How to get rows between time intervals - postgresql

I have delivery slots that has a from column (datetime).
Delivery slots are stored as 1 hour to 1 hour and 30 minute intervals, daily.
i.e. 3.00am-4.30am, 6.00am-7.30am, 9.00am-10.30am and so forth
id | from
------+---------------------
1 | 2016-01-01 03:00:00
2 | 2016-01-01 04:30:00
3 | 2016-01-01 06:00:00
4 | 2016-01-01 07:30:00
5 | 2016-01-01 09:00:00
6 | 2016-01-01 10:30:00
7 | 2016-01-01 12:00:00
8 | 2016-01-02 03:00:00
9 | 2016-01-02 04:30:00
10 | 2016-01-02 06:00:00
11 | 2016-01-02 07:30:00
12 | 2016-01-02 09:00:00
13 | 2016-01-02 10:30:00
14 | 2016-01-02 12:00:00
I’m trying to get all delivery_slots between the hours of 3.00am - 4.30 am. Ive got the following so far:
SELECT * FROM delivery_slots WHERE EXTRACT(HOUR FROM delivery_slots.from) >= 3 AND EXTRACT(MINUTE FROM delivery_slots.from) >= 0 AND EXTRACT(HOUR FROM delivery_slots.from) <= 4 AND EXTRACT(MINUTE FROM delivery_slots.from) <= 30;
Which kinda works. Kinda, because it is only returning delivery slots that have minutes of 00.
Thats because of the last where condition (EXTRACT(MINUTE FROM delivery_slots.from) <= 30)
To give you an idea, of what I am trying to expect:
id | from
-------+---------------------
1 | 2016-01-01 03:00:00
2 | 2016-01-01 04:30:00
8 | 2016-01-02 03:00:00
9 | 2016-01-02 04:30:00
15 | 2016-01-03 03:00:00
16 | 2016-01-03 04:30:00
etc...
Is there a better way to go about this?

Try this: (not tested)
SELECT * FROM delivery_slots WHERE delivery_slots.from::time >= '03:00:00' AND delivery_slots.from::time <= '04:30:00'
Hope this helps.
Cheers.

The easiest way to do this, in my mind, is to cast the from column as a type time and do a where >= and <=, like so
select * from testing where (date::time >= '3:00'::time and date::time <= '4:30'::time);

Related

Is there a way to do a selective sum using a time interval in Postgres?

I have two tables, the first table has columns: id, start_time, and end_time. The second table has columns: id, timestamp, value. Is there a way to make a sum of table 2 based on the conditions in table 1?
Table 1:
id
start_date
end_date
5
2000-01-01 01:00:00
2000-01-05 02:45:00
5
2000-01-10 01:00:00
2000-01-15 02:45:00
6
2000-01-01 01:00:00
2000-01-05 02:45:00
6
2000-01-11 01:00:00
2000-01-12 02:45:00
6
2000-01-15 01:00:00
2000-01-20 02:45:00
Table 2:
id
timestamp
value
5
2000-01-01 05:00:00
1
5
2000-01-01 06:00:00
2
6
2000-01-01 05:00:00
1
6
2000-01-11 05:00:00
2
6
2000-01-15 05:00:00
2
6
2000-01-15 05:30:00
2
Desired result:
id
start_date
end_date
Sum
5
2000-01-01 01:00:00
2000-01-05 02:45:00
3
5
2000-01-10 01:00:00
2000-01-15 02:45:00
null
6
2000-01-01 01:00:00
2000-01-05 02:45:00
1
6
2000-01-11 01:00:00
2000-01-12 02:45:00
2
6
2000-01-15 01:00:00
2000-01-20 02:45:00
4
Try this :
SELECT a.id, a.start_date, a.end_date, sum(b.value) AS sum
FROM table1 AS a
LEFT JOIN table2 AS b
ON b.id = a.id
AND b.timestamp >= a.start_date
AND b.timestamp < a.end_date
GROUP BY a.id, a.start_date, a.end_date

