Develop Reference

efficient way to calculate between columns with conditions

I have a dataframe looks like
Cnt_A Cnt_B Cnt_C Cnt_D
ID_1 0 1 3 0
ID_2 1 0 0 0
ID_3 5 2 0 8
...
I'd like to count columns that are not zero and put the result into new column like this,
Total_Not_Zero_Cols Cnt_A Cnt_B Cnt_C Cnt_D
ID_1 2 0 1 3 0
ID_2 1 1 0 0 0
ID_3 3 5 2 0 8
...
I did loop to get the result, but it took very long time (of course).
I can't figure out the most efficient way to calculate between columns with condition :(
Thank you in advance

Check if each value not equals to 0 then sum on columns axis:
df['Total_Not_Zero_Cols'] = df.ne(0).sum(axis=1)
print(df)
# Output
Cnt_A Cnt_B Cnt_C Cnt_D Total_Not_Zero_Cols
ID_1 0 1 3 0 2
ID_2 1 0 0 0 3
ID_3 5 2 0 8 1

Use ne to generate a DataFrame of booleans with True for non-zeros values, then aggregate the rows as integers using sum:
df['Total_Not_Zero_Cols'] = df.ne(0).sum(axis=1)

Numpy based -
Use -
np.sum(df!=0, axis=1)
Output
ID_1 2
ID_2 1
ID_3 3
dtype: int64

Count number of columns above a date

I have a pandas dataframe with several columns and I would like to know the number of columns above the date 2016-12-31 . Here is an example:
ID
Bill
Date 1
Date 2
Date 3
Date 4
Bill 2
4
6
2000-10-04
2000-11-05
1999-12-05
2001-05-04
8
6
8
2016-05-03
2017-08-09
2018-07-14
2015-09-12
17
12
14
2016-11-16
2017-05-04
2017-07-04
2018-07-04
35
And I would like to get this column
Count
0
2
3

Just create the mask and call sum on axis=1
date = pd.to_datetime('2016-12-31')
(df[['Date 1','Date 2','Date 3','Date 4']]>date).sum(1)
OUTPUT:
0 0
1 2
2 3
dtype: int64
If needed, call .to_frame('count') to create datarame with column as count
(df[['Date 1','Date 2','Date 3','Date 4']]>date).sum(1).to_frame('Count')
Count
0 0
1 2
2 3

Use df.filter to filter the Date* columns + .sum(axis=1)
(df.filter(like='Date') > '2016-12-31').sum(axis=1).to_frame(name='Count')
Result:
Count
0 0
1 2
2 3

You can do:
df['Count'] = (df.loc[:, [x for x in df.columns if 'Date' in x]] > '2016-12-31').sum(axis=1)
Output:
ID Bill Date 1 Date 2 Date 3 Date 4 Bill 2 Count
0 4 6 2000-10-04 2000-11-05 1999-12-05 2001-05-04 8 0
1 6 8 2016-05-03 2017-08-09 2018-07-14 2015-09-12 17 2
2 12 14 2016-11-16 2017-05-04 2017-07-04 2018-07-04 35 3
We select columns with 'Date' in the name. It's better when we have lots of columns like these and don't want to put them one by one. Then we compare it with lookup date and sum 'True' values.

how to groupby and aggregate dynamic columns in pandas

I have following dataframe in pandas
code tank nozzle_1 nozzle_2 nozzle_var nozzle_sale
123 1 1 1 10 10
123 1 2 2 12 10
123 2 1 1 10 10
123 2 2 2 12 10
123 1 1 1 10 10
123 2 2 2 12 10
Now, I want to generate cumulative sum of all the columns grouping over tank and take out the last observation. nozzle_1 and nozzle_2 columns are dynamic, it could be nozzle_3, nozzle_4....nozzle_n etc. I am doing following in pandas to get the cumsum
## Below code for calculating cumsum of dynamic columns nozzle_1 and nozzle_2
cols= df.columns[df.columns.str.contains(pat='nozzle_\d+$', regex=True)]
df.assign(**df.groupby('tank')[cols].agg(['cumsum'])\
.pipe(lambda x: x.set_axis(x.columns.map('_'.join), axis=1, inplace=False)))
## nozzle_sale_cumsum is static column
df[nozzle_sale_cumsum] = df.groupby('tank')['nozzle_sale'].cumsum()
From above code I will get cumsum of following columns
tank nozzle_1 nozzle_2 nozzle_var nozzle_1_cumsum nozzle_2_cumsum nozzle_sale_cumsum
1 1 1 10 1 1 10
1 2 2 12 3 3 20
2 1 1 10 1 1 10
2 2 2 12 3 3 20
1 1 1 10 4 4 30
2 2 2 12 5 5 30
Now, I want to get last values of all 3 cumsum columns grouping over tank. I can do it with following code in pandas, but it is hard coded with column names.
final_df= df.groupby('tank').agg({'nozzle_1_cumsum':'last',
'nozzle_2_cumsum':'last',
'nozzle_sale_cumsum':'last',
}).reset_index()
Problem with above code is nozzle_1_cumsum and nozzle_2_cumsum is hard coded which is not the case. How can I do this in pandas with dynamic columns.

