Remove duplicated considering one element from a list of lists python

Question

I basically have a list of list like this:

[['1348559717', '11'],
 ['1348559717', '15'],
 ['1348562275', '16'],
 ['1348562275', '20'],
 ['1348562284', '17'],
 ['1348562284', '18'],
 ['1348562290', '19'],
 ['1349346149', '15'],
 ['1349348467', '14'],
 ['1350001260', '17']]

I would like to remove the lists which contains duplicated values in index [0]. Also, i always need to have the list which its index [1] = '20', in this case ['1348562275', '20']. So my wanted list would be:

[['1348559717', '11'],
 ['1348562275', '20'],
 ['1348562284', '17'],
 ['1348562290', '19'],
 ['1349346149', '15'],
 ['1349348467', '14'],
 ['1350001260', '17']]

Does anyone have any idea how could i do it?

Sash Sinha · Accepted Answer · 2021-04-02 15:33:48Z

1

You could use a dictionary then convert it back to a list after processing:

from pprint import PrettyPrinter

lst = [['1348559717', '11'],
       ['1348559717', '15'],
       ['1348562275', '16'],
       ['1348562275', '20'],
       ['1348562284', '17'],
       ['1348562284', '18'],
       ['1348562290', '19'],
       ['1349346149', '15'],
       ['1349348467', '14'],
       ['1350001260', '17']]

id_to_num = {}
for id_, num in lst:
    if id_ not in id_to_num or num == '20':
        id_to_num[id_] = num

new_lst = [[id_, num] for id_, num in id_to_num.items()]

PrettyPrinter().pprint(new_lst)

Output:

[['1348559717', '11'],
 ['1348562275', '20'],
 ['1348562284', '17'],
 ['1348562290', '19'],
 ['1349346149', '15'],
 ['1349348467', '14'],
 ['1350001260', '17']]

edited Apr 2, 2021 at 15:33

answered Apr 2, 2021 at 15:25

Sash Sinha

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2 Comments

Emerson Oliveira Over a year ago

Nice! That's what i was looking for. Btw, a question a bit off topic, why use pprint instead of print?

Sash Sinha Over a year ago

So it actually looks your expected output/is not all printed on one line. Like If you just use print the output won't have new lines between the elements.

Rahul Sambyal · Accepted Answer · 2021-04-02 15:39:33Z

1

You can probably optimize it more but here is the idea :

items = [['1348559717', '11'],
 ['1348559717', '15'],
 ['1348562275', '16'],
 ['1348562275', '20'],
 ['1348562284', '17'],
 ['1348562284', '18'],
 ['1348562290', '19'],
 ['1349346149', '15'],
 ['1349348467', '14'],
 ['1350001260', '17']]

    items_dict = {}
    for i in items:
      if i[0] not in items_dict or i[1] == '20':
        items_dict[i[0]] = i[1]
    items_list = []
    for k,v in items_dict.items():
      items_list.append([k, v])
    print(items_list)

Output :

[['1348559717', '11'],
 ['1348562275', '20'],
 ['1348562284', '17'],
 ['1348562290', '19'],
 ['1349346149', '15'],
 ['1349348467', '14'],
 ['1350001260', '17']]

answered Apr 2, 2021 at 15:39

Rahul Sambyal

513 bronze badges

Comments

bluefcat · Accepted Answer · 2021-04-02 15:43:55Z

1

How about use dict type like this?

some_list = [...] # your list what basically have 
some_dict = {item[0]: item[1] for item in some_list}

#always need to have the list which its index [1] = '20'
for k, v in [i for i in a if i[1] == '20']:
    some_dict[k] = v

some_list = [[k, v] for k, v in some_dict.items()]

Output:

[['1348559717', '15'], 
 ['1348562275', '20'], 
 ['1348562284', '18'], 
 ['1348562290', '19'], 
 ['1349346149', '15'], 
 ['1349348467', '14'], 
 ['1350001260', '17']]

I hope you can tell the rules about delete duplicated value in more detail.

edited Apr 2, 2021 at 15:43

answered Apr 2, 2021 at 15:35

bluefcat

213 bronze badges

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