Data from last 12 months each month with trailing 12 months

This is TSQL and I'm trying to calculate repeat purchase rate for last 12 months. This is achieved by looking at sum of customers who have bought more than 1 time last 12 months and the total number of customers last 12 months.
The SQL code below will give me just that; but i would like to dynamically do this for the last 12 months. This is the part where i'm stuck and not should how to best achieve this.
Each month should include data going back 12 months. I.e. June should hold data between June 2018 and June 2018, May should hold data from May 2018 till May 2019.
[Order Date] is a normal datefield (yyyy-mm-dd hh:mm:ss)
DECLARE #startdate1 DATETIME
DECLARE #enddate1 DATETIME
SET #enddate1 = DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE())-1, 0) -- Starting June 2018
SET #startdate1 = DATEADD(mm,DATEDIFF(mm,0,GETDATE())-13,0) -- Ending June 2019
;
with dataset as (
select [Phone No_] as who_identifier,
count(distinct([Order No_])) as mycount
from [MyCompany$Sales Invoice Header]
where [Order Date] between #startdate1 and #enddate1
group by [Phone No_]
),
frequentbuyers as (
select who_identifier, sum(mycount) as frequentbuyerscount
from dataset
where mycount > 1
group by who_identifier),
allpurchases as (
select who_identifier, sum(mycount) as allpurchasescount
from dataset
group by who_identifier
)
select sum(frequentbuyerscount) as frequentbuyercount, (select sum(allpurchasescount) from allpurchases) as allpurchasecount
from frequentbuyers
I'm hoping to achieve end result looking something like this:
...Dec, Jan, Feb, March, April, May, June each month holding both values for frequentbuyercount and allpurchasescount.
Here is the code. I made a little modification for the frequentbuyerscount and allpurchasescount. If you use a sumif like expression you don't need a second cte.
if object_id('tempdb.dbo.#tmpMonths') is not null drop table #tmpMonths
create table #tmpMonths ( MonthID datetime, StartDate datetime, EndDate datetime)
declare #MonthCount int = 12
declare #Month datetime = DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0)
while #MonthCount > 0 begin
insert into #tmpMonths( MonthID, StartDate, EndDate )
select #Month, dateadd(month, -12, #Month), #Month
set #Month = dateadd(month, -1, #Month)
set #MonthCount = #MonthCount - 1
end
;with dataset as (
select m.MonthID as MonthID, [Phone No_] as who_identifier,
count(distinct([Order No_])) as mycount
from [MyCompany$Sales Invoice Header]
inner join #tmpMonths m on [Order Date] between m.StartDate and m.EndDate
group by m.MonthID, [Phone No_]
),
buyers as (
select MonthID, who_identifier
, sum(iif(mycount > 1, mycount, 0)) as frequentbuyerscount --sum only if count > 1
, sum(mycount) as allpurchasescount
from dataset
group by MonthID, who_identifier
)
select
b.MonthID
, max(tm.StartDate) StartDate, max(tm.EndDate) EndDate
, sum(b.frequentbuyerscount) as frequentbuyercount
, sum(b.allpurchasescount) as allpurchasecount
from buyers b inner join #tmpMonths tm on tm.MonthID = b.MonthID
group by b.MonthID
Be aware, that the code was tested only syntax-wise.
After the test data, this is the result:
MonthID | StartDate | EndDate | frequentbuyercount | allpurchasecount
-----------------------------------------------------------------------------
2018-08-01 | 2017-08-01 | 2018-08-01 | 340 | 3702
2018-09-01 | 2017-09-01 | 2018-09-01 | 340 | 3702
2018-10-01 | 2017-10-01 | 2018-10-01 | 340 | 3702
2018-11-01 | 2017-11-01 | 2018-11-01 | 340 | 3702
2018-12-01 | 2017-12-01 | 2018-12-01 | 340 | 3703
2019-01-01 | 2018-01-01 | 2019-01-01 | 340 | 3703
2019-02-01 | 2018-02-01 | 2019-02-01 | 2 | 8
2019-03-01 | 2018-03-01 | 2019-03-01 | 2 | 3
2019-04-01 | 2018-04-01 | 2019-04-01 | 2 | 3
2019-05-01 | 2018-05-01 | 2019-05-01 | 2 | 3
2019-06-01 | 2018-06-01 | 2019-06-01 | 2 | 3
2019-07-01 | 2018-07-01 | 2019-07-01 | 2 | 3

