How to find indexes of first unique value from a list of dictionary using a key

Question

Assuming a list of dictionaries

l=[
    {"id":1, "score":80, "remarks":"B" },
    {"id":2, "score":80, "remarks":"A" },
    {"id":1, "score":80, "remarks":"C" },
    {"id":3, "score":80, "remarks":"B" },
    {"id":1, "score":80, "remarks":"F" },
]

I would like to find the indexes of the first unique value given a key. So given the list above i am expecting a result of

using_id = [0,1,3]
using_score = [0]
using_remarks = [0,1,2,4]

What makes it hard form me is the list has type of dictionary, if it were numbers i could just use this line

indexes = [l.index(x) for x in sorted(set(l))]

using set() on a list of dictionary throws an error TypeError: unhashable type: 'dict'. The constraints are: Must only use modules that came default with python3.10,the code should be scalable friendly as the length of the list will reach into the hundreds, as few lines as possible is a bonus too :)

Of course there is the brute force method, but can this be made to be more efficient, or use lesser lines of code ?

unique_items = []
unique_index = []


for index, item in enumerate(l, start=0):
    if item["remarks"] not in unique_items:
        unique_items.append(item["remarks"])
        unique_index.append(index)

print(unique_items)
print(unique_index)

Why would using_remarks = [0,1,3,4]? Surely it should be [0,1,2,4] — DarkKnight, Sep 16 '22 at 08:07
Are the values guaranteed to be [hashable](/q/14535730/90527)? — outis, Sep 16 '22 at 08:12
@outis yes i think so, the value type will remain the same on all the items in the list. so "remarks" will always be a string. The only types i am using in my dictionary is string, number, and boolean — Jake quin, Sep 16 '22 at 08:16
Please [edit] clarifications into to the question (as explained