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Calculate streak in pandas without apply - python

I have a DataFrame like this:
date | type | column1
----------------------------
2019-01-01 | A | 1
2019-02-01 | A | 1
2019-03-01 | A | 1
2019-04-01 | A | 0
2019-05-01 | A | 1
2019-06-01 | A | 1
2019-07-01 | B | 1
2019-08-01 | B | 1
2019-09-01 | B | 0
I want to have a column called "streak" that has a streak, but grouped by column "type":
date | type | column1 | streak
-------------------------------------
2019-01-01 | A | 1 | 1
2019-02-01 | A | 1 | 2
2019-03-01 | A | 1 | 3
2019-04-01 | A | 0 | 0
2019-05-01 | A | 1 | 1
2019-06-01 | A | 1 | 2
2019-07-01 | B | 1 | 1
2019-08-01 | B | 1 | 2
2019-09-01 | B | 0 | 0
I managed to do it like that:
def streak(df):
grouper = (df.column1 != df.column1.shift(1)).cumsum()
df['streak'] = df.groupby(grouper).cumsum()['column1']
return df
df = df.groupby(['type']).apply(streak)
But I'm wondering if it's possible to do it inline without using a groupby and apply, because my DataFrame contains about 100M rows and it takes several hours to process.
Any ideas on how to optimize this for speed?

You want the cumsum of 'column1' grouping by 'type' + the cumsum of a Boolean Series which resets the grouping at every 0.
df['streak'] = df.groupby(['type', df.column1.eq(0).cumsum()]).column1.cumsum()
date type column1 streak
0 2019-01-01 A 1 1
1 2019-02-01 A 1 2
2 2019-03-01 A 1 3
3 2019-04-01 A 0 0
4 2019-05-01 A 1 1
5 2019-06-01 A 1 2
6 2019-07-01 B 1 1
7 2019-08-01 B 1 2
8 2019-09-01 B 0 0

IIUC, this is what you need.
m = df.column1.ne(df.column1.shift()).cumsum()
df['streak'] =df.groupby([m , 'type'])['column1'].cumsum()
Output
date type column1 streak
0 1/1/2019 A 1 1
1 2/1/2019 A 1 2
2 3/1/2019 A 1 3
3 4/1/2019 A 0 0
4 5/1/2019 A 1 1
5 6/1/2019 A 1 2
6 7/1/2019 B 1 1
7 8/1/2019 B 1 2
8 9/1/2019 B 0 0