How about:
df.filter(regex='_cumsum').groupby(df['tank']).last()
Output:
nozzle_1_cumsum nozzle_2_cumsum nozzle_sale_cumsum
tank
1 4 4 30
2 5 5 30
You can also replace df.filter(...) by, e.g., df.iloc[:,-3:] or df[col_names].

Subtracting Rows based on ID Column - Pandas

I have a dataframe which looks like this:
UserId Date_watched Days_not_watch
1 2010-09-11 5
1 2010-10-01 8
1 2010-10-28 1
2 2010-05-06 12
2 2010-05-18 5
3 2010-08-09 10
3 2010-09-25 5
I want to find out the no. of days the user gave as a gap, so I want a column for each row for each user and my dataframe should look something like this:
UserId Date_watched Days_not_watch Gap(2nd watch_date - 1st watch_date - days_not_watch)
1 2010-09-11 5 0 (First gap will be 0 for all users)
1 2010-10-01 8 15 (11th Sept+5=16th Sept; 1st Oct - 16th Sept=15days)
1 2010-10-28 1 9
2 2010-05-06 12 0
2 2010-05-18 5 0 (because 6th May+12 days=18th May)
3 2010-08-09 10 0
3 2010-09-25 4 36
3 2010-10-01 2 2
I have mentioned the formula for calculating the Gap beside the column name of the dataframe.

Here is one approach using groupby + shift:
# sort by date first
df['Date_watched'] = pd.to_datetime(df['Date_watched'])
df = df.sort_values(['UserId', 'Date_watched'])
# calculate groupwise start dates, shifted
grp = df.groupby('UserId')
starts = grp['Date_watched'].shift() + \
pd.to_timedelta(grp['Days_not_watch'].shift(), unit='d')
# calculate timedelta gaps
df['Gap'] = (df['Date_watched'] - starts).fillna(pd.Timedelta(0))
# convert to days and then integers
df['Gap'] = (df['Gap'] / pd.Timedelta('1 day')).astype(int)
print(df)
UserId Date_watched Days_not_watch Gap
0 1 2010-09-11 5 0
1 1 2010-10-01 8 15
2 1 2010-10-28 1 19
3 2 2010-05-06 12 0
4 2 2010-05-18 5 0
5 3 2010-08-09 10 0
6 3 2010-09-25 5 37

Drop rows after maximum value in a grouped Pandas dataframe

I've got a date-ordered dataframe that can be grouped. What I am attempting to do is groupby a variable (Person), determine the maximum (weight) for each group (person), and then drop all rows that come after (date) the maximum.
Here's an example of the data:
df = pd.DataFrame({'Person': 1,1,1,1,1,2,2,2,2,2],'Date': '1/1/2015','2/1/2015','3/1/2015','4/1/2015','5/1/2015','6/1/2011','7/1/2011','8/1/2011','9/1/2011','10/1/2011'], 'MonthNo':[1,2,3,4,5,1,2,3,4,5], 'Weight':[100,110,115,112,108,205,210,211,215,206]})
Date MonthNo Person Weight
0 1/1/2015 1 1 100
1 2/1/2015 2 1 110
2 3/1/2015 3 1 115
3 4/1/2015 4 1 112
4 5/1/2015 5 1 108
5 6/1/2011 1 2 205
6 7/1/2011 2 2 210
7 8/1/2011 3 2 211
8 9/1/2011 4 2 215
9 10/1/2011 5 2 206
Here's what I want the result to look like:
Date MonthNo Person Weight
0 1/1/2015 1 1 100
1 2/1/2015 2 1 110
2 3/1/2015 3 1 115
5 6/1/2011 1 2 205
6 7/1/2011 2 2 210
7 8/1/2011 3 2 211
8 9/1/2011 4 2 215
I think its worth noting, there can be disjoint start dates and the maximum may appear at different times.
My idea was to find the maximum for each group, obtain the MonthNo the maximum was in for that group, and then discard any rows with MonthNo greater Max Weight MonthNo. So far I've been able to obtain the max by group, but cannot get past doing a comparison based on that.
Please let me know if I can edit/provide more information, haven't posted many questions here! Thanks for the help, sorry if my formatting/question isn't clear.

Using idxmax with groupby
df.groupby('Person',sort=False).apply(lambda x : x.reset_index(drop=True).iloc[:x.reset_index(drop=True).Weight.idxmax()+1,:])
Out[131]:
Date MonthNo Person Weight
Person
1 0 1/1/2015 1 1 100
1 2/1/2015 2 1 110
2 3/1/2015 3 1 115
2 0 6/1/2011 1 2 205
1 7/1/2011 2 2 210
2 8/1/2011 3 2 211
3 9/1/2011 4 2 215

You can use groupby.transform with idxmax. The first 2 steps may not be necessary depending on how your dataframe is structured.
# convert Date to datetime
df['Date'] = pd.to_datetime(df['Date'])
# sort by Person and Date to make index usable for next step
df = df.sort_values(['Person', 'Date']).reset_index(drop=True)
# filter for index less than idxmax transformed by group
df = df[df.index <= df.groupby('Person')['Weight'].transform('idxmax')]
print(df)
Date MonthNo Person Weight
0 2015-01-01 1 1 100
1 2015-02-01 2 1 110
2 2015-03-01 3 1 115
5 2011-06-01 1 2 205
6 2011-07-01 2 2 210
7 2011-08-01 3 2 211
8 2011-09-01 4 2 215