['name'] == 'granite', 'r', 'y'))

          name taste
          0 apple y
          1 orange y
          2 granite r


          Or,



          df.assign(taste=np.where(df['name'] == 'granite', 'r', 'y'))
          name taste
          0 apple y
          1 orange y
          2 granite r


          Or, more simply, for in-place assignment,



          df['taste'] = np.where(df['name'] == 'granite', 'r', 'y')
          df

          name taste
          0 apple y
          1 orange y
          2 granite r





          share|improve this answer


























          • Perfect! Thanks for the answer and the explanation of why apply() worked in the initial formulation.

            – Scott Wilson
            Mar 29 at 17:38











          • @ScottWilson awesome ... Please accept one of our answers if your question was solved. Click the grey check to the left of the answer to toggle it green. Thanks.

            – cs95
            Mar 29 at 17:55













          2














          2










          2









          When calling Series.apply, the lambda receives each row value (i.e., a scalar value). With assign, the lambda receives the entire DataFrame. Understanding this means you can now do something such as



          df.assign(taste=lambda x: np.where(x['name'] == 'granite', 'r', 'y'))

          name taste
          0 apple y
          1 orange y
          2 granite r


          Or,



          df.assign(taste=np.where(df['name'] == 'granite', 'r', 'y'))
          name taste
          0 apple y
          1 orange y
          2 granite r


          Or, more simply, for in-place assignment,



          df['taste'] = np.where(df['name'] == 'granite', 'r', 'y')
          df

          name taste
          0 apple y
          1 orange y
          2 granite r





          share|improve this answer














          When calling Series.apply, the lambda receives each row value (i.e., a scalar value). With assign, the lambda receives the entire DataFrame. Understanding this means you can now do something such as



          df.assign(taste=lambda x: np.where(x['name'] == 'granite', 'r', 'y'))

          name taste
          0 apple y
          1 orange y
          2 granite r


          Or,



          df.assign(taste=np.where(df['name'] == 'granite', 'r', 'y'))
          name taste
          0 apple y
          1 orange y
          2 granite r


          Or, more simply, for in-place assignment,



          df['taste'] = np.where(df['name'] == 'granite', 'r', 'y')
          df

          name taste
          0 apple y
          1 orange y
          2 granite r






          share|improve this answer













          share|improve this answer




          share|improve this answer










          answered Mar 28 at 22:06









          cs95cs95

          174k34 gold badges265 silver badges328 bronze badges




          174k34 gold badges265 silver badges328 bronze badges















          • Perfect! Thanks for the answer and the explanation of why apply() worked in the initial formulation.

            – Scott Wilson
            Mar 29 at 17:38











          • @ScottWilson awesome ... Please accept one of our answers if your question was solved. Click the grey check to the left of the answer to toggle it green. Thanks.

            – cs95
            Mar 29 at 17:55

















          • Perfect! Thanks for the answer and the explanation of why apply() worked in the initial formulation.

            – Scott Wilson
            Mar 29 at 17:38











          • @ScottWilson awesome ... Please accept one of our answers if your question was solved. Click the grey check to the left of the answer to toggle it green. Thanks.

            – cs95
            Mar 29 at 17:55
















          Perfect! Thanks for the answer and the explanation of why apply() worked in the initial formulation.

          – Scott Wilson
          Mar 29 at 17:38





          Perfect! Thanks for the answer and the explanation of why apply() worked in the initial formulation.

          – Scott Wilson
          Mar 29 at 17:38













          @ScottWilson awesome ... Please accept one of our answers if your question was solved. Click the grey check to the left of the answer to toggle it green. Thanks.

          – cs95
          Mar 29 at 17:55





          @ScottWilson awesome ... Please accept one of our answers if your question was solved. Click the grey check to the left of the answer to toggle it green. Thanks.

          – cs95
          Mar 29 at 17:55













          2


















          assign is not the function you should using with condition assignment



          df['taste']=np.where(df['name'].eq('granite'),'rocky','yummy')
          df
          Out[513]:
          name taste
          0 apple yummy
          1 orange yummy
          2 granite rocky





          share|improve this answer


























          • Thanks! The important point is to use np.where() in place of if/else. The answer below shows that assign() can be used too.

            – Scott Wilson
            Mar 29 at 17:40















          2


















          assign is not the function you should using with condition assignment



          df['taste']=np.where(df['name'].eq('granite'),'rocky','yummy')
          df
          Out[513]:
          name taste
          0 apple yummy
          1 orange yummy
          2 granite rocky





          share|improve this answer


























          • Thanks! The important point is to use np.where() in place of if/else. The answer below shows that assign() can be used too.

            – Scott Wilson
            Mar 29 at 17:40













          2














          2










          2









          assign is not the function you should using with condition assignment



          df['taste']=np.where(df['name'].eq('granite'),'rocky','yummy')
          df
          Out[513]:
          name taste
          0 apple yummy
          1 orange yummy
          2 granite rocky





          share|improve this answer














          assign is not the function you should using with condition assignment



          df['taste']=np.where(df['name'].eq('granite'),'rocky','yummy')
          df
          Out[513]:
          name taste
          0 apple yummy
          1 orange yummy
          2 granite rocky






          share|improve this answer













          share|improve this answer




          share|improve this answer










          answered Mar 28 at 22:05









          WeNYoBenWeNYoBen

          166k10 gold badges59 silver badges95 bronze badges




          166k10 gold badges59 silver badges95 bronze badges















          • Thanks! The important point is to use np.where() in place of if/else. The answer below shows that assign() can be used too.

            – Scott Wilson
            Mar 29 at 17:40

















          • Thanks! The important point is to use np.where() in place of if/else. The answer below shows that assign() can be used too.

            – Scott Wilson
            Mar 29 at 17:40
















          Thanks! The important point is to use np.where() in place of if/else. The answer below shows that assign() can be used too.

          – Scott Wilson
          Mar 29 at 17:40





          Thanks! The important point is to use np.where() in place of if/else. The answer below shows that assign() can be used too.

          – Scott Wilson
          Mar 29 at 17:40


















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          은진 송씨 목차 역사 본관 분파 인물 조선 왕실과의 인척 관계 집성촌 항렬자 인구 같이 보기 각주 둘러보기 메뉴은진 송씨세종실록 149권, 지리지 충청도 공주목 은진현