Average number of actions per day of the week using pandas - python

Assume I have a count of the number of event per hour as follows:
np.random.seed(42)
idx = pd.date_range('2017-01-01', '2017-01-14', freq='1H')
df = pd.DataFrame(np.random.choice([1,2,3,4,5,6], size=idx.shape[0]), index=idx, columns=['count'])
df.head()
Out[3]:
count
2017-01-01 00:00:00 4
2017-01-01 01:00:00 5
2017-01-01 02:00:00 3
2017-01-01 03:00:00 5
2017-01-01 04:00:00 5
If I want to know the total number of events per day of the week, I can do either:
df.pivot_table(values='count', index=df.index.dayofweek, aggfunc='sum')
or
df.groupby(df.index.dayofweek).sum()
Both yields:
Out[4]:
count
0 161
1 170
2 164
3 133
4 169
5 98
6 172
However, if I want to compute the average number of events per weekday, the following
df.pivot_table(values='count', index=df.index.dayofweek, aggfunc='mean') # [#1]
is wrong!! This approach uses the sum (as computed above), and divides it by the number of hours that appeared in each day of the week.
The workaround I found is:
df_by_day = df.resample('1d').sum()
df_by_day.pivot_table(values='count', index=df_by_day.index.dayofweek, aggfunc='mean')
That is, first resample to days, and then pivot it. Somehow the approach in [#1] feel natural to me. Is there a more pythonic way to achieve what I want? Why does without resampling the mean is wrongly computed?

Resample first using df.resample and then df.groupby:
df = df.resample('1d').sum()
print(df)
count
2017-01-01 92
2017-01-02 86
2017-01-03 86
2017-01-04 90
2017-01-05 64
2017-01-06 82
2017-01-07 97
2017-01-08 80
2017-01-09 75
2017-01-10 84
2017-01-11 74
2017-01-12 69
2017-01-13 87
2017-01-14 1
out = df.groupby(df.index.dayofweek)['count'].mean()
print(out)
1 85.0
2 82.0
3 66.5
4 84.5
5 49.0
6 86.0
Name: count, dtype: float64

Related

Groupby ID, sort by Time, and divide last by first

I have the following dataframe:
ID ..... Quantity Time
54 100 2020-01-01 00:00:04
55 100 2020-01-01 00:00:04
54 88 2020-01-01-00:00:05
54 66 2020-01-01 00:00:06
55 100 2020-01-01 00:00:07
55 88 2020-01-01 00:00:07
I would like to group the dataframe (sorted by time!) by ID and then take the quantity of the last row and divide it by the first row per ID.
The result should look like this:
ID ..... Quantity Time Result
54 100 2020-01-01 00:00:04
54 88 2020-01-01-00:00:05
54 66 2020-01-01 00:00:06 0.66
55 100 2020-01-01 00:00:04
55 100 2020-01-01 00:00:07
55 88 2020-01-01 00:00:07 0.88
So far I used the following code to get the first and the last row for every ID.
g = df.sort_values(by=['Time']).groupby('ID')
df_new=(pd.concat([g.head(1), g.tail(1)])
.sort_values(by='ID')
.reset_index(drop=True))
and then I used the following code to get the Result of the division:
df_new['Result'] = df_new['Quantity'].iloc[1::2].div(df_new['Quantity'].shift())
The problem is: the dataframe stays not sorted by time. It is really important that I take (timewise) the last quantity per ID and divide it by the first quantity (in time) per ID.
Thanks for any hints where I need to change the code!
There are not pairs ID values, but triples, so first convert column to datetime if necessary by to_datetime, then sorting per 2 columns by DataFrame.sort_values and last use second or third solution from previous answer:
df['Time'] = pd.to_datetime(df['Time'])
df = df.sort_values(['ID','Time'])
first = df.groupby('ID')['Quantity'].transform('first')
df['Result'] = df.drop_duplicates('ID', keep='last')['Quantity'].div(first)
print (df)
ID Quantity Time Result
0 54 100 2020-01-01 00:00:04 NaN
2 54 88 2020-01-01 00:00:05 NaN
3 54 66 2020-01-01 00:00:06 0.66
1 55 100 2020-01-01 00:00:04 NaN
4 55 100 2020-01-01 00:00:07 NaN
5 55 88 2020-01-01 00:00:07 0.88
Convert Time to datetime :
df["Time"] = pd.to_datetime(df["Time"])
Sort on ID and Time :
df = df.sort_values(["ID", "Time"])
Group on ID:
grouping = df.groupby("ID").Quantity
Get results for the division of last by first:
result = grouping.last().div(grouping.first()).array
Now, you can assign the results back to the original dataframe :
df.loc[df.Quantity.eq(grouping.transform("last")), "Result"] = result
df
ID Quantity Time Result
0 54 100 2020-01-01 00:00:04 NaN
2 54 88 2020-01-01 00:00:05 NaN
3 54 66 2020-01-01 00:00:06 0.66
1 55 100 2020-01-01 00:00:04 NaN
4 55 100 2020-01-01 00:00:07 NaN
5 55 88 2020-01-01 00:00:07 0.88

