Reputation: 135
I got this sample DF :
df = pd.DataFrame({'CUSTOM_CRITERIA':[1111,22222,1111,1212,1212,3333,5555, 1111],
'AD_UNIT_NAME':['inp2_l_d', 'inp1', 'pixel_d', 'inp2_l_d', 'anchor_m','anchor_m','anchor_m','inp2_l_d'],
'TOTAL_CODE_SERVED_COUNT':[10, 20, 10, 12, 18,500,100,50]})
I need to get for each custom_criteria the max total_code_served_count by condition of which has more code served -> anchor_m [total_code served] OR inp2_l_d[total_code served] + pixel_d[total_code served] for each CUSTOM_CRITERIA
My current solution looks like this:
data_dict = clean_data.to_dict(orient='records')
for item in data_dict:
desktop_impression_max_calculated = sum([d['TOTAL_CODE_SERVED_COUNT'] for d in data_dict if d['CUSTOM_CRITERIA'] == item['CUSTOM_CRITERIA'] and ('inp2_l_d' in d['AD_UNIT_NAME'].lower() or 'pixel_d' in d['AD_UNIT_NAME'].lower())])
mobile_impression_max_calculated = sum([d['TOTAL_CODE_SERVED_COUNT'] for d in data_dict if d['CUSTOM_CRITERIA'] == item['CUSTOM_CRITERIA'] and 'anchor_m' in d['AD_UNIT_NAME'].lower()])
item['IMPRESSIONS_MAX'] = max(desktop_impression_max_calculated,mobile_impression_max_calculated)
clean_data = pd.DataFrame(data_dict)
agg_map = {'IMPRESSIONS_MAX': 'first' }
clean_data = clean_data.groupby('CUSTOM_CRITERIA').agg(agg_map).reset_index()
this takes a long time to run when a high amount of data is present due to N^2 complexity. I'm sure there is a better and simpler way to do it with pandas.
Upvotes: 3
Views: 70
Reputation: 71689
You can create two masked columns by multiplying the values in TOTAL_CODE_SERVED_COUNT
column by the boolean masks m1
and m2
, then groupby
these masked columns on CUSTOM_CRITERIA
and aggregate using sum
, finally take the max
along axis=1
to get the final result:
m1 = df['AD_UNIT_NAME'].str.contains(r'(?i)inp2_l_d|pixel_d')
m2 = df['AD_UNIT_NAME'].str.contains(r'(?i)anchor_m')
pd.DataFrame((df['TOTAL_CODE_SERVED_COUNT'].values * [m1, m2]).T)\
.groupby(df['CUSTOM_CRITERIA']).sum().max(1).reset_index(name='IMPRESSIONS_MAX')
CUSTOM_CRITERIA IMPRESSIONS_MAX
0 1111 70
1 1212 18
2 3333 500
3 5555 100
4 22222 0
Upvotes: 4