I have this table:
type id qty hour date
a manual 1 2 2020-05-06
a manual 2 3 2020-05-06
a manual 4 6 2020–05-06
b manual 2 4 2020-05-03
b manual 1 2 2020-05-03
b manual 4 5 2020-05-03
I need to get the productivity by dividing qty and hour, my query is:
select type, date, qty/hour as prod,
from table1
where id = ‘manual’
group by type, date
with the query above, I got:
type id qty hour date prod
a manual 1 2 2020-05-06 0.5
a manual 2 3 2020-05-06 0.67
a manual 4 6 2020–05-06 0.67
b manual 2 4 2020-05-03 0.5
b manual 1 2 2020-05-03 0.5
b manual 4 5 2020-05-03 0.8
Next, I need to get the average productivity based on the type and also date, the expected result:
type date avg
a 2020-05-06 0.613
b 2020-05-03 0.6
How should I write my query to get the expected result?
If I understand correctly, you want:
select type, date, avg(qty/hour) as prod,
from table1
where id = 'manual'
group by type, date
I don't know how you generated your current intermediate results, but grouping by type and date must generate a result set of only 2 records, based on your sample input data.
Related
I have a table that currently looks like this (simplified to illustate my issue):
Thing| Date
1 2022-12-12
2 2022-11-05
3 2022-11-18
4 2022-12-01
1 2022-11-02
2 2022-11-21
5 2022-12-03
5 2022-12-08
2 2022-11-18
1 2022-11-20
I would like to generate the following:
Thing| 2022-11 | 2022-12
1 2 1
2 3 0
3 1 0
4 0 1
5 0 2
I'm new to SQL and can't quite figure this out - would I use some sort of FOR loop equivalent in my SELECT clause? I'm happy to figure out the exact syntax myself, I just need someone to point me in the right direction.
Thank you!
You may use conditional aggregation as the following:
Select Thing,
Count(Case When Date Between '2022-11-01' And '2022-11-30' Then 1 End) As '2022-11',
Count(Case When Date Between '2022-12-01' And '2022-12-31' Then 1 End) As '2022-12'
From table_name
Group By Thing
Order By Thing
See a demo.
The count function counts only the not null values, so for each row not matching the condition inside the count function a null value is returned, hence not counted.
(using sql or pandas)
I want to delete records if the Date difference between two records is less than 30 days.
But first record of ID must be remained.
#example
ROW ID DATE
1 A 2020-01-01 -- first
2 A 2020-01-03
3 A 2020-01-31
4 A 2020-02-05
5 A 2020-02-28
6 A 2020-03-09
7 B 2020-03-06 -- first
8 B 2020-05-07
9 B 2020-06-02
#expected results
ROW ID DATE
1 A 2020-01-01
4 A 2020-02-05
6 A 2020-03-09
7 B 2020-03-06
8 B 2020-05-07
ROW 2,3 are within 30 days from ROW 1
ROW 5 is within 30 days from ROW 4
ROW 9 is within 30 days from ROW 8
To cope with your task it is not possible to call any
vectorized methods.
The cause is that after a row is recognized as a duplicate, then
this row "does not count" when you check further rows.
E.g. after rows 2020-01-03 and 2020-01-31 were deleted (as
"too close" to the previous row) then 2020-02-05 row should be
left, because now the distance to the previous row (2020-01-01)
is big enough.
So I came up with a solution based on a "function with memory":
def isDupl(elem):
if isDupl.prev is None:
isDupl.prev = elem
return False
dDiff = (elem - isDupl.prev).days
rv = dDiff <= 30
if not rv:
isDupl.prev = elem
return rv
This function should be invoked for each DATE in the
current group (with same ID) but before that isDupl.prev
must be set to None.
So the function to apply to each group of rows is:
def isDuplGrp(grp):
isDupl.prev = None
return grp.DATE.apply(isDupl)
And to get the expected result, run:
df[~(df.groupby('ID').apply(isDuplGrp).reset_index(level=0, drop=True))]
(you may save it back to df).
The result is:
ROW ID DATE
0 1 A 2020-01-01
3 4 A 2020-02-05
5 6 A 2020-03-09
6 7 B 2020-03-06
7 8 B 2020-05-07
And finally, a remark about the other solution:
It contains rows:
3 4 A 2020-02-05
4 5 A 2020-02-28
which are only 23 days apart, so this solution is wrong.
The same pertains to rows:
5 A 2020-02-28
6 A 2020-03-09
which are also too close in time.
You can try this:
Convert date to datetime64
Get the first date from each group df.groupby('ID')['DATE'].transform('first')
Add a filter to keep only dates greater than 30 days
Append the first date of each group to the dataframe
Code:
df['DATE'] = pd.to_datetime(df['DATE'])
df1 = df[(df['DATE'] - df.groupby('ID')['DATE'].transform('first')) >= pd.Timedelta(30, unit='D')]
df1 = df1.append(df.groupby('ID', as_index=False).agg('first')).sort_values(by=['ID', 'DATE'])
print(df1)
ROW ID DATE
0 1 A 2020-01-01
2 3 A 2020-01-31
3 4 A 2020-02-05
4 5 A 2020-02-28
5 6 A 2020-03-09
1 7 B 2020-03-06
7 8 B 2020-05-07
8 9 B 2020-06-02
I`m trying to calculate the sum of one field for a specific period of time, after grouping function is applied.