Monthly aggregated values, pandas dataframe

A sample CSV data in which the first column is a time stamp (date + time):
2018-01-01 10:00:00,23,43
2018-01-02 11:00:00,34,35
2018-01-05 12:00:00,25,4
2018-01-10 15:00:00,22,96
2018-01-01 18:00:00,24,53
2018-03-01 10:00:00,94,98
2018-04-20 10:00:00,90,9
2018-04-10 10:00:00,45,51
2018-01-01 10:00:00,74,44
2018-12-01 10:00:00,76,87
2018-11-01 10:00:00,76,87
2018-12-12 10:00:00,87,90
I already wrote some codes to do the monthly aggregated values task while waiting for someone to give me some suggestions.
Thanks #moys, anyway!
import pandas as pd
df = pd.read_csv('Sample.txt', header=None, names = ['Timestamp', 'Value 1', 'Value 2'])
df1['Timestamp'] = pd.to_datetime(df1['Timestamp'])
df1['Monthly'] = df1['Timestamp'].dt.to_period('M')
grouper = pd.Grouper(key='Monthly')
df2 = df1.groupby(grouper)['Value 1', 'Value 2'].sum().reset_index()
The output is:
Monthly Value 1 Value 2
0 2018-01 202 275
1 2018-03 94 98
2 2018-04 135 60
3 2018-12 163 177
4 2018-11 76 87
What if there's a dataset with more columns, how to motified the my code to make it automatically working on the dataset which has more columns?
2018-02-01 10:00:00,23,43,32
2018-02-02 11:00:00,34,35,43
2018-03-05 12:00:00,25,4,43
2018-02-10 15:00:00,22,96,24
2018-05-01 18:00:00,24,53,98
2018-02-01 10:00:00,94,98,32
2018-02-20 10:00:00,90,9,24
2018-07-10 10:00:00,45,51,32
2018-01-01 10:00:00,74,44,34
2018-12-04 10:00:00,76,87,53
2018-12-02 10:00:00,76,87,21
2018-12-12 10:00:00,87,90,98
You can do something like below
df.groupby(pd.to_datetime(df['date']).dt.month).sum().reset_index()
Output Here, 'date' column is the month number.
date val1 val2
0 1 202 275
1 3 94 98
2 4 135 60
3 11 76 87
4 12 163 177

