Aggregation in array element - python

Question

I have a OP : {'2017-05-06': [3, 7, 8],'2017-05-07': [3, 9, 10],'2017-05-08': [4]}

from the OP I just want another OP :

{'2017-05-06': [15, 11, 10],'2017-05-07': [19, 13, 12],'2017-05-08': [4]}

which means:
Ncleand is 2017-05-06
element total is 18 so '2017-05-06': [3 -18, 7-18, 8-18] = '2017-05-06': [15, 11, 10]
likewise all elements data.
So final output is {'2017-05-06': [15, 11, 10],'2017-05-07': [19, 13, 12],'2017-05-08': [4]}

How to do this?

Note : I am using python 3.6.2 and pandas 0.22.0

code so far :

import pandas as pd
dfs = pd.read_excel('ff2.xlsx', sheet_name=None)
dfs1 = {i:x.groupby(pd.to_datetime(x['date']).dt.strftime('%Y-%m-%d'))['duration'].sum() for i, x in dfs.items()}
d = pd.concat(dfs1).groupby(level=1).apply(list).to_dict()
actuald = pd.concat(dfs1).div(80).astype(int)
sum1 = actuald.groupby(level=1).transform('sum')
m = actuald.groupby(level=1).transform('size') > 1
cleand = sum1.sub(actuald).where(m, actuald).groupby(level=1).apply(list).to_dict()
print (cleand)

From the cleand I want to do this?

Why doesn't [4] become [0]? The element total of [4] is 4, and 4 - 4 is 0. — Aran-Fey
– Aran-Fey, Commented Mar 26, 2018 at 9:03
Also, please post your code. We expect help seekers to make an attempt to solve the problem, and show proof of that attempt. — Aran-Fey
– Aran-Fey, Commented Mar 26, 2018 at 9:05

Ohad Eytan · Accepted Answer · 2018-03-26 09:54:45Z

2

In a compact (but somehow inefficient) way:

>>> op = {'2017-05-06': [3, 7, 8],'2017-05-07': [3, 9, 10],'2017-05-08': [4]}
>>> { x:[sum(y)-i for i in y] if len(y)>1 else y for x,y in op.items() }

#output:
{'2017-05-06': [15, 11, 10], '2017-05-07': [19, 13, 12], '2017-05-08': [4]}

edited Mar 26, 2018 at 9:54

answered Mar 26, 2018 at 9:13

Ohad Eytan

8,5341 gold badge24 silver badges32 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

MMMMS Over a year ago

can we make if 2017-05-08': [0] then 2017-05-08': [1]? I mean make 1 where 0 in the element?

Ohad Eytan Over a year ago

Sure. That would be: { x:[sum(y)-i for i in y] if len(y)>1 else [1] if y == [0] else y for x,y in op.items() }

juanpa.arrivillaga Over a year ago

This code needlessly computes the sum again each iteration over each of the lists. Don't try to cram everything into a comprehension or one-liner at the expense of readability and in this case, performance.

MMMMS Over a year ago

But where there is [0,0,0] then its printing [0,0,0] instead of [1,1,1]

Veera Balla Deva · Accepted Answer · 2018-03-26 09:08:22Z

2

def get_list_manipulation(list_):
    subtraction = list_
    if len(list_) != 1:
        total = sum(list_)
        subtraction = [total-val for val in list_]
    return subtraction

for key, values in data.items():
    data[key] = get_list_manipulation(values)
>>>{'2017-05-06': [15, 11, 10], '2017-05-07': [19, 13, 12], '2017-05-08': [4]}

answered Mar 26, 2018 at 9:08

Veera Balla Deva

7886 silver badges19 bronze badges

Collectives™ on Stack Overflow

Aggregation in array element - python

2 Answers 2

4 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

4 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related