1. groupby.head(1)

The relevant groupby method to drop duplicates in each group is groupby.head(1). Note that it is important to pass 1 to select the first row of each date-cid pair.

df1 = df.groupby(['date', 'cid']).head(1)

2. duplicated() is more flexible

Another method is to use duplicated() to create a boolean mask and filter.

df3 = df[~df.duplicated(['date', 'cid'])]

An advantage of this method over drop_duplicates() is that is can be chained with other boolean masks to filter the dataframe more flexibly. For example, to select the unique cids in Nevada for each date, use:

df_nv = df[df['state'].eq('NV') & ~df.duplicated(['date', 'cid'])]

3. groupby.sample(1)

Another method to select a unique row from each group to use groupby.sample(). Unlike the previous methods mentioned, it selects a row from each group randomly (whereas the others only keep the first row from each group).

df4 = df.groupby(['date', 'cid']).sample(n=1)

You can verify that df1, df2 (ayhan's output) and df3 all produce the very same output and df4 produces an output where size and nunique of cid match for each date (as required in the OP). In short, the following returns True.

w, x, y, z = [d.groupby('date')['cid'].agg(['size', 'nunique']) for d in (df1, df2, df3, df4)]
w.equals(x) and w.equals(y) and w.equals(z)   # True

and w, x, y, z all look like the following:

       size  nunique
date        
2005      7        3
2006    237       10
2007   3610      227
2008   1318       52
2009   2664      142
2010    997       57
2011   6390      219
2012   2904       99
2013   7875      238
2014   3979      146

You can try to sort the values first and then groupby.tail

df.sort_values('return').groupby(['ticker','year']).tail(1)

  ticker  year  return
0   aapl  1999       1
1   aapl  2000       3

Rather than removing duplicates during the pivot table process, use the df.drop_duplicates() function to selectively drop duplicates.

For example if you are pivoting using these index='c0' and columns='c1' then this simple step yields the correct counts.

In this example the 5th row is a duplicate of the 4th (ignoring the non-pivoted c2 column

import pandas as pd
data = {'c0':[0,1,0,1,1], 'c1':[0,0,1,1,1], 'person':[0,0,1,1,1], 'c_other':[1,2,3,4,5]}
df = pd.DataFrame(data)
df2 = df.drop_duplicates(subset=['c0','c1','person'])
pd.pivot_table(df2, index='c0',columns='c1',values='person', aggfunc='count')

This correctly outputs

c1  0  1
c0      
0   1  1
1   1  1

Alternative solution with diff and sum.

data['days'] = data.sort_values('date').groupby('id').date.diff()
data = data.groupby(['id', 'jtitle']).agg({'days': 'sum', 'date': 'first'}).reset_index()

# to filter to only more than 0 days
data[data.days.dt.days > 0]

Result

     id jtitle      days       date
0  2000   job1 3570 days 2021-01-01
1  2001   job1 1096 days 2021-02-17
2  3000   job3 7356 days 2021-01-01
3  3300   job5  134 days 2022-01-22

Group By returns key, value pairs where key is the identifier of the group and the value is the group itself, i.e. a subset of an original df that matched the key.

In your example week_grouped = df.groupby('week') is set of groups (pandas.core.groupby.DataFrameGroupBy object) which you can explore in detail as follows:

for k, gr in week_grouped:
    # do your stuff instead of print
    print(k)
    print(type(gr)) # This will output <class 'pandas.core.frame.DataFrame'>
    print(gr)
    # You can save each 'gr' in a csv as follows
    gr.to_csv('{}.csv'.format(k))

Or alternatively you can compute aggregation function on your grouped object

result = week_grouped.sum()
# This will be already one row per key and its aggregation result
result.to_csv('result.csv') 

In your example you need to assign the function result to some variable as by default pandas objects are immutable.

some_variable = week_grouped.sum() 
some_variable.to_csv('week_grouped.csv') # This will work

basically result.csv and week_grouped.csv are meant to be same