Pandas - Compute data from a column to another

Considering the following dataframe:
df = pd.read_json("""{"week":{"0":1,"1":1,"2":1,"3":1,"4":1,"5":1,"6":2,"7":2,"8":2,"9":2,"10":2,"11":2,"12":3,"13":3,"14":3,"15":3,"16":3,"17":3},"extra_hours":{"0":"01:00:00","1":"00:00:00","2":"01:00:00","3":"01:00:00","4":"00:00:00","5":"01:00:00","6":"01:00:00","7":"01:00:00","8":"01:00:00","9":"01:00:00","10":"00:00:00","11":"01:00:00","12":"01:00:00","13":"02:00:00","14":"01:00:00","15":"02:00:00","16":"00:00:00","17":"00:00:00"},"extra_hours_over":{"0":null,"1":null,"2":null,"3":null,"4":null,"5":null,"6":null,"7":null,"8":null,"9":null,"10":null,"11":null,"12":null,"13":null,"14":null,"15":null,"16":null,"17":null}}""")
df.tail(6)
week extra_hours extra_hours_over
12 3 01:00:00 NaN
13 3 02:00:00 NaN
14 3 01:00:00 NaN
15 3 02:00:00 NaN
16 3 00:00:00 NaN
17 3 00:00:00 NaN
Now, in every week, the maximum amount of extra_hours is 4h, meaning I have to subtract 30min blocks from extra_hour column, and fill the extra_hour_over column, so that in every week, total sum of extra_hour has a maximum of 4h.
So, given the example dataframe, a possible solution (for week 3) would be like this:
week extra_hours extra_hours_over
12 3 01:00:00 00:00:00
13 3 01:30:00 00:30:00
14 3 00:30:00 00:30:00
15 3 01:00:00 01:00:00
16 3 00:00:00 00:00:00
17 3 00:00:00 00:00:00
I would need to aggregate total extra_hours per week, check in which days it passes 4h, and then randomly subtract half-hour chunks.
What would be the easiest/most direct way to achieve this?
Here goes one attempt for what you seem to be asking. The idea is simple, although the code fairly verbose:
1) Create some helper variables (minutes, extra_minutes, total for the week)
2) Loop through a temporary dataset that will contain only while sum is > 240 minutes.
3) In the loop, use random.choice to select a time to remove 30 min from.
4) Apply the changes to minutes and extra minutes
The code:
df = pd.read_json("""{"week":{"0":1,"1":1,"2":1,"3":1,"4":1,"5":1,"6":2,"7":2,"8":2,"9":2,"10":2,"11":2,"12":3,"13":3,"14":3,"15":3,"16":3,"17":3},"extra_hours":{"0":"01:00:00","1":"00:00:00","2":"01:00:00","3":"01:00:00","4":"00:00:00","5":"01:00:00","6":"01:00:00","7":"01:00:00","8":"01:00:00","9":"01:00:00","10":"00:00:00","11":"01:00:00","12":"01:00:00","13":"02:00:00","14":"01:00:00","15":"02:00:00","16":"00:00:00","17":"00:00:00"},"extra_hours_over":{"0":null,"1":null,"2":null,"3":null,"4":null,"5":null,"6":null,"7":null,"8":null,"9":null,"10":null,"11":null,"12":null,"13":null,"14":null,"15":null,"16":null,"17":null}}""")
df['minutes'] = pd.DatetimeIndex(df['extra_hours']).hour * 60 + pd.DatetimeIndex(df['extra_hours']).minute
df['extra_minutes'] = 0
df['tot_time'] = df.groupby('week')['minutes'].transform('sum')
while not df[df['tot_time'] > 240].empty:
mask = df[(df['minutes']>=30)&(df['tot_time']>240)].groupby('week').apply(lambda x: np.random.choice(x.index)).values
df.loc[mask,'minutes'] -= 30
df.loc[mask,'extra_minutes'] += 30
df['tot_time'] = df.groupby('week')['minutes'].transform('sum')
df['extra_hours_over'] = df['extra_minutes'].apply(lambda x: pd.Timedelta(minutes=x))
df['extra_hours'] = df['minutes'].apply(lambda x: pd.Timedelta(minutes=x))
df.drop(['minutes','extra_minutes'], axis=1).tail(6)
Out[1]:
week extra_hours extra_hours_over tot_time
12 3 00:30:00 00:30:00 240
13 3 01:30:00 00:30:00 240
14 3 00:30:00 00:30:00 240
15 3 01:30:00 00:30:00 240
16 3 00:00:00 00:00:00 240
17 3 00:00:00 00:00:00 240
Note: Because I am using np.random.choice, the same observation can be picked twice, which will make that observation change by a chunk of more than 30 min.

