Resolving 'list indices must be integers or slices, not str' in Python











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list1 one is a dict of dfs. data1 is a lone df.



I want to create another dict of dfs(d) based off a crosstab which will categorize each column in each df in 'list1' on specific questions in 'data1'. This is what I've tried:



d={}
for df in list1:
for col in list1[df]:
d[df]= pd.DataFrame(pd.concat([pd.crosstab([df][col],data1['column1']),
pd.crosstab([df][col],data1['column2'])],
axis=1))


I then get 'TypeError: list indices must be integers or slices, not str'.



I appreciate that there have been many questions on this before on stackoverflow, however I have struggled to apply any of the solutions to my problem.










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    up vote
    0
    down vote

    favorite












    list1 one is a dict of dfs. data1 is a lone df.



    I want to create another dict of dfs(d) based off a crosstab which will categorize each column in each df in 'list1' on specific questions in 'data1'. This is what I've tried:



    d={}
    for df in list1:
    for col in list1[df]:
    d[df]= pd.DataFrame(pd.concat([pd.crosstab([df][col],data1['column1']),
    pd.crosstab([df][col],data1['column2'])],
    axis=1))


    I then get 'TypeError: list indices must be integers or slices, not str'.



    I appreciate that there have been many questions on this before on stackoverflow, however I have struggled to apply any of the solutions to my problem.










    share|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      list1 one is a dict of dfs. data1 is a lone df.



      I want to create another dict of dfs(d) based off a crosstab which will categorize each column in each df in 'list1' on specific questions in 'data1'. This is what I've tried:



      d={}
      for df in list1:
      for col in list1[df]:
      d[df]= pd.DataFrame(pd.concat([pd.crosstab([df][col],data1['column1']),
      pd.crosstab([df][col],data1['column2'])],
      axis=1))


      I then get 'TypeError: list indices must be integers or slices, not str'.



      I appreciate that there have been many questions on this before on stackoverflow, however I have struggled to apply any of the solutions to my problem.










      share|improve this question















      list1 one is a dict of dfs. data1 is a lone df.



      I want to create another dict of dfs(d) based off a crosstab which will categorize each column in each df in 'list1' on specific questions in 'data1'. This is what I've tried:



      d={}
      for df in list1:
      for col in list1[df]:
      d[df]= pd.DataFrame(pd.concat([pd.crosstab([df][col],data1['column1']),
      pd.crosstab([df][col],data1['column2'])],
      axis=1))


      I then get 'TypeError: list indices must be integers or slices, not str'.



      I appreciate that there have been many questions on this before on stackoverflow, however I have struggled to apply any of the solutions to my problem.







      python pandas






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      edited Nov 11 at 10:12









      Kanak

      2,63621223




      2,63621223










      asked Nov 11 at 10:00









      zepaft

      233




      233
























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          df iterates over the keys of list1 when you do for df in list1. So it is a string. Later, when you pass [df][col], it creates a list of one element which is a key, from which you try to get a string column. I suppose you meant d[df] instead.



          For clarity purpose I would also suggest renaming list1 and changing the for loop in for df in <new_name>.keys()






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            up vote
            0
            down vote













            df iterates over the keys of list1 when you do for df in list1. So it is a string. Later, when you pass [df][col], it creates a list of one element which is a key, from which you try to get a string column. I suppose you meant d[df] instead.



            For clarity purpose I would also suggest renaming list1 and changing the for loop in for df in <new_name>.keys()






            share|improve this answer

























              up vote
              0
              down vote













              df iterates over the keys of list1 when you do for df in list1. So it is a string. Later, when you pass [df][col], it creates a list of one element which is a key, from which you try to get a string column. I suppose you meant d[df] instead.



              For clarity purpose I would also suggest renaming list1 and changing the for loop in for df in <new_name>.keys()






              share|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                df iterates over the keys of list1 when you do for df in list1. So it is a string. Later, when you pass [df][col], it creates a list of one element which is a key, from which you try to get a string column. I suppose you meant d[df] instead.



                For clarity purpose I would also suggest renaming list1 and changing the for loop in for df in <new_name>.keys()






                share|improve this answer












                df iterates over the keys of list1 when you do for df in list1. So it is a string. Later, when you pass [df][col], it creates a list of one element which is a key, from which you try to get a string column. I suppose you meant d[df] instead.



                For clarity purpose I would also suggest renaming list1 and changing the for loop in for df in <new_name>.keys()







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 11 at 10:09









                Arthur Havlicek

                750411




                750411






























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