Pandas boolean condition from nested list of dictionaries - pandas

Question

[{'id': 123,
'type': 'salary', #Parent node
'tx': 'house',
'sector': 'EU',
'transition': [{'id': 'hash', #Child node
'id': 123,
'type': 'salary',
'tx': 'house' }]},
{'userid': 123,
'type': 'salary', #Parent node
'tx': 'office',
'transition': [{'id': 'hash', # Child node
'id': 123,
'type': 'salary',
'tx': 'office'}]}]
As a pandas column ('info') I have some information stored as a nested list of dictionaries like the example above.
What I'm trying to do is a boolean condition whether this list has the following attributes:
More than one 'type' == 'salary' in any of all parents nodes
Field 'tx' is different in any of all parents nodes with 'type' == 'salary'
So far I've tried to flatten a list and filter but it is not solving the first and seconds nodes
a = df.iloc[0].info
values = [item for sublist in [[list(i.values()) for i in a]][0]for item in sublist]

Answer 1

If you want to one line solution, you can use:
df['check'] = df['info'].apply(lambda x: True if sum([1 if i['type']=='salary' else 0 for i in x]) > 1 and [i['tx'] for i in x if i['type']=='salary'].count([i['tx'] for i in x if i['type']=='salary'][0]) != len([i['tx'] for i in x if i['type']=='salary']) else False)
or (expanded):
def check(x):
total_salary = sum([1 if i['type']=='salary' else 0 for i in x]) # get count of "type": "salary" matches
tx_list = [i['tx'] for i in x if i['type']=='salary'] # get tx values when type==salary
tx_check = tx_list.count(tx_list[0]) != len(tx_list) # check all values are same in tx_list
if total_salary > 1 and tx_check:
return True
else:
return False
df['check'] = df['info'].apply(check)