My dataset look like this:
Date Company Country Sold
01.01.2020 A BE 1
02.01.2020 A BE 0
03.01.2020 A BE 1
03.01.2020 A BE 1
04.01.2020 A BE 1
05.01.2020 B DE 1
06.01.2020 B DE 0
I would like to add a new column per each row, that calculates the sum of Sold (per each group "Company, Country" for the last 7 days - not including the current day
Date Company Country Sold LastWeek_Count
01.01.2020 A BE 1 0
02.01.2020 A BE 0 1
03.01.2020 A BE 1 1
03.01.2020 A BE 1 1
04.01.2020 A BE 1 3
05.01.2020 B DE 1 0
06.01.2020 B DE 0 1
I tried the following, but it is also including the current date, and it gives differnt values for the same date, i.e 03.01.2020
df['LastWeek_Count'] = df.groupby(['Company', 'Country']).rolling(7, on ='Date')['Sold'].sum().reset_index()
Is there a buildin function in pandas that I can use to perform these calculations?
You can use a .rolling window of 8 and then subtract the sum of the Date (for each grouped row) to effectively get the previous 7 days. For this sample data, we should also pass min_periods=1 (otherwise you will get NaN values, but for your actual dataset, you will need to decide what you want to do with windows that are < 8).
Then from the .rolling window of 8, simply do another .groupby of the relevant columns but also include Date this time, and take the max value of the newly created LastWeek_Count column. You need to take the max, because you have multiple records per day, so by taking the max, you are taking the total aggregated amount per Date.
Then, create a series that takes the grouped by sum per Date. In the final step subtract the sum by date from the rolling 8-day max, which is a workaround to how you can get the sum of the previous 7 days, as there is not a parameter for an offset with .rolling:
df['Date'] = pd.to_datetime(df['Date'], dayfirst=True)
df['LastWeek_Count'] = df.groupby(['Company', 'Country']).rolling(8, min_periods=1, on='Date')['Sold'].sum().reset_index()['Sold']
df['LastWeek_Count'] = df.groupby(['Company', 'Country', 'Date'])['LastWeek_Count'].transform('max')
s = df.groupby(['Company', 'Country', 'Date'])['Sold'].transform('sum')
df['LastWeek_Count'] = (df['LastWeek_Count']-s).astype(int)
Out[17]:
Date Company Country Sold LastWeek_Count
0 2020-01-01 A BE 1 0
1 2020-01-02 A BE 0 1
2 2020-01-03 A BE 1 1
3 2020-01-03 A BE 1 1
4 2020-01-04 A BE 1 3
5 2020-01-05 B DE 1 0
6 2020-01-06 B DE 0 1
One way would be to first consolidate the Sold value of each group (['Date', 'Company', 'Country']) on a single line using a temporary DF.
After that, apply your .groupby with .rolling with an interval of 8 rows.
After calculating the sum, subtract the value of each line with the value in Sold column and add that column in the original DF with .merge
#convert Date column to datetime
df['Date'] = pd.to_datetime(df['Date'], format='%d.%m.%Y')
#create a temporary DataFrame
df2 = df.groupby(['Date', 'Company', 'Country'])['Sold'].sum().reset_index()
#calc the lastweek
df2['LastWeek_Count'] = (df2.groupby(['Company', 'Country'])
.rolling(8, min_periods=1, on = 'Date')['Sold']
.sum().reset_index(drop=True)
)
#subtract the value of 'lastweek' from the current 'Sold'
df2['LastWeek_Count'] = df2['LastWeek_Count'] - df2['Sold']
#add th2 new column in the original DF
df.merge(df2.drop(columns=['Sold']), on = ['Date', 'Company', 'Country'])
#output:
Date Company Country Sold LastWeek_Count
0 2020-01-01 A BE 1 0.0
1 2020-01-02 A BE 0 1.0
2 2020-01-03 A BE 1 1.0
3 2020-01-03 A BE 1 1.0
4 2020-01-04 A BE 1 3.0
5 2020-01-05 B DE 1 0.0
6 2020-01-06 B DE 0 1.0
import pandas as pd
def nearest(items, pivot):
return min(items, key=lambda x: abs(x - pivot))
df = pd.read_csv("C:/Files/input.txt", dtype=str)
duplicatesDf = df[df.duplicated(subset=['CLASS_ID', 'START_TIME', 'TEACHER_ID'], keep=False)]
duplicatesDf['START_TIME'] = pd.to_datetime(duplicatesDf['START_TIME'], format='%Y/%m/%d %H:%M:%S.%f')
print duplicatesDf
print df['START_TIME'].dt.date
df:
ID,CLASS_ID,START_TIME,TEACHER_ID,END_TIME
1,123,2020/06/01 20:47:26.000,o1,2020/06/02 00:00:00.000
2,123,2020/06/01 20:47:26.000,o1,2020/06/04 20:47:26.000
3,789,2020/06/01 20:47:26.000,o3,2020/06/03 14:47:26.000
4,789,2020/06/01 20:47:26.000,o3,2020/06/03 14:40:00.000
5,456,2020/06/01 20:47:26.000,o5,2020/06/08 20:00:26.000
So, I've got a dataframe like mentioned above. As you can see, I have multiple records with the same CLASS_ID,START_DATE and TEACHER_ID. Whenever, multiple records like these are present, I would like to retain only 1 record based on the condition that, the retained record should have its END_DATE nearest to its START_DATE(by minute level precision).