Create calendar year variable from daily date

I have a Stata elapsed date variable (excerpt below) for each patient patid. But I believe I need to generate a new variable if I were to make use of only the year element within that date, that is, not just change the display format.
clear
input long patid float date
1015 18766
1018 13135
1020 13325
1025 14384
1029 14514
1050 13501
1070 14523
1071 14878
1090 14701
1092 14159
end
format %td date
How do I generate a year variable for all dates within the same year? That is all days from 1st January to 31st December of the same year?
That calls for just the function year(), which together with similar stuff is prominently documented at help datetime.
clear
input long patid float date
1015 18766
1018 13135
1020 13325
1025 14384
1029 14514
1050 13501
1070 14523
1071 14878
1090 14701
1092 14159
end
format %td date
gen year = year(date)
list
+--------------------------+
| patid date year |
|--------------------------|
1. | 1015 19may2011 2011 |
2. | 1018 18dec1995 1995 |
3. | 1020 25jun1996 1996 |
4. | 1025 20may1999 1999 |
5. | 1029 27sep1999 1999 |
|--------------------------|
6. | 1050 18dec1996 1996 |
7. | 1070 06oct1999 1999 |
8. | 1071 25sep2000 2000 |
9. | 1090 01apr2000 2000 |
10. | 1092 07oct1998 1998 |
+--------------------------+

Pandas: Combine resampling with groupby and calculate time differences

I am doing data analysis with trading data. I would like to use Pandas in order to examine the times when the traders are active.
In particular, I try to extract the difference in minutes between the dates of every first trade of every trader for each day and cumulate it to a monthly basis
The data looks like this:
Timestamp (Datetime) | Buyer | Volume
--------------------------------------
2012-01-01 09:00:00 | John | 10
2012-01-01 10:00:00 | Mark | 10
2012-01-01 16:00:00 | Mark | 10
2012-01-01 11:00:00 | Kevin | 10
2012-02-01 10:00:00 | Mark | 10
2012-02-01 09:00:00 | John | 10
2012-02-01 17:00:00 | Mark | 10
Right now I use resampling to retrieve the first trade on a daily basis. However, I want to group also by the buyer to calculate the differences in their trading dates. Like this
Timestamp (Datetime) | Buyer | Volume
--------------------------------------
2012-01-01 09:00:00 | John | 10
2012-01-01 10:00:00 | Mark | 10
2012-01-01 11:00:00 | Kevin | 10
2012-01-02 10:00:00 | Mark | 10
2012-01-02 09:00:00 | John | 10
Overall I am looking to calculate the differences in minutes between the first trades on a daily basis for each trader.
Update
For example in the case of John on the 2012-01-01: Dist = 60 (Diff John-Mark) + 120 (Diff John-Kevin) = 180
I would highly appreciate if anyone has an idea how to do this.
Thank you
Your original frame (the resampled one)
In [71]: df_orig
Out[71]:
buyer date volume
0 John 2012-01-01 09:00:00 10
1 Mark 2012-01-01 10:00:00 10
2 Kevin 2012-01-01 11:00:00 10
3 Mark 2012-01-02 10:00:00 10
4 John 2012-01-02 09:00:00 10
Set the index to the date column, keeping the date column in place
In [75]: df = df_orig.set_index('date',drop=False)
Create this aggregation function
def f(frame):
frame.sort('date',inplace=True)
frame['start'] = frame.date.iloc[0]
return frame
Groupby the single date
In [74]: x = df.groupby(pd.TimeGrouper('1d')).apply(f)
Create the differential in minutes
In [86]: x['diff'] = (x.date-x.start).apply(lambda x: float(x.item().total_seconds())/60)
In [87]: x
Out[87]:
buyer date volume start diff
date
2012-01-01 2012-01-01 09:00:00 John 2012-01-01 09:00:00 10 2012-01-01 09:00:00 0
2012-01-01 10:00:00 Mark 2012-01-01 10:00:00 10 2012-01-01 09:00:00 60
2012-01-01 11:00:00 Kevin 2012-01-01 11:00:00 10 2012-01-01 09:00:00 120
2012-01-02 2012-01-02 09:00:00 John 2012-01-02 09:00:00 10 2012-01-02 09:00:00 0
2012-01-02 10:00:00 Mark 2012-01-02 10:00:00 10 2012-01-02 09:00:00 60
Here's the explanation. We use the TimeGrouper to have the grouping by date, where a frame is passed to the function f. This function, then uses the first date of the day (the sort is necessary here). You subtract this from the date on the entry to get a timedelta64, which is then massaged to minutes (this is a bit hacky right now because of some numpy issues, should be more natural in 0.12)
Thanks for you update, I originally thought you wanted the diff per buyer, not from the first buyer, but that's just a minor tweak.
Update:
To track the buyer name as well (which corresponds to the start date), just include
it in the function f
def f(frame):
frame.sort('date',inplace=True)
frame['start'] = frame.date.iloc[0]
frame['start_buyer'] = frame.buyer.iloc[0]
return frame
Then can groupby on this at the end:
In [14]: x.groupby(['start_buyer']).sum()
Out[14]:
diff
start_buyer
John 240

