python, sort descending dataframe with pandas

Question

I'm trying to sort a dataframe by descending. I put 'False' in the ascending argument, but my order is still ascending.

My code is:

from pandas import DataFrame
import pandas as pd

d = {'one':[2,3,1,4,5],
     'two':[5,4,3,2,1],
     'letter':['a','a','b','b','c']}

df = DataFrame(d)

test = df.sort(['one'], ascending=[False])

but the output is

  letter  one  two
2      b    1    3
0      a    2    5
1      a    3    4
3      b    4    2
4      c    5    1

Answer 1

New syntax (either):

 test = df.sort_values(['one'], ascending=[False])
 test = df.sort_values(['one'], ascending=[0])

Answer 2

Edit: This is out of date, see @Merlin's answer.

[False] , being a nonempty list , is not the same as False . You should write:

test = df.sort('one', ascending=False)

Answer 3

For pandas 0.17 and above, use this :

test = df.sort_values('one', ascending=False)

Since 'one' is a series in the pandas data frame, hence pandas will not accept the arguments in the form of a list.

Answer 4

from pandas import DataFrame
import pandas as pd

d = {'one':[2,3,1,4,5],
 'two':[5,4,3,2,1],
 'letter':['a','a','b','b','c']}

df = DataFrame(d)

test = df.sort_values(['one'], ascending=False)
test

Answer 5

from pandas import DataFrame import pandas as pd

d = {'one':[2,3,1,4,5], 'two':[5,4,3,2,1], 'letter':['a','a','b','b','c']}

df = DataFrame(d)

#check below statement

df.groupby(["one"]).count().sort_values("one", ascending=False)

Answer 6

https://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.sort_values.html

I don't think you should ever provide the False value in square brackets (ever), also the column values when they are more than one, then only they are provided as a list! Not like ['one'] .

test = df.sort_values(by='one', ascending = False)