Related

I have a dataframe that counts the number of times an event has occured per user per day. Users may have 0 events per day and (since the table is an aggregate from a raw event log) rows with 0 events are missing from the dataframe. I would like to add these missing rows and group the data by week so that each user has one entry per week (including 0 if applicable).
Here is an example of my input:
import numpy as np
import pandas as pd
np.random.seed(42)
df = pd.DataFrame({
"person_id": np.arange(3).repeat(5),
"date": pd.date_range("2022-01-01", "2022-01-15", freq="d"),
"event_count": np.random.randint(1, 7, 15),
})
# end of each week
# Note: week 2022-01-23 is not in df, but should be part of the result
desired_index = pd.to_datetime(["2022-01-02", "2022-01-09", "2022-01-16", "2022-01-23"])
df
| | person_id | date | event_count |
|---:|------------:|:--------------------|--------------:|
| 0 | 0 | 2022-01-01 00:00:00 | 4 |
| 1 | 0 | 2022-01-02 00:00:00 | 5 |
| 2 | 0 | 2022-01-03 00:00:00 | 3 |
| 3 | 0 | 2022-01-04 00:00:00 | 5 |
| 4 | 0 | 2022-01-05 00:00:00 | 5 |
| 5 | 1 | 2022-01-06 00:00:00 | 2 |
| 6 | 1 | 2022-01-07 00:00:00 | 3 |
| 7 | 1 | 2022-01-08 00:00:00 | 3 |
| 8 | 1 | 2022-01-09 00:00:00 | 3 |
| 9 | 1 | 2022-01-10 00:00:00 | 5 |
| 10 | 2 | 2022-01-11 00:00:00 | 4 |
| 11 | 2 | 2022-01-12 00:00:00 | 3 |
| 12 | 2 | 2022-01-13 00:00:00 | 6 |
| 13 | 2 | 2022-01-14 00:00:00 | 5 |
| 14 | 2 | 2022-01-15 00:00:00 | 2 |
This is how my desired result looks like:
| | person_id | level_1 | event_count |
|---:|------------:|:--------------------|--------------:|
| 0 | 0 | 2022-01-02 00:00:00 | 9 |
| 1 | 0 | 2022-01-09 00:00:00 | 13 |
| 2 | 0 | 2022-01-16 00:00:00 | 0 |
| 3 | 0 | 2022-01-23 00:00:00 | 0 |
| 4 | 1 | 2022-01-02 00:00:00 | 0 |
| 5 | 1 | 2022-01-09 00:00:00 | 11 |
| 6 | 1 | 2022-01-16 00:00:00 | 5 |
| 7 | 1 | 2022-01-23 00:00:00 | 0 |
| 8 | 2 | 2022-01-02 00:00:00 | 0 |
| 9 | 2 | 2022-01-09 00:00:00 | 0 |
| 10 | 2 | 2022-01-16 00:00:00 | 20 |
| 11 | 2 | 2022-01-23 00:00:00 | 0 |
I can produce it using:
(
df
.groupby(["person_id", pd.Grouper(key="date", freq="w")]).sum()
.groupby("person_id").apply(
lambda df: (
df
.reset_index(drop=True, level=0)
.reindex(desired_index, fill_value=0))
)
.reset_index()
)
However, according to the docs of reindex, I should be able to use it with level=1 as a kwarg directly and without having to do another groupby. However, when I do this I get an "inner join" of the two indices instead of an "outer join":
result = (
df
.groupby(["person_id", pd.Grouper(key="date", freq="w")]).sum()
.reindex(desired_index, level=1)
.reset_index()
)
| | person_id | date | event_count |
|---:|------------:|:--------------------|--------------:|
| 0 | 0 | 2022-01-02 00:00:00 | 9 |
| 1 | 0 | 2022-01-09 00:00:00 | 13 |
| 2 | 1 | 2022-01-09 00:00:00 | 11 |
| 3 | 1 | 2022-01-16 00:00:00 | 5 |
| 4 | 2 | 2022-01-16 00:00:00 | 20 |
Why is that, and how am I supposed to use df.reindex correctly?
I have found a similar SO question on reindexing a multi-index level, but the accepted answer there uses df.unstack, which doesn't work for me, because not every level of my desired index occurs in my current index (and vice versa).

You need reindex by both levels of MultiIndex:
mux = pd.MultiIndex.from_product([df['person_id'].unique(), desired_index],
names=['person_id','date'])
result = (
df
.groupby(["person_id", pd.Grouper(key="date", freq="w")]).sum()
.reindex(mux, fill_value=0)
.reset_index()
)
print (result)
person_id date event_count
0 0 2022-01-02 9
1 0 2022-01-09 13
2 0 2022-01-16 0
3 0 2022-01-23 0
4 1 2022-01-02 0
5 1 2022-01-09 11
6 1 2022-01-16 5
7 1 2022-01-23 0
8 2 2022-01-02 0
9 2 2022-01-09 0
10 2 2022-01-16 20
11 2 2022-01-23 0

I have a df that looks like this:
date | user_id | purchase
2020-01-01 | 1 | 10
2020-10-01 | 1 | 12
2020-15-01 | 1 | 5
2020-11-01 | 2 | 500 ...
Now, I want to add an n_day retention flag for each user_id in my df. The expected output should look like:
date | user_id | purchase | 3D_retention (did user purchase within next 3 days)
2020-01-01 | 1 | 10 | 0 (because there was no purchase on/before 2020-04-01 after 2020-01-01
2020-10-01 | 1 | 12 | 1 (because there was a purchase on 2020-11-01 which was within 3 days from 2020-10-01
2020-11-01 | 1 | 5 | 0
What is the best way of doing this in pandas?

