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Calculate difference between values with same key index

Hello: I have this DataFrame (sample)
User
Timestamp
Production
A
2020-01-01
5
A
2020-06-01
7
A
2020-12-01
15
B
2020-01-01
2
B
2020-06-01
7
B
2020-12-01
9
So, I need to calculate the difference between the values and append to the DataFrame. The resulting column i'm trying to obtain should be as follows: the production column for all the periods per user.
The resulting table would be as follows (production column omitted due to problem in table editor):
User
Timestamp
Difference
A
2020-01-01
5
A
2020-06-01
2
A
2020-12-01
8
B
2020-01-01
2
B
2020-06-01
5
B
2020-12-01
2
So i tried with the .diff() function but obviously it doesn't recognize when the index is different. I then tried with a groupby() applied on the user column to then compute the diff function, but I get the same problem:
df['Difference'] = df.groupby('User')['Production'].diff()
Can someone help me out?
Thanks!
EDIT:
Made a step ahead, but still trying to figure it out:
i wrote this:
grouped = df.groupby('User')
diff = lambda x: x['Production'].shift(+1) - x['Production']
daf['diff'] = grouped.apply(diff).reset_index(0, drop=True).fillna(df['Production'])
This does the difference the way I want It, but still messes up when the User identifier changes.

Pandas total count each day

I have a large dataset (df) with lots of columns and I am trying to get the total number of each day.
|datetime|id|col3|col4|col...
1 |11-11-2020|7|col3|col4|col...
2 |10-11-2020|5|col3|col4|col...
3 |09-11-2020|5|col3|col4|col...
4 |10-11-2020|4|col3|col4|col...
5 |10-11-2020|4|col3|col4|col...
6 |07-11-2020|4|col3|col4|col...
I want my result to be something like this
|datetime|id|col3|col4|col...|Count
6 |07-11-2020|4|col3|col4|col...| 1
3 |5|col3|col4|col...| 1
2 |10-11-2020|5|col3|col4|col...| 1
4 |4|col3|col4|col...| 2
1 |11-11-2020|7|col3|col4|col...| 1
I tried to use resample like this df = df.groupby(['id','col3', pd.Grouper(key='datetime', freq='D')]).sum().reset_index() and this is my result. I am still new to programming and Pandas but I have read up on pandas docs and am still unable to do it.
|datetime|id|col3|col4|col...
6 |07-11-2020|4|col3|1|0.0
3 |07-11-2020|5|col3|1|0.0
2 |10-11-2020|5|col3|1|0.0
4 |10-11-2020|4|col3|2|0.0
1 |11-11-2020|7|col3|1|0.0

try this:
df = df.groupby(['datetime','id','col3']).count()

If you want the count values for all columns based only on the date, then:
df.groupby('datetime').count()
And you'll get a DataFrame who has the date time as the index and the column cells representing the number of entries for that given index.

Python pandas: Get first values of group

I have a list of recorded diagnoses like this:
df = pd.DataFrame({
"DiagnosisTime": ["2017-01-01 08:23:00", "2017-01-01 08:23:00", "2017-01-01 08:23:03", "2017-01-01 08:27:00", "2019-12-31 20:19:39", "2019-12-31 20:19:39"],
"ID": [1,1,1,1,2,2]
})
There are multiple subjects that can be identified by an ID. For each subject there may be one or more diagnosis. Each diagnosis may be comprised of multiple entries (as multiple things are recoreded (not in this example)).
The individual diagnoses (with multiple rows) can (to some extend) be identified by the DiagnosisTime. However, sometimes there is a little delay during the writing of data for one diagnosis so that I want to allow a small tolerance of a few seconds when grouping by DiagnosisTime.
In this example I want a result as follows:
There are two diagnoses for ID 1: rows 0, 1, 2 and row 3. Note the slightly different DiagnosisTime in row 2 compared to 0 and 1. ID 2 is comprised of 1 diagnosis comprised of rows 4 and 5.
For each ID I want to set the counter back to 1 (or 0 if thats easier).
This is how far I've come:
df["DiagnosisTime"] = pd.to_datetime(df["DiagnosisTime"])
df["diagnosis_number"] = df.groupby([pd.Grouper(freq='5S', key="DiagnosisTime"), 'ID']).ngroup()
I think I successfully identified diagnoses within one ID (not entirely sure about the Grouper), but I don't know how to reset the counter.
If that is not possible I would also be satisfied with a function which returns all records of one ID that have the lowest diagnosis_number within that group.

You can add lambda function with GroupBy.transform and factorize:
df["diagnosis_number"] = (df.groupby('ID')['diagnosis_number']
.transform(lambda x: pd.factorize(x)[0]) + 1)
print (df)
DiagnosisTime ID diagnosis_number
0 2017-01-01 08:23:00 1 1
1 2017-01-01 08:23:00 1 1
2 2017-01-01 08:23:03 1 1
3 2017-01-01 08:27:00 1 2
4 2019-12-31 20:19:39 2 1
5 2019-12-31 20:19:39 2 1

