'TypeError: can only concatenate str (not "numpy.float64") to str data set question

Please help Ive spent three hours on stack now i tried ''.join, str, removing the "+" for "," and nothing works to remove this error! The last 2 comments is where the error happens! #Description: Build a anime recommendation using python #Store the data df = pd.read_csv('animes.csv') #Show the first 3 rows of data df.head(2)

    #Count of the number of rows/animes in the data set and the number oof colums
    df.shape
    
    
    #List of wanted columns for anime recommendations
    columns =['title','synopsis','genre','aired','episodes']
    
    #Data updated
    df[columns].head(3)
    
    #Missing values check
    df[columns].isnull().values.any()
    
    #Create a funtion to combine the values of the new columns into a single string
    def get_new_features(data):
      new_features =[]
      for i in range (0, data.shape[0]):
         new_features.append(data['title'][i]+''+data['synopsis'][i]+''+data['genre'][i]+''+data['aired'][i]+''+data['episodes'][i])
    
         return new_features
     
    
    
     #Create a column to hold combine strings
     df ['new_features'] = get_new_features(df)
    
     #show data
     df.head(4)

    --------------------------------------------------------------------------
    TypeError                                 Traceback (most recent call last)
    <ipython-input-97-d5676456ab85> in <module>()
          1 #Create a column to hold combine strings
    ----> 2 df ['new_features'] = get_new_features(df)
          3 
          4 #show data
          5 df.head(4)
    
    <ipython-input-95-842623950c0e> in get_new_features(data)
          3   new_features =[]
          4   for i in range (0, data.shape[0]):
    ----> 5      new_features.append(data['title'][i]+''+data['synopsis'][i]+''+data['genre'][i]+''+data['aired'][i]+''+data['episodes'][i])
          6 
          7      return new_features
    
    TypeError: can only concatenate str (not "numpy.float64") to str


Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source