CODEWITHSUNDEEP
pythonpython-3.xpandas
Josiah Hulsey
Josiah Hulsey

Reputation: 499

Pandas .shift function yields a "'float' object has no attribute 'shift'", 'occurred at index 0')

I am trying to create a new column in a pandas dataframe using a very complex if statement (I have simplified it for the sake of clarity below). I keep getting the error: ("'float' object has no attribute 'shift'", 'occurred at index 0'). I have looked around stack/the internet and have not come up with a great answer for my solution. Some answers involve taking the .shift out of a function, however, I need to have it within a function due to the complex nature of the if statement I am writing.

I have attached an image below detailing what I ultimately want the function to do. I believe it explains it better than I could describe it with words. Any help or guidance would be greatly appreciated.

Please let me know if you have any questions or if I can clarify anything!

Ultimate end result desired

Code example

df=pd.read_csv(file)

def ubk (df):
    x = df['k_calc'].shift(1)  
    if x <90 :
        return 1
    elif x >90:
        return 2

df['test'] = df.apply(ubk,axis = 1)

Upvotes: 1

Views: 5490

Answers (2)

Nakor
Nakor

Reputation: 1514

Why don't you just do this:

df['test'] = 1+(df['k_calc'].shift(1)>=90).astype(int)

The error you get is because you might be misunderstanding what apply does. When you do df.apply(ubk,axis = 1), pandas will apply ubk to every row in your dataframe. As a result, in your function call, df is not your original dataframe but one of its rows. So when you do x = df['k_calc'].shift(1), since df['k_calc'] is a single entry (a float), pandas complains: he doesn't know any shift() method for float.

Upvotes: 1

Andy L.
Andy L.

Reputation: 25269

You may pass additional parameter to apply if you want. In this case you may pass the main df and your ubk handles/processes it as you want. I don't know exact purpose of your ubk, so I just modify ubk to accomplish what you describe for column test. It seems your logic is not efficient, but you may have your own reason to use it. So, it is up to you.

sample data:

In [301]: df
Out[301]:
   lowest_low   k_calc    d_cal
0        9.07  75.0000      NaN
1        9.07  79.7297      NaN
2        9.07  92.5675      NaN
3        9.07  66.2116  78.3772

function and call apply to create test columns with condition: if previous cell of k_calc < 90 returns 1, > 90 returns 2 as follows

def ubk (s, m_df):
    x = m_df['k_calc'].shift(1)[s.name]  
    if x <90 :
        return 1
    elif x >90:
        return 2

df['test'] = df.apply(ubk, axis=1, args=(df,))

Out[304]:
   lowest_low   k_calc    d_cal  test
0        9.07  75.0000      NaN   NaN
1        9.07  79.7297      NaN   1.0
2        9.07  92.5675      NaN   1.0
3        9.07  66.2116  78.3772   2.0

Upvotes: 3

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