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Groups > comp.lang.python > #85578 > unrolled thread

Group by interval time

Started bycharles.sartori@gmail.com
First post2015-02-12 04:34 -0800
Last post2015-02-12 06:54 -0800
Articles 5 — 4 participants

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  Group by interval time charles.sartori@gmail.com - 2015-02-12 04:34 -0800
    Re: Group by interval time MRAB <python@mrabarnett.plus.com> - 2015-02-12 13:17 +0000
    Re: Group by interval time alister <alister.nospam.ware@ntlworld.com> - 2015-02-12 13:29 +0000
    Re: Group by interval time Peter Otten <__peter__@web.de> - 2015-02-12 14:29 +0100
      Re: Group by interval time charles.sartori@gmail.com - 2015-02-12 06:54 -0800

#85578 — Group by interval time

Fromcharles.sartori@gmail.com
Date2015-02-12 04:34 -0800
SubjectGroup by interval time
Message-ID<6f6f567d-7e25-4a74-96ef-d20cb44a9646@googlegroups.com>
Hello there!

I`m trying to group by a list of Row() objects in 12days interval and sum(). values. Here is an example of the list

[Row(time=datetime.datetime(2013, 1, 1, 0, 0), sum=4676557380615), Row(time=datetime.datetime(2013, 1, 2, 0, 0), sum=6549630855895), Row(time=datetime.datetime(2013, 1, 3, 0, 0), sum=6549630855895), ...]

Row() objects has two vars: row.time and row.sum


The result that I`m looking for is: 
[[datetime.datetime(2013, 1, 1, 0, 0), value],
[datetime.datetime(2013, 1, 12, 0, 0), value],
[datetime.datetime(2013, 1, 24, 0, 0), value]
]
Where value is the sum() of all row.sum in that interval.

I`m trying to use itertools.groupby by I could not get it to work yet.

Thnak you.

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#85580

FromMRAB <python@mrabarnett.plus.com>
Date2015-02-12 13:17 +0000
Message-ID<mailman.18689.1423747073.18130.python-list@python.org>
In reply to#85578
On 2015-02-12 12:34, charles.sartori@gmail.com wrote:
> Hello there!
>
> I`m trying to group by a list of Row() objects in 12days interval and sum(). values. Here is an example of the list
>
> [Row(time=datetime.datetime(2013, 1, 1, 0, 0), sum=4676557380615), Row(time=datetime.datetime(2013, 1, 2, 0, 0), sum=6549630855895), Row(time=datetime.datetime(2013, 1, 3, 0, 0), sum=6549630855895), ...]
>
> Row() objects has two vars: row.time and row.sum
>
>
> The result that I`m looking for is:
> [[datetime.datetime(2013, 1, 1, 0, 0), value],
> [datetime.datetime(2013, 1, 12, 0, 0), value],
> [datetime.datetime(2013, 1, 24, 0, 0), value]
> ]
> Where value is the sum() of all row.sum in that interval.
>
> I`m trying to use itertools.groupby by I could not get it to work yet.
>
Try grouping by the number of days since datetime.datetime(2013, 1, 1, 
0, 0), integer-divided by 12.

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#85581

Fromalister <alister.nospam.ware@ntlworld.com>
Date2015-02-12 13:29 +0000
Message-ID<mbi9rt$o7c$1@speranza.aioe.org>
In reply to#85578
On Thu, 12 Feb 2015 04:34:03 -0800, charles.sartori wrote:

> Hello there!
> 
> I`m trying to group by a list of Row() objects in 12days interval and
> sum(). values. Here is an example of the list
> 
> [Row(time=datetime.datetime(2013, 1, 1, 0, 0), sum=4676557380615),
> Row(time=datetime.datetime(2013, 1, 2, 0, 0), sum=6549630855895),
> Row(time=datetime.datetime(2013, 1, 3, 0, 0), sum=6549630855895), ...]
> 
> Row() objects has two vars: row.time and row.sum
> 
> 
> The result that I`m looking for is:
> [[datetime.datetime(2013, 1, 1, 0, 0), value],
> [datetime.datetime(2013, 1, 12, 0, 0), value],
> [datetime.datetime(2013, 1, 24, 0, 0), value]
> ]
> Where value is the sum() of all row.sum in that interval.
> 
> I`m trying to use itertools.groupby by I could not get it to work yet.
> 
> Thnak you.

what is the code you have tried & what errors are you getting?



-- 
life, n.:
	A whim of several billion cells to be you for a while.

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#85582

FromPeter Otten <__peter__@web.de>
Date2015-02-12 14:29 +0100
Message-ID<mailman.18690.1423747817.18130.python-list@python.org>
In reply to#85578
charles.sartori@gmail.com wrote:

> Hello there!
> 
> I`m trying to group by a list of Row() objects in 12days interval and
> sum(). values. Here is an example of the list
> 
> [Row(time=datetime.datetime(2013, 1, 1, 0, 0), sum=4676557380615),
> [Row(time=datetime.datetime(2013, 1, 2, 0, 0), sum=6549630855895),
> [Row(time=datetime.datetime(2013, 1, 3, 0, 0), sum=6549630855895), ...]
> 
> Row() objects has two vars: row.time and row.sum
> 
> 
> The result that I`m looking for is:
> [[datetime.datetime(2013, 1, 1, 0, 0), value],
> [datetime.datetime(2013, 1, 12, 0, 0), value],
> [datetime.datetime(2013, 1, 24, 0, 0), value]
> ]
> Where value is the sum() of all row.sum in that interval.
> 
> I`m trying to use itertools.groupby by I could not get it to work yet.

If the data is sorted by time then you can use groupby, otherwise consider 
putting it into a (default)dict. Here is an example for both methods:

import datetime
import random
from itertools import groupby
from collections import defaultdict, namedtuple

BASEDATE = datetime.datetime(2015, 1, 1)
INTERVAL = 12 # time interval in days
DAYS = datetime.timedelta(days=INTERVAL)

def make_sample_rows():
    random.seed(42)
    Row = namedtuple("Row", "time sum")
    return [
        Row(BASEDATE + datetime.timedelta(days=random.randrange(-20, 80)),
            random.randrange(300))
        for i in range(30)]

def get_key(row):
    offset = (row.time - BASEDATE).days // INTERVAL
    return BASEDATE + datetime.timedelta(days=offset)

def format_time(time):
    return time.strftime("%Y-%m-%d")

if __name__ == "__main__":
    rows = make_sample_rows()

    # with groupby()
    for key, group in groupby(sorted(rows), key=get_key):
        print("{} - {}".format(format_time(key), format_time(key+DAYS)))
        print("-" * 23)

        group = list(group)
        for row in group:
            print("{} {:4}".format(format_time(row.time), row.sum))
        print("{:>15}".format("----"))
        print("{:15}".format(sum(row.sum for row in group)))
        print("")

    # with defaultdict
    d = defaultdict(int)
    for row in rows:
        d[get_key(row)] += row.sum

    for time, sum in sorted(d.items()):
        print("{} {:4}".format(format_time(time), sum))

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#85585

Fromcharles.sartori@gmail.com
Date2015-02-12 06:54 -0800
Message-ID<07d750f9-04be-457c-9f29-bd4db700aff7@googlegroups.com>
In reply to#85582
Thank you Peter, I was doing wrong at get_key function...

Thnak you so much!

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