'DataFrame' object has no attribute 'sort'

Question

I face some problem here, in my python package I have install numpy, but I still have this error 'DataFrame' object has no attribute 'sort'

Anyone can give me some idea..

This is my code :

final.loc[-1] =['', 'P','Actual']
final.index = final.index + 1  # shifting index
final = final.sort()
final.columns=[final.columns,final.iloc[0]]
final = final.iloc[1:].reset_index(drop=True)
final.columns.names = (None, None)

Answer 1

sort() was deprecated for DataFrames in favor of either:

• sort_values() to sort by column(s)
• sort_index() to sort by the index

sort() was deprecated (but still available) in Pandas with release 0.17 (2015-10-09) with the introduction of sort_values() and sort_index() . It was removed from Pandas with release 0.20 (2017-05-05).

Answer 2

Pandas Sorting 101

sort has been replaced in v0.20 by DataFrame.sort_values and DataFrame.sort_index . Aside from this, we also have argsort .

Here are some common use cases in sorting, and how to solve them using the sorting functions in the current API. First, the setup.

# Setup
np.random.seed(0)
df = pd.DataFrame({'A': list('accab'), 'B': np.random.choice(10, 5)})    
df                                                                                                                                        
   A  B
0  a  7
1  c  9
2  c  3
3  a  5
4  b  2

Sort by Single Column

For example, to sort df by column "A", use sort_values with a single column name:

df.sort_values(by='A')

   A  B
0  a  7
3  a  5
4  b  2
1  c  9
2  c  3

If you need a fresh RangeIndex, use DataFrame.reset_index .

Sort by Multiple Columns

For example, to sort by both col "A" and "B" in df , you can pass a list to sort_values :

df.sort_values(by=['A', 'B'])

   A  B
3  a  5
0  a  7
4  b  2
2  c  3
1  c  9

Sort By DataFrame Index

df2 = df.sample(frac=1)
df2

   A  B
1  c  9
0  a  7
2  c  3
3  a  5
4  b  2

You can do this using sort_index :

df2.sort_index()

   A  B
0  a  7
1  c  9
2  c  3
3  a  5
4  b  2

df.equals(df2)                                                                                                                            
# False
df.equals(df2.sort_index())                                                                                                               
# True

Here are some comparable methods with their performance:

%timeit df2.sort_index()                                                                                                                  
%timeit df2.iloc[df2.index.argsort()]                                                                                                     
%timeit df2.reindex(np.sort(df2.index))                                                                                                   

605 µs ± 13.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
610 µs ± 24.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
581 µs ± 7.63 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Sort by List of Indices

For example,

idx = df2.index.argsort()
idx
# array([0, 7, 2, 3, 9, 4, 5, 6, 8, 1])

This "sorting" problem is actually a simple indexing problem. Just passing integer labels to iloc will do.

df.iloc[idx]

   A  B
1  c  9
0  a  7
2  c  3
3  a  5
4  b  2