CODEWITHSUNDEEP
pythondatetimepandasgroupingpython-datetime
Ana
Ana

Reputation: 1586

How to group rows within a time period using Python

I have a DataFrame of some transactions. I want to group these transactions with respect to their item and time column values: the goal is to group items that are within 1 hour of each other. So we start a new group at the time of the next observation that wasn't within an hour of the observation prior (See column start time in DataFrame B).

Here is the data: I want to convert A to B.

A=
item    time             result
A   2016-04-18 13:08:25  Y
A   2016-04-18 13:57:05  N
A   2016-04-18 14:00:12  N
A   2016-04-18 23:45:50  Y
A   2016-04-20 16:53:48  Y
A   2016-04-20 17:11:47  N
B   2016-04-18 15:24:48  N
C   2016-04-23 13:20:44  N
C   2016-04-23 14:02:23  Y


B=
item    start time            end time      Ys  Ns  total count
A   2016-04-18 13:08:25 2016-04-18 14:08:25 1   2   3
A   2016-04-18 23:45:50 2016-04-18 00:45:50 1   0   1
A   2016-04-20 16:53:48 2016-04-20 17:53:48 1   1   2
B   2016-04-18 15:24:48 2016-04-18 16:24:48 0   1   1
C   2016-04-23 13:20:44 2016-04-23 14:20:44 1   1   2

Here is what I did:

grouped = A.groupby('item')
A['end'] = (grouped['time'].transform(lambda grp: grp.min()+pd.Timedelta(hours=1)))
A2 = A.loc[(A['time'] <= A['end'])]

This gives me one group per day: the transaction within 1 hour of the first transaction. So, I'm missing other transactions in the same day but more than 1 hour apart from the first. My struggle is how to get those groups. I can then use pd.crosstab to get the details I want from the result column.

Another idea I have is to sort A by item and time, and then go row by row. If the time is within 1 hour of the previous row, it adds to that group, otherwise, it creates a new group.

Upvotes: 3

Views: 1906

Answers (3)

Stefan
Stefan

Reputation: 42885

1) Set up a window_end column for later use with .groupby(), and define .get_windows() to check, for each item group, if a row fits the current current 1hr window, or do nothing and keep the initialized value. Apply to all item groups:

df['window_end'] = df.time + pd.Timedelta('1H')

def get_windows(data):
    window_end = data.iloc[0].window_end
    for index, row in data.iloc[1:].iterrows():
        if window_end > row.time:
            df.loc[index, 'window_end'] = window_end
        else:
            window_end = row.window_end

df.groupby('item').apply(lambda x: get_windows(x))

2) Use windows and item with .groupby() and return .value_counts() as transposed DataFrame, clean up index, and add total:

df = df.groupby(['window_end', 'item']).result.apply(lambda x: x.value_counts().to_frame().T)
df = df.fillna(0).astype(int).reset_index(level=2, drop=True)
df['total'] = df.sum(axis=1)

to get:

                            N  Y  total
window_end          item               
2016-04-18 14:08:25 A    A  2  1      3
2016-04-18 16:24:48 B    B  1  0      1
2016-04-19 00:45:50 A    A  0  1      1
2016-04-20 17:53:48 A    A  1  1      2
2016-04-23 14:20:44 C    C  1  1      2

Upvotes: 1

piRSquared
piRSquared

Reputation: 294218

Setup

import pandas as pd
from StringIO import StringIO

text = """item    time             result
A   2016-04-18 13:08:25  Y
A   2016-04-18 13:57:05  N
A   2016-04-18 14:00:12  N
A   2016-04-18 23:45:50  Y
A   2016-04-20 16:53:48  Y
A   2016-04-20 17:11:47  N
B   2016-04-18 15:24:48  N
C   2016-04-23 13:20:44  N
C   2016-04-23 14:02:23  Y
"""

df = pd.read_csv(StringIO(text), delimiter="\s{2,}", parse_dates=[1], engine='python')

Solution

I needed to create a few process functions:

def set_time_group(df):
    cur_time = pd.NaT
    for index, row in df.iterrows():
        if pd.isnull(cur_time):
            cur_time = row.time
        delta = row.time - cur_time
        if delta.seconds / 3600. < 1:
            df.loc[index, 'time_ref'] = cur_time
        else:
            df.loc[index, 'time_ref'] = row.time
            cur_time = row.time
    return df

def summarize_results(df):
    df_ = df.groupby('result').count().iloc[:, 0]
    df_.loc['total count'] = df_.sum()
    return df_

dfg1 = df.groupby('item').apply(set_time_group)
dfg2 = dfg1.groupby(['item', 'time_ref']).apply(summarize_results)
df_f = dfg2.unstack().fillna(0)

Demonstration

print df_f

result                      N    Y  total count
item time_ref                                  
A    2016-04-18 13:08:25  2.0  1.0          3.0
     2016-04-18 23:45:50  0.0  1.0          1.0
     2016-04-20 16:53:48  1.0  1.0          2.0
B    2016-04-18 15:24:48  1.0  0.0          1.0
C    2016-04-23 13:20:44  1.0  1.0          2.0

Upvotes: 1

MaxU - stand with Ukraine
MaxU - stand with Ukraine

Reputation: 210822

inspired (+1) by Stefan's solution I came to this one:

B = (A.groupby(['item', A.groupby('item')['time']
                         .diff().fillna(0).dt.total_seconds()//60//60
               ],
               as_index=False)['time'].min()
)


B[['N','Y']] = (A.groupby(['item', A.groupby('item')['time']
                                    .diff().fillna(0).dt.total_seconds()//60//60
                          ])['result']
                 .apply(lambda x: x.value_counts().to_frame().T).fillna(0)
                 .reset_index()[['N','Y']]
)

Output:

In [178]: B
Out[178]:
  item                time    N    Y
0    A 2016-04-18 13:08:25  3.0  1.0
1    A 2016-04-18 23:45:50  0.0  1.0
2    A 2016-04-20 16:53:48  0.0  1.0
3    B 2016-04-18 15:24:48  1.0  0.0
4    C 2016-04-23 13:20:44  1.0  1.0

PS the idea is to use A.groupby('item')['time'].diff().fillna(0).dt.total_seconds()//60//60 as a part of grouping:

In [179]: A.groupby('item')['time'].diff().fillna(0).dt.total_seconds()//60//60
Out[179]:
0     0.0
1     0.0
2     0.0
3     9.0
4    41.0
5     0.0
6     0.0
7     0.0
8     0.0
Name: time, dtype: float64

Upvotes: 1

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