function to calculate aggregate sum count in postgresql

Is there a function that calculates the total count of the complete month like below? I am not sure if postgres. I am looking for the grand total value.
2012-08=# select date_trunc('day', time), count(distinct column) from table_name group by 1 order by 1;
date_trunc | count
---------------------+-------
2012-08-01 00:00:00 | 22
2012-08-02 00:00:00 | 34
2012-08-03 00:00:00 | 25
2012-08-04 00:00:00 | 30
2012-08-05 00:00:00 | 27
2012-08-06 00:00:00 | 31
2012-08-07 00:00:00 | 23
2012-08-08 00:00:00 | 28
2012-08-09 00:00:00 | 28
2012-08-10 00:00:00 | 28
2012-08-11 00:00:00 | 24
2012-08-12 00:00:00 | 36
2012-08-13 00:00:00 | 28
2012-08-14 00:00:00 | 23
2012-08-15 00:00:00 | 23
2012-08-16 00:00:00 | 30
2012-08-17 00:00:00 | 20
2012-08-18 00:00:00 | 30
2012-08-19 00:00:00 | 20
2012-08-20 00:00:00 | 24
2012-08-21 00:00:00 | 20
2012-08-22 00:00:00 | 17
2012-08-23 00:00:00 | 23
2012-08-24 00:00:00 | 25
2012-08-25 00:00:00 | 35
2012-08-26 00:00:00 | 18
2012-08-27 00:00:00 | 16
2012-08-28 00:00:00 | 11
2012-08-29 00:00:00 | 22
2012-08-30 00:00:00 | 26
2012-08-31 00:00:00 | 17
(31 rows)
--------------------------------
Total | 12345
As best I can guess from your question and comments you want sub-totals of the distinct counts by month. You can't do this with group by date_trunc('month',time) because that'll do a count(distinct column) that's distinct across all days.
For this you need a subquery or CTE:
WITH day_counts(day,day_col_count) AS (
select date_trunc('day', time), count(distinct column)
from table_name group by 1
)
SELECT 'Day', day, day_col_count
FROM day_counts
UNION ALL
SELECT 'Month', date_trunc('month', day), sum(day_col_count)
FROM day_counts
GROUP BY 2
ORDER BY 2;
My earlier guess before comments was: Group by month?
select date_trunc('month', time), count(distinct column)
from table_name
group by date_trunc('month', time)
order by time
Or are you trying to include running totals or subtotal lines? For running totals you need to use sum as a window function. Subtotals are just a pain, as SQL doesn't really lend its self to them; you need to UNION two queries then wrap them in an outer ORDER BY.
select
date_trunc('day', time)::text as "date",
count(distinct column) as count
from table_name
group by 1
union
select
'Total',
count(distinct column)
from table_name
group by 1, date_trunc('month', time)
order by "date" = 'Total', 1