CODEWITHSUNDEEP
pythonpandas
jason m
jason m

Reputation: 6835

Splitting pandas dataframe into many chunks

Let's say I have a dataframe with the following structure:

    observation
d1  1
d2  1
d3  -1
d4  -1
d5  -1
d6  -1
d7  1
d8  1
d9  1
d10 1
d11 -1
d12 -1
d13 -1  
d14 -1
d15 -1
d16 1
d17 1
d18 1
d19 1
d20 1

Where d1:d20 is some datetime index (generalized here).

If I wanted to split d1:d2, d3:d6, d7:d10, etc into their own respective "chunks", how would I do that pythonically?

Note:

df1 = df[(df.observation==1)]
df2 = df[(df.observation==-1)]

is not what I want.

I can think of brute force ways, which would work, but are not wildly elegant.

Upvotes: 2

Views: 1600

Answers (2)

boot-scootin
boot-scootin

Reputation: 12515

Here's an example using real date.datetime objects as indices.

import pandas as pd
import numpy as np
import datetime
import random

df = pd.DataFrame({'x': np.random.randn(40)}, index = [date.fromordinal(random.randint(start_date, end_date)) for i in range(40)])

def filter_on_datetime(df, year = None, month = None, day = None):
    if all(d is not None for d in {year, month, day}):
        idxs = [idx for idx in df.index if idx.year == year and idx.month == month and idx.day == day]
    elif year is not None and month is not None and day is None:
        idxs = [idx for idx in df.index if idx.year == year and idx.month == month]
    elif year is not None and month is None and day is None:
        idxs = [idx for idx in df.index if idx.year == year]
    elif year is None and month is not None and day is not None:
        idxs = [idx for idx in df.index if idx.month == month and idx.day == day]
    elif year is None and month is None and day is not None:
        idxs = [idx for idx in df.index if idx.day == day]
    elif year is None and month is not None and day is None:
        idxs = [idx for idx in df.index if idx.month == month]
    elif year is not None and month is None and day is not None:
        idxs = [idx for idx in df.index if idx.year == year and idx.day == day] 
    else:
        idxs = df.index
    return df.ix[idxs]

Running this:

>>> print(filter_on_datetime(df = df, year = 2016, month = 2))
                   x
2016-02-01 -0.141557
2016-02-03  0.162429
2016-02-05  0.703794
2016-02-07 -0.184492
2016-02-09 -0.921793
2016-02-12  1.593838
2016-02-17  2.784899
2016-02-19  0.034721
2016-02-26 -0.142299

Upvotes: 0

akuiper
akuiper

Reputation: 215137

You can create a group variable based on the cumsum() of the diff() of the observation column where if the diff() is not equal to zero, assign a True value, thus every time a new value appears, a new group id will be created with the cumsum(), and then you can either apply standard analysis after groupby() with df.groupby((df.observation.diff() != 0).cumsum())...(other chained analysis here) or split them into smaller data frames with list-comprehension:

lst = [g for _, g in df.groupby((df.observation.diff() != 0).cumsum())]

lst[0]
# observation
#d1         1
#d2         1

lst[1]
# observation
#d3        -1
#d4        -1
#d5        -1
#d6        -1
...

Index chunks here:

[i.index for i in lst]

#[Index(['d1', 'd2'], dtype='object'),
# Index(['d3', 'd4', 'd5', 'd6'], dtype='object'),
# Index(['d7', 'd8', 'd9', 'd10'], dtype='object'),
# Index(['d11', 'd12', 'd13', 'd14', 'd15'], dtype='object'),
# Index(['d16', 'd17', 'd18', 'd19', 'd20'], dtype='object')]

Upvotes: 6

Related Questions