Pandas get unique monthly data based on date range

I have something like the following dataframe:
d=pd.DataFrame()
d['id']=['a','a','a','b','b','c']
d['version_start']=['2017-01-01','2017-02-12','2017-03-25','2017-01-01','2017-6-15','2017-01-22']
d['version_end']=['2017-02-11','2017-03-24','2017-08-01','2017-06-14','2018-01-01','2018-01-01']
d['version_start']=pd.to_datetime(d.version_start)
d['version_end']=pd.to_datetime(d.version_end)
d['values']=[10,15,20,5,6,200]
print d
id version_start version_end values
0 a 2017-01-01 2017-02-11 10
1 a 2017-02-12 2017-03-24 15
2 a 2017-03-25 2017-08-01 20
3 b 2017-01-01 2017-06-14 5
4 b 2017-06-15 2018-01-01 6
5 c 2017-01-22 2018-01-01 200
The version start and version end represent for each ID, the date range for which that row can be considered valid. For example, the total values for a given date would be the records for which that date is between the version start and version end.
I am looking to get for a set of dates (the first of the month for each month in 2017) the sum of the "values" field. I can do this by looping through each month as follows:
df=pd.DataFrame()
for month in pd.date_range('2017-01-01','2018-01-01',freq='MS'):
s = d[(d.version_start<=month)&(d.version_end>month)]
s['month']=month
s=s.set_index(['month','id'])[['values']]
df=df.append(s)
print df.groupby(level='month')['values'].sum()
2017-01-01 15
2017-02-01 215
2017-03-01 220
2017-04-01 225
2017-05-01 225
2017-06-01 225
2017-07-01 226
2017-08-01 206
2017-09-01 206
2017-10-01 206
2017-11-01 206
2017-12-01 206
Name: values, dtype: int64
Is there a more elegant / efficient solution that doesn't require looping through this list of dates?
d.version_start=d.version_start+ pd.offsets.MonthBegin(0)
d.version_end=d.version_end+ pd.offsets.MonthBegin(0)
d['New']=d[['version_start','version_end']].apply(lambda x : pd.date_range(start=x.version_start,end=x.version_end,freq='MS').tolist(),1)
d.set_index(['id','version_start','version_end','values']).New.apply(pd.Series).stack().reset_index('values').groupby(0)['values'].sum()
Out[845]:
0
2017-01-01 15
2017-02-01 215
2017-03-01 230
2017-04-01 240
2017-05-01 225
2017-06-01 225
2017-07-01 231
2017-08-01 226
2017-09-01 206
2017-10-01 206
2017-11-01 206
2017-12-01 206
2018-01-01 206
Name: values, dtype: int64
I keep thinking there should be a way more elegant way to do this, but for now:
s = pd.Series(0, index=pd.date_range('2017-01-01','2018-01-01',freq='MS'))
for _id, start, end, values in d.itertuples(index=False):
s[start:end] += values
this returns the proper series, and works with any series for that matter.
If you want version_end day to be excluded, a quick fix is to add this line before the for cycle (only works if you are using 'MS' as frequency):
d.version_end = d.version_end.apply(lambda t: t.replace(day=2))
I think the idea of using explicit indexing is cleaner than conditional indexing based on comparisons between dates, which at scale is terribly slow (timestamps are a valid alternative if you are forced to do this on huge arrays).

Query same time value every day in Pandas timeseries

I would like to get the 07h00 value every day, from a multiday DataFrame that has 24 hours of minute data in it each day.
import numpy as np
import pandas as pd
aframe = pd.DataFrame([np.arange(10000), np.arange(10000) * 2]).T
aframe.index = pd.date_range("2015-09-01", periods = 10000, freq = "1min")
aframe.head()
Out[174]:
0 1
2015-09-01 00:00:00 0 0
2015-09-01 00:01:00 1 2
2015-09-01 00:02:00 2 4
2015-09-01 00:03:00 3 6
2015-09-01 00:04:00 4 8
aframe.tail()
Out[175]:
0 1
2015-09-07 22:35:00 9995 19990
2015-09-07 22:36:00 9996 19992
2015-09-07 22:37:00 9997 19994
2015-09-07 22:38:00 9998 19996
2015-09-07 22:39:00 9999 19998
In this 10 000 row DataFrame spanning 7 days, how would I get the 7am value each day as efficiently as possible? Assume I might have to do this for very large tick databases so I value speed and low memory usage highly.
I know I can index with strings such as:
aframe.ix["2015-09-02 07:00:00"]
Out[176]:
0 1860
1 3720
Name: 2015-09-02 07:00:00, dtype: int64
But what I need is basically a wildcard style query for example
aframe.ix["* 07:00:00"]
You can use indexer_at_time:
>>> locs = aframe.index.indexer_at_time('7:00:00')
>>> aframe.iloc[locs]
0 1
2015-09-01 07:00:00 420 840
2015-09-02 07:00:00 1860 3720
2015-09-03 07:00:00 3300 6600
2015-09-04 07:00:00 4740 9480
2015-09-05 07:00:00 6180 12360
2015-09-06 07:00:00 7620 15240
2015-09-07 07:00:00 9060 18120
There's also indexer_between_time if you need select all indices that lie between two particular time of day.
Both of these methods return the integer locations of the desired values; the corresponding rows of the Series or DataFrame can be fetched with iloc, as shown above.

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