Get first two items from the first sublist if first element of the sublist is unique in Python

Question

I have a list:

df = [['apple', 'red', '0.2'], ['apple', 'green', '8.9'], ['apple', 'brown', '2.9'], 
      ['guava', 'green', '1.9'], ['guava', 'yellow', '4.9'], ['guava', 'light green', '2.3']]

From here I want to only get the first 2 items from the first distinct sublist given the condition that the value of the first sublist is unique.

Expected output:

df = [['apple', 'red'], ['guava', 'green']]

Code till now:

dummy_list = []

for item in df:
    if item[0] not in dummy_list:        
        dummy_list.append(item[:2])

This is not working and appending all the elements. Any help on this please

Answer 1

Or smarter : use a dict and setdefault to add the mapping only for the first

result = {}
for value in df:
    result.setdefault(value[0], value[:2])
result = list(result.values())

print(result)

Or you could keep a count of the added keys to avoid repeating them (in a separate list)

keys = set()
result = []
for value in df:
    if value[0] not in keys:
        result.append(value[:2])
        keys.add(value[0])

print(result) # [['apple', 'red'], ['guava', 'green']]

Answer 2

You can use itertools.groupby and for the key use operator.itemgetter:

from itertools import groupby
from operator import itemgetter

df = [['apple', 'red', '0.2'], ['apple', 'green', '8.9'], ['apple', 'brown', '2.9'], 
      ['guava', 'green', '1.9'], ['guava', 'yellow', '4.9'], ['guava', 'light green', '2.3']]

df1 = [next(g)[:2] for k, g in groupby(df, key=itemgetter(0))]

FYI itemgetter(0) is the same as lambda x: x[0] so you could use that too.