i modified the date to be as yyyy-mm-dd format
date user_id purchase
0 2020-01-01 1 10
1 2020-01-10 1 12
2 2020-01-15 1 5
3 2020-01-11 2 500
df['date']=pd.to_datetime(df['date'])
next_purchase_days =6
df['retention']=df.groupby('user_id')['date'].transform(lambda x: ((x.shift(-1) - x).dt.days< next_purchase_days).astype(int) )
df
df
date user_id purchase retention
0 2020-01-01 1 10 0
1 2020-01-10 1 12 1
2 2020-01-15 1 5 0
3 2020-01-11 2 500 0

I have a dataframe that looks something like:
+---+----+---------------+------------+------------+
| | id | date1 | date2 | days_ahead |
+---+----+---------------+------------+------------+
| 0 | 1 | 2021-10-21 | 2021-10-24 | 3 |
| 1 | 1 | 2021-10-22 | NaN | NaN |
| 2 | 1 | 2021-11-16 | 2021-11-24 | 8 |
| 3 | 2 | 2021-10-22 | 2021-10-24 | 2 |
| 4 | 2 | 2021-10-22 | 2021-10-24 | 2 |
| 5 | 3 | 2021-10-26 | 2021-10-31 | 5 |
| 6 | 3 | 2021-10-30 | 2021-11-04 | 5 |
| 7 | 3 | 2021-11-02 | NaN | NaN |
| 8 | 3 | 2021-11-04 | 2021-11-04 | 0 |
| 9 | 4 | 2021-10-28 | NaN | NaN |
+---+----+---------------+------------+------------+
I am trying to fill the missing data with the days_ahead median of each id group,
For example:
Median of id 1 = 5.5 which rounds to 6
filled value of date2 at index 1 should be 2021-10-28
Similarly, for id 3 Median = 5
filled value of date2 at index 7 should be 2021-11-07
And,
for id 4 Median = NaN
filled value of date2 at index 9 should be 2021-10-28
I Tried
df['date2'].fillna(df.groupby('id')['days_ahead'].transform('median'), inplace = True)
But this fills with int values.
Although, I can use lambda and apply methods to identify int and turn it to date, How do I directly use groupby and fillna together?

You can round values with convert to_timedelta, add to date1 with fill_valueparameter and replace missing values:
df['date1'] = pd.to_datetime(df['date1'])
df['date2'] = pd.to_datetime(df['date2'])
td = pd.to_timedelta(df.groupby('id')['days_ahead'].transform('median').round(), unit='d')
df['date2'] = df['date2'].fillna(df['date1'].add(td, fill_value=pd.Timedelta(0)))
print (df)
id date1 date2 days_ahead
0 1 2021-10-21 2021-10-24 3.0
1 1 2021-10-22 2021-10-28 NaN
2 1 2021-11-16 2021-11-24 8.0
3 2 2021-10-22 2021-10-24 2.0
4 2 2021-10-22 2021-10-24 2.0
5 3 2021-10-26 2021-10-31 5.0
6 3 2021-10-30 2021-11-04 5.0
7 3 2021-11-02 2021-11-07 NaN
8 3 2021-11-04 2021-11-04 0.0
9 4 2021-10-28 2021-10-28 NaN

I'm trying to calculate running difference on the date column depending on "event column".
So, to add another column with date difference between 1 in event column (there only 0 and 1).
Spo far I came to this half-working crappy solution
Dataframe:
df = pd.DataFrame({'date':[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17],'event':[0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0],'duration':None})
Code:
x = df.loc[df['event']==1, 'date']
k = 0
for i in range(len(x)):
df.loc[k:x.index[i], 'duration'] = x.iloc[i] - k
k = x.index[i]
But I'm sure there is a more elegant solution.
Thanks for any advice.
Output format:
+------+-------+----------+
| date | event | duration |
+------+-------+----------+
| 1 | 0 | 3 |
| 2 | 0 | 3 |
| 3 | 1 | 3 |
| 4 | 0 | 6 |
| 5 | 0 | 6 |
| 6 | 0 | 6 |
| 7 | 0 | 6 |
| 8 | 0 | 6 |
| 9 | 1 | 6 |
| 10 | 0 | 4 |
| 11 | 0 | 4 |
| 12 | 0 | 4 |
| 13 | 1 | 4 |
| 14 | 0 | 2 |
| 15 | 1 | 2 |
+------+-------+----------+