Pandas: using groupby and nunique taking time into account

I have a dataframe in this form:
A B time
1 2 2019-01-03
1 3 2018-04-05
1 4 2020-01-01
1 4 2020-02-02
where A and B contain some integer identifiers.
I want to measure the number of different identifiers each A has interacted with. To do this I usually simply do
df.groupby('A')['B'].nunique()
I now have to do a slightly different thing: each identifier has a date assigned (different for each identifier), that splits its interactions in 2 parts: the ones happening before that date, and the ones happening after that date. The same operation previously done (counting number of unique B interacted with ) needs to be done for both parts separately.
For example, if the date for A=1 was 2018-07-01, the output would be
A before after
1 1 2
In the real data, A contains millions of different identifiers, each with its unique date assigned.
EDITED
To be more clear I added a line to df. I want to count the number of different values of B each A interacts with before and after the date

I would convert A into dates, compare those with df['time'] and then groupby().value_counts():
(df['A'].map(date_dict)
.gt(df['time'])
.groupby(df['A'])
.value_counts()
.unstack()
.rename({False:'after',True:'before'}, axis=1)
)
Output:
after before
A
1 2 1

pandas - How to check consecutive order of dates and copy groups of them?

At first I have two problems, the first will follow now:
I a dataframe df with many times the same userid and along with it a date and some unimportant other columns:
userid date
0 243 2014-04-01
1 234 2014-12-01
2 234 2015-11-01
3 589 2016-07-01
4 589 2016-03-01
I am currently trying to groupby them by userid and sort the dates descending and cut out the twelve oldest. My code looks like this:
df = df.groupby(['userid'], group_keys=False).agg(lambda x: x.sort_values(['date'], ascending=False, inplace=False).head(12))
And I get this error:
ValueError: cannot copy sequence with size 6 to array axis with dimension 12
At the moment my aim is to avoid to split the dataframe in individual ones.
My second problem is more complex:
I try to find out if the sorted dates (respectively per group of userids) are monthly consecutive. This means if there is an date for one group of userid, for example userid: 234 and date: 2014-04-01, the next entry below must be userid: 234 and date:2014-03-01. There is no focus on the day, only the year and month are important.
And only this consecutive 12 dates should be copied in another dataframe.
A second dataframe df2 contains the same userid, but they are unique and another column is 'code'. Here is an example:
userid code
0 433805 1
24 5448 0
48 3434 1
72 34434 1
96 3202 1
120 23766 1
153 39457 0
168 4113 1
172 3435 5
374 34093 1
I summarize: I try to check if there are 12 consecutive months per userid and copy every correct sequence in another dataframe. For this I have also compare the 'code' from df2.
This is a version of my code:
df['YearMonthDiff'] = df['date'].map(lambda x: 1000*x.year + x.month).diff()
df['id_before'] = df['userid'].shift()
final_df = pd.DataFrame()
for group in df.groupby(['userid'], group_keys=False):
fi = group[1]
if (fi['userid'] <> fi['id_before']) & group['YearMonthDiff'].all(-1.0) & df.loc[fi.userid]['code'] != 5:
final_df.append(group['userid','date', 'consum'])
At first calculated from the date an integer and made diff(). On other posts I saw they shift the column to compare the values from the current row and the row before. Then I made groupby(userid) to iterate over the single groups. Now it's extra ugly I tried to find the beginning of such an userid-group, try to check if there are only consecutive months and the correct 'code'. And at least I append it on the final dataframe.
On of the biggest problems is to compare the row with the following row. I can iterate over them with iterrow(), but I cannot compare them without shift(). There exits a calendar function, but on these I will take a look on the weekend. Sorry for the mess I am new to pandas.
Has anyone an idea how to solve my problem?

for your first problem, try this
df.groupby(by='userid').apply(lambda x: x.sort_values(by='date',ascending=False).iloc[[e for e in range(12) if e <len(x)]])

Using groupby and nlargest, we get the index values of those largest dates. Then we use .loc to get just those rows
df.loc[df.groupby('userid').date.nlargest(12).index.get_level_values(1)]
Consider the dataframe df
dates = pd.date_range('2015-08-08', periods=10)
df = pd.DataFrame(dict(
userid=np.arange(2).repeat(4),
date=np.random.choice(dates, 8, False)
))
print(df)
date userid
0 2015-08-12 0 # <-- keep
1 2015-08-09 0
2 2015-08-11 0
3 2015-08-15 0 # <-- keep
4 2015-08-13 1
5 2015-08-10 1
6 2015-08-17 1 # <-- keep
7 2015-08-16 1 # <-- keep
We'll keep the latest 2 dates per user id
df.loc[df.groupby('userid').date.nlargest(2).index.get_level_values(1)]
date userid
0 2015-08-12 0
3 2015-08-15 0
6 2015-08-17 1
7 2015-08-16 1

Maybe someone is interested, I solved my second problem thus:
I cast the date to an int, calculated the difference and I shift the userid one row, like in my example. And then follows this... found a solution on stackoverflow
gr_ob = df.groupby('userid')
gr_dict = gr_ob.groups
final_df = pd.DataFrame(columns=['userid', 'date', 'consum'])
for group_name in gr_dict.keys():
new_df = gr_ob.get_group(group_name)
if (new_df['userid'].iloc[0] <> new_df['id_before'].iloc[0]) & (new_df['YearMonthDiff'].iloc[1:] == -1.0).all() & (len(new_df) == 12):
final_df = final_df.append(new_df[['userid', 'date', 'consum']])