Home > Blockchain >  Calculating handling time out of overlapping intervals
Calculating handling time out of overlapping intervals

Time:10-20

I have a raw data exported and transformed a bit from salesforce below;

df = pd.DataFrame(columns=['contact_start','name', 'aht'], 
                  data=[['2021-09-27 09:58:00','Venus','180'],
                        ['2021-09-27 10:00:00','Venus','240'],
                        ['2021-09-27 11:05:00','Venus','60'],
                        ['2021-09-27 10:55:00','Mars','30'],
                        ['2021-09-27 10:56:00','Mars','30']])

original data

using these codes below

df["contact_start"] = pd.to_datetime(df["contact_start"], format = "%Y-%m-%d %H:%M:%S",errors='coerce')
df["date"] = df["contact_start"].dt.strftime('%Y-%m-%d')
df['aht']=pd.to_datetime(df["aht"], unit='s').dt.strftime("%H:%M:%S")
df['contact_finish'] = pd.to_timedelta(df['aht'])   pd.to_datetime(df['contact_start'])
df['contact_finish'] = df['contact_finish'].astype('datetime64[s]')

I transform this into :

transform

but my final goal is to deal with overlapping and I ran out of ideas how to make it happen.

the outcome should be like this in below:

df = pd.DataFrame(columns=['date','name', 'total_duration_sec'], 
                  data=[['2021-09-27','Venus','420'], 
                        ['2021-09-27','Mars','60']])

outcome

I guess it looks simple but in fact it is really not. I would appreciate any help.

CodePudding user response:

I think you could create a time difference in seconds between successive contact_start per name

upper_seconds = (
    df.sort_values(['name','contact_start'])
      .groupby('name')['contact_start'].diff(-1)
      .dt.total_seconds().abs())

print(upper_seconds.sort_index())
# 0     120.0
# 1    3900.0
# 2       NaN
# 3      60.0
# 4       NaN
# Name: contact_start, dtype: float64

Now you can use this as a upper clip on aht then groupby name and date and sum.

res = (
    df['aht'].astype(int)
      .clip(upper=upper_seconds)
      .groupby([df['name'], df['date']]).sum()
      .reset_index(name='total_duration_sec')
)
print(res)
    name        date  total_duration_sec
0   Mars  2021-09-27                  60
1  Venus  2021-09-27                 420

Note that I used first two lines you already wrote to have the good type.

df["contact_start"] = pd.to_datetime(df["contact_start"], 
                                     format = "%Y-%m-%d %H:%M:%S",errors='coerce')
df["date"] = df["contact_start"].dt.strftime('%Y-%m-%d')

CodePudding user response:

You can make your existing code work by adding these lines to your code:

overlapped = pd.Series(df.groupby(['name']).apply(lambda x: (x['contact_finish'] - x['contact_start'].shift(-1)).dt.total_seconds().shift()).droplevel(0), name='overlapped')
overlapped = overlapped.mask(overlapped<0, 0).fillna(0)

df['date'] = df['contact_start'].dt.date
df = df.groupby(['date', 'name']).apply(lambda x: (((x['contact_finish'] - x['contact_start']).dt.seconds) - overlapped).sum()).reset_index(name='total_duration_sec')

OUTPUT:

         date   name  total_duration_sec
0  2021-09-27   Mars                60.0
1  2021-09-27  Venus               420.0
  • Related