Using your initial dataframe:
df = pd.DataFrame({'date':[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17],'event':[0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0],'duration':None})
Add an index-like column to mark where the transitions occur (you could also base this on the date column if it is unique):
df = df.reset_index().rename(columns={'index':'idx'})
df.loc[df['event']==0, 'idx'] = np.nan
df['idx'] = df['idx'].fillna(method='bfill')
Then, use a groupby() to count the records, and backfill them to match your structure:
df['duration'] = df.groupby('idx')['event'].count()
df['duration'] = df['duration'].fillna(method='bfill')
# Alternatively, the previous two lines can be combined as pointed out by OP
# df['duration'] = df.groupby('idx')['event'].transform('count')
df = df.drop(columns='idx')
print(df)
date event duration
0 1 0 2.0
1 2 1 2.0
2 3 0 3.0
3 4 0 3.0
4 5 1 3.0
5 6 0 5.0
6 7 0 5.0
7 8 0 5.0
8 9 0 5.0
9 10 1 5.0
10 11 0 6.0
11 12 0 6.0
12 13 0 6.0
13 14 0 6.0
14 15 0 6.0
15 16 1 6.0
16 17 0 NaN
It ends up as a float value because of the NaN in the last row. This approach works well in general if there are obvious "groups" of things to count.
As an alternative, because the dates are already there as integers you can look at the differences in dates directly:
df = pd.DataFrame({'date':[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17],'event':[0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0]})
tmp = df[df['event']==1].copy()
tmp['duration'] = (tmp['date'] - tmp['date'].shift(1)).fillna(tmp['date'])
df = pd.merge(df, tmp[['date','duration']], on='date', how='left').fillna(method='bfill')

I have a df that looks like this:
Video | Start | End | Duration |
vid1 |2018-10-02 16:00:29 |2018-10-02 20:07:05 | 246 |
vid2 |2018-10-04 16:03:08 |2018-10-04 16:10:11 | 7 |
vid3 |2018-10-04 10:13:40 |2018-10-06 12:07:38 | 113 |
What I want to do is resample dataframe by 10 minutes by start column and assign 1 if the video lasted in that timestamp and 0 if not.
The desired output is:
Start | vid1 | vid2 | vid3 |
2018-10-02 16:00:00| 1 | 0 | 0 |
2018-10-02 16:10:00| 1 | 0 | 0 |
...
2018-10-04 16:10:00| 0 | 1 | 0 |
2018-10-04 16:20:00| 0 | 0 | 1 |
The output is presented only for visualization the output, hence, it can contain errors.
The problem is that I can not resample dataframe in a way to make a desired crosstab output.

Try this:
df.apply(lambda x: pd.Series(x['Video'],
index=pd.date_range(x['Start'].floor('10T'),
x['End'].ceil('10T'),
freq='10T')), axis=1)\
.stack().str.get_dummies().reset_index(level=0, drop=True)
Output:
vid1 vid2 vid3
2018-10-02 16:00:00 1 0 0
2018-10-02 16:10:00 1 0 0
2018-10-02 16:20:00 1 0 0
2018-10-02 16:30:00 1 0 0
2018-10-02 16:40:00 1 0 0
... ... ... ...
2018-10-06 11:30:00 0 0 1
2018-10-06 11:40:00 0 0 1
2018-10-06 11:50:00 0 0 1
2018-10-06 12:00:00 0 0 1
2018-10-06 12:10:00 0 0 1
[330 rows x 3 columns]