rename function with agg

[Jacques2101]

I just update to 1.1.3 and it seems that there is a problem (bug?) to rename function after an aggregate call. All the functions call are named so I can't use rename as previously.
How can I do to rename my function ?
I cannot do anymore

 .rename(columns={'<lambda_0>': 'function 0',
                  '<lambda_1>': 'function 1',....

(this is of course an example of functions names)

import pandas as pd
pdf = pd.DataFrame([[4, 9]] * 3, columns=['A', 'B'])
pdf.agg(["sum", lambda df: df.sum(), lambda df: df.sum()])

Output

		A	B
sum		12	27
<lambda>	12	27
<lambda>	12	27

Expected Output

		A	B
sum		12	27
<lambda_0>	12	27
<lambda_1>	12	27

Output of pd.show_versions()

INSTALLED VERSIONS

commit : db08276
python : 3.8.5.final.0
python-bits : 64
OS : Darwin
OS-release : 19.6.0
Version : Darwin Kernel Version 19.6.0: Mon Aug 31 22:12:52 PDT 2020; root:xnu-6153.141.2~1/RELEASE_X86_64
machine : x86_64
processor : i386
byteorder : little
LC_ALL : en_US.UTF-8
LANG : en_US.UTF-8
LOCALE : en_US.UTF-8

pandas : 1.1.3
numpy : 1.19.1
pytz : 2020.1
dateutil : 2.8.1
pip : 20.2.3
setuptools : 50.3.0.post20201006
Cython : None
pytest : None
hypothesis : None
sphinx : None
blosc : None
feather : None
xlsxwriter : None
lxml.etree : 4.5.2
html5lib : 1.1
pymysql : None
psycopg2 : None
jinja2 : 2.11.2
IPython : 7.18.1
pandas_datareader: 0.9.0
bs4 : 4.9.1
bottleneck : None
fsspec : None
fastparquet : None
gcsfs : None
matplotlib : 3.3.1
numexpr : None
odfpy : None
openpyxl : 3.0.5
pandas_gbq : None
pyarrow : 0.15.1
pytables : None
pyxlsb : None
s3fs : None
scipy : 1.5.0
sqlalchemy : None
tables : None
tabulate : 0.8.7
xarray : 0.16.1
xlrd : 1.2.0
xlwt : None
numba : None

[paste the output of pd.show_versions() here leaving a blank line after the details tag]

[TomAugspurger]

I believe the old behavior was considered a bug (xref #36189). I didn't follow that issue though, so can you confirm that @charlesdong1991?

[charlesdong1991]

Thanks @Jacques2101 for the issue post!!

Yes, the old behaviour was considered as a regression bug, see the discussion in #36189 (comment)

so to summarize, the index shouldn't be changed (at least for now no discussion to make it public), and such change (lambda_0) was shortly exposed to the public and has been fixed in 1.1.3

I hereby close the issue, but please feel free to ask if you have any doubts/questions!

[Jacques2101]

ok. It was great to have an index to change the name after...
I tried to change my code to (this is a toy data. Mine are more complicated) :

def func(df):
    return pd.Series([lambda x: x.sum(), 
                     lambda x: x.min()], 
                     index=['sum', 'min'])


pdf = pd.DataFrame([[4, 9]] * 3, columns=['A', 'B'])
pdf.apply(func)  

but now I have as output this. How to have the value of the function ?
Thx

A	B
<function func.. at 0x7f96bbd9...	<function func.. at 0x7f96b51c...
<function func.. at 0x7f96b51c...	<function func.. at 0x7f96b51c...

[charlesdong1991]

probably you've simplifed a lot for the example, based on your example, it looks like you don't need to put a function for this, what if you remove the function, and apply directly with series, something like:

s = pd.Series([lambda x: x.sum(), lambda x: x.min()], index=['sum', 'min'])
pdf = pd.DataFrame([[4, 9]] * 3, columns=['A', 'B'])
pdf.apply(s)

i think this should return result?

[Jacques2101]

indeed I have a result this time but I lost the name of the function and it was my objective.

A	B

lambda> 120 270
lambda> 4 9

[Jacques2101]

my data and functions are more like that :

def perf(prices, n):
    '''
    Calculates the annualized performance of a price series.
    '''
    prices_w = prices.resample('W').last()
    return 100*(prices_w.iloc[-1]/prices_w.iloc[-52*n])**(1/n)-100


func = pd.Series([lambda x: perf(x, 1),
                  lambda x: perf(x, 3)],
                 index=['f1', 'f2'])

ndata = 10000
dates = pd.date_range('20130101', periods=ndata)
df = pd.DataFrame(np.random.randint(low=1, high=100, size=(ndata, 7)), index=dates, columns=list('ABCDEFG')).T

df.apply(func, axis=1)

I have data with your solution but I lost again the name function (here f1, f2)

[charlesdong1991]

thanks for your follow-up. @Jacques2101

indeed I have a result this time but I lost the name of the function and it was my objective.

it is not supported to have the renamings by providing index names in series

I have data with your solution but I lost again the name function (here f1, f2)

as said, it is not supported this way. A quick solution would be below which I think is as simple as df.rename.

import pandas as pd
import numpy as np


def perf(prices, n):
    '''
    Calculates the annualized performance of a price series.
    '''
    return 100*(prices.iloc[-1]/prices.iloc[-n])**(1/n)-100


func = pd.Series([lambda x: perf(x, 1),
                  lambda x: perf(x, 10)],
                 index=['f1', 'f2'])


dates = pd.date_range('20130101', periods=1000)
df = pd.DataFrame(np.random.randint(low=1, high=100, size=(1000, 7)), index=dates, columns=list('ABCDEFG')).T

new_df = df.apply(func, axis=1)
new_df.columns = ['new_name1', 'new_name2']

[Jacques2101]

it works. It was simpler than I thought...thx
can I ask you an other thing :
in the 'perf' function I calculate the performance (based on weekly data whereas my data is daily) but I would like also have a function that calculate the calendar performance. I mean the performance between the last day of 2018 and last day of 2019 (the year is an input of my function).
Do you have any idea ?