In this case,
for CLASS_ID 123, the record with ID 1 will be retained, as its END_DATE 2020/06/02 00:00:00.000 is nearest to its START_DATE 2020/06/01 20:47:26.000 as compared to record with ID 2 whose END_DATE is 2020/06/04 20:47:26.000. Similarly for CLASS_ID 789, record with ID 4 will be retained.
Hence the expected output will be:
ID,CLASS_ID,START_TIME,TEACHER_ID,END_TIME
1,123,2020/06/01 20:47:26.000,o1,2020/06/02 00:00:00.000
4,789,2020/06/01 20:47:26.000,o3,2020/06/03 14:40:00.000
5,456,2020/06/01 20:47:26.000,o5,2020/06/08 20:00:26.000
I've been going through the following links,
https://stackoverflow.com/a/32237949,
https://stackoverflow.com/a/33043374
to find a solution but have unfortunately reached an impasse.
Hence, would some kind soul mind helping me out a bit. Many thanks.
IIUC, we can use .loc and idxmin() after creating a condtional column to measure the elapsed time between the start and the end, we will apply idxmin() as a groupby operation on your CLASS_ID column.
df.loc[
df.assign(mins=(df["END_TIME"] - df["START_TIME"]))
.groupby("CLASS_ID")["mins"]
.idxmin()
]
ID CLASS_ID START_TIME TEACHER_ID END_TIME
0 1 123 2020-06-01 20:47:26 o1 2020-06-02 00:00:00
4 5 456 2020-06-01 20:47:26 o5 2020-06-08 20:00:26
3 4 789 2020-06-01 20:47:26 o3 2020-06-03 14:40:00
in steps.
Time Delta.
print(df.assign(mins=(df["END_TIME"] - df["START_TIME"]))[['CLASS_ID','mins']])
CLASS_ID mins
0 123 0 days 03:12:34
1 123 3 days 00:00:00
2 789 1 days 18:00:00
3 789 1 days 17:52:34
4 456 6 days 23:13:00
minimum index from time delta column whilst grouping with CLASS_ID
print(df.assign(mins=(df["END_TIME"] - df["START_TIME"]) )
.groupby("CLASS_ID")["mins"]
.idxmin())
CLASS_ID
123 0
456 4
789 3
Name: mins, dtype: int64
I have a PostgreSQL database and I have a table that I am looking to query to determine which presses have been updated between the first cycle created_timestamp and the most recent cycle created_timestamp. Here is an example of the table, which is called event_log_summary.
press_id cycle_number created_timestamp
1 1 2020-02-07 16:07:52
1 2 2020-02-07 16:07:53
1 3 2020-02-07 16:07:54
1 4 2020-04-01 13:23:10
2 1 2020-01-13 8:33:23
2 2 2020-01-13 8:33:24
2 3 2020-01-13 8:33:25
3 1 2020-02-21 18:45:44
3 2 2020-02-21 18:45:45
3 3 2020-02-26 14:22:12
This is the query that I used to get me a three column output of press_id, mincycle, max_cycle, but then I want to compare the maxcycle created_timestamp to the mincycle created_timestamp and see if there is at least x amount of time between the two, say at least 1 day, I am unsure about how to implement that.
SELECT
press_id,
MIN(cycle_number) AS minCycle,
MAX(cycle_number) AS maxCycle
FROM
event_log_detail
GROUP BY
press_id
I have tried different things like using WHERE (MAX(cycle_number) - MIN(cycle_number > 1), but I am pretty new to SQL and don't quite fully know how to implement this. The output I am looking for, would have a difference of at least one day would be the following:
press_id
1
3
Presses 1 and 3 have their maximum cycle created_timestamp at least 1-day difference than their minimum cycle created_timestamp. I am just looking for the press_ids whose first cycle and the last cycle have a difference of at least 1 day, I don't need any other information on the output, just one column with the press_ids. Any help would be appreciated. Thanks.
You can use a HAVING clause:
select press_id,
max(created_timestamp) - min(created_timestamp) as diff
from event_log_detail
group by press_id
having max(created_timestamp) > min(created_timestamp) + interval '1 day';