How can i get the minimum item more efficiently?

Question

I have a list consist of tuple items like (id, cost, clicks, views) as below:

statistic_data_list = [(12324, 9, 6, 9), (12325, 11, 5, 3), (12326, 10, 7, 2)] 

And I want to get the item's id which meet following conditions:

• 1 when not all cost of item is equal to 0, get the item's id which cost is lowest.
• 2 when all cost of item is equal 0, then if not all clicks of item is equal to 0, get the item's id which clicks is lowest.
• 3 when all clicks of item is equal to 0, then if not all views of item is equal to 0, get the item's id which views is lowest.

# (1)
#  input:  
[(12324, 9, 6, 9), (12325, 11, 5, 3), (12326, 10, 7, 2)]
#  expected result: 
12324 # (whose cost is lowest) 

# (2)
#  input:  
[(12324, 0, 6, 9), (12325, 0, 5, 3), (12326, 0, 7, 2)]
#  expected result:
12325 #  (whose clicks is lowest when all cost is 0)

# (3)
#  input:  
[(12324, 0, 0, 9), (12325, 0, 0, 3), (12326, 0, 0, 2)]
#  expected result: 
12326  #  (whose views is lowest when all cost is 0 also clicks)

How can I get the specified item's id more efficiently?

# My attemp so far

cost_clicks_views_list = [(12324, 9, 6, 9), (12325, 11, 5, 3), (12326, 10, 7, 2)]

len_cost_not_0 = len(list(filter(lambda item: item[1], cost_clicks_views_list)))
len_clicks_not_0 = len(list(filter(lambda item: item[2], cost_clicks_views_list)))
len_views_not_0 = len(list(filter(lambda item: item[3], cost_clicks_views_list)))


if len_cost_not_0:
    min_cost_id_list = [ item[0] for item in cost_clicks_views_list if item[1]==min([i[1] for i in cost_clicks_views_list]) ]
    print(min_cost_id_list) # [(12324]

else:
    if len_clicks_not_0:
        min_clicks_id_list = [item[0] for item in cost_clicks_views_list if item[2] == min([i[2] for i in cost_clicks_views_list])]
        print(min_clicks_id_list) # [(12325]

    else:
        if len_views_not_0:
            min_views_id_list = [item[0] for item in cost_clicks_views_list if item[3] == min([i[3] for i in cost_clicks_views_list])]
            print(min_views_id_list)  # [12326]

Any commentary is very welcome. great thanks.

Answer 1

You can use list comprehension for checking these. When at least one cost among all items is not equal to zero, to get the item's id whose cost is lowest, try this :

sdl = [(12324, 10, 0.6, 9), (12325, 11, 0.5, 3), (12326, 10, 0.7, 2)]
a = [j[0] for j in sdl if j[1]==min([k[1] for k in sdl if all([True if i[1] != 0 else False for i in sdl])])]

OUTPUT :

a = [12324, 12326]

Here all three items have non-zero cost and lowest cost is 10 corresponding to which there are two ids 12324 and 12326 .