Compare Same and Different in Two Columns of Dataframe

UPDATE: A comment by @rickhg12hs points out that my earlier answers (see below), while detecting when a person switches to a new cell phone service, do not handle the case of a person switching back to a previous service.

To handle this possibility, we must use something like the logic in another answer (@Pranav Hosangadi), though I would do it slightly differently:

df['new'] = ( df
    .groupby('address_id', sort=False)
    .apply(lambda x: x.business != x.business.shift().bfill())
    .reset_index(0).business )

Input:

   address_id business
0           1  verizon
1           1  verizon
2           1  comcast
3           2   sprint
4           2      att
5           2   sprint

Output:

   address_id business    new
0           1  verizon  False
1           1  verizon  False
2           1  comcast   True
3           2   sprint  False
4           2      att   True
5           2   sprint   True

Performance comparison: Here is test code for 600k rows and 5 columns, with results showing that PREVIOUS UPDATE takes about 0.1 seconds to identify 333815 rows for which new==True, while UPDATE takes about 35 seconds to find 335334 True rows, reflecting about 0.5% additional rows for which a person has switched cell phone services and then switched back.

rng = np.random.default_rng()
details = {
'address_id': rng.integers(1,100_000, size=600_000),
'business': [['verizon','comcast','sprint','att'][i] for i in rng.integers(0,3, size=600_000)],
'foo': 1,
'bar': 2
}
df = pd.DataFrame(details)
 
print('groupby() ...')
start = datetime.now()
x = ( df
    .groupby('address_id', sort=False) )
print(f'... complete after {datetime.now() - start} time elapsed.')

print('apply() ...')
start = datetime.now()
x = ( x
    .apply(lambda x: x.business != x.business.shift().bfill()) )
print(f'... complete after {datetime.now() - start} time elapsed.')

print('reset_index() ...')
start = datetime.now()
df['new'] = ( x
    .reset_index(0).business )
print(f'... complete after {datetime.now() - start} time elapsed.')

print(df)
print('rows with "new" == True', df.new.sum())

df = pd.DataFrame(details)
print('PREVIOUS UPDATE() ...')
start = datetime.now()
df['new'] = df.address_id.map(df.groupby('address_id').first().business) != df.business
print(f'... complete after {datetime.now() - start} time elapsed.')
print('rows with "new" == True', df.new.sum())

Results:

groupby() ...
... complete after 0:00:00 time elapsed.
apply() ...
... complete after 0:00:33.541322 time elapsed.
reset_index() ...
... complete after 0:00:00.040942 time elapsed.
        address_id business  foo  bar    new
0            20223   sprint    1    2  False
1            29297  comcast    1    2  False
2            92489  comcast    1    2  False
3            29297  verizon    1    2   True
4            98901  comcast    1    2  False
...            ...      ...  ...  ...    ...
599995       29823  comcast    1    2   True
599996       39328  comcast    1    2   True
599997       27594  comcast    1    2  False
599998       14903   sprint    1    2   True
599999       87375  verizon    1    2   True

[600000 rows x 5 columns]
rows with "new" == True 335334

PREVIOUS UPDATE() ...
... complete after 0:00:00.097930 time elapsed.
rows with "new" == True 333815

PREVIOUS UPDATE: Here is an even simpler way than my original answer using join() (see below) to do what your question asks:

df['new'] = df.address_id.map(df.groupby('address_id').first().business) != df.business

Explanation:

• Use groupby() and first() to create a dataframe whose business column contains the first one encountered for each address_id
• Use Series.map() to transform the original dataframe's address_id column into this first business value
• Add column new which is True only if this new business differs from the original business column.

ORIGINAL SOLUTION:

Here is a simple way to do what you've asked using groupby() and join():

df = df.join(df.groupby('address_id').first(), on='address_id', rsuffix='_first')
df = df.assign(new=df.business != df.business_first).drop(columns='business_first')f

Output:

   address_id business    new
0           1  verizon  False
1           1  verizon  False
2           1  comcast   True
3           2   sprint  False
4           2      att   True

Explanation:

• Use groupby() and first() to create a dataframe whose business column contains the first one encountered for each address_id
• Use join() to add a column business_first to df containing the corresponding first business for each address_id
• Use assign() to add a column new containing a boolean indicating whether the row contains a new business with an existing address_id
• Use drop() to eliminate the business_first column.

First, groupby on address_id.

groups = df.groupby("address_id")

Then, iterate over the groups, and find where the value of business changes:

for address_id, grp_data in groups:
    changed = grp_data['business'].ne(grp_data['business'].shift().bfill())
   
    df.loc[grp_data.index, "changed"] = changed

.shift().bfill() shifts all data one index over (0 -> 1, 1 -> 2, and so on), and then backfills the first value. For example:

>>> df["business"]
0    verizon
1    verizon
2    comcast
3     sprint
4        att
Name: business, dtype: object

>>> df["business"].shift()
0        NaN
1    verizon
2    verizon
3    comcast
4     sprint
Name: business, dtype: object

>>> df["business"].shift().bfill()
0    verizon
1    verizon
2    verizon
3    comcast
4     sprint
Name: business, dtype: object

Running the loop makes the following dataframe:

   address_id business changed
0           1  verizon   False
1           1  verizon   False
2           1  comcast    True
3           2   sprint   False
4           2      att    True