Python list: Group by most often seen

Question

I have a 2d list [index, value] with repeated indexes. I need to select unique indexes with most often occurring value or if values seen equal number of times – the last one.

[
  [0,-1],
  [1, 0],
  [1, 1],
  [2, 1],
  [2,-1],
  [2, 1],
]
 =>
[
  [0,-1],
  [1, 1], # last seen
  [2, 1], # most often seen
]

I can use numpy or any other popular lib instead if it makes it easier

Asked By: Boppity Bop

||

Source

Answer 1

You can do like this,

from itertools import groupby
from collections import Counter

result = []
for index, lst in groupby(l, key=lambda x:x[0]):
     lst = [i[1] for i in lst]
     if len(lst) == len(set(lst)) or len(set(Counter(lst).values())) == 1:
         item = lst[-1]
     else:
         item = max(set(lst), key=lst.count)
     result.append([index, item])

In [160]: result
Out[160]: [[0, -1], [1, 1], [2, 1]]

len(lst) == len(set(lst)) -> Idenity if the list doesn’t have any replication.
len(set(Counter(lst).values())) == 1 -> Handing the special condition mentioned by @sajad.

Answered By: Rahul K P

Answer 2

long_list=[[0,-1],[0,-1],[1, 0],[1, 1],[2, 1],[2,-1],[2, 1],]
short_list=[]
for element in long_list:
    if element not in short_list:
        short_list.append(element)
print(short_list)

Output: [[0, -1], [1, 0], [1, 1], [2, 1], [2, -1]]

Answered By: Saleque Ahmed

Answer 3

from collections import defaultdict

input = [
  [0,-1],
  [1, 0],
  [1, 1],
  [2, 1],
  [2,-1],
  [2, 1],
]

out = []
counter_map = defaultdict(lambda: defaultdict(int))

for index, value in input:
    counter_map[index][value] += 1
    
for index, value_count_maps in counter_map.items():
    best = None
    for value, count in value_count_maps.items():
        if best == None or value >= best[0]:
            best = [value, count]
    out.append(best)
        
print(out) # [[0, -1], [1, 1], [2, -1]]

Answered By: Arnoux

Answer 4

I would keep it simple instead, and note that the ordering in the answer matters; so it’s necessary to preserve it along with other bookkeeping structures.

from collections import defaultdict

l = [
  [2, 1],
  [2, 1],
  [2, 2],
  [2, 2],
  [2,-1],
  [2,-1]
]

result = []

groups = defaultdict(list)
for index, value in l:
    groups[index].append(value) # keep the list (ordered series) of values for each index
    
for index, group in groups.items():
    best_count = 0
    best_value = None
    counts = defaultdict(int)
    for value in group:
        counts[value] += 1 # count each value for the index
        # we look for the most frequest value, and in case of ties, 
        # we prefer the one which has the last occurrence in the 
        # series (list) of values
        if counts[value] >= best_count: 
            best_value = value
            best_count = counts[value]
    result.append([index, best_value])

print(result)

Answered By: Sajad

Python list: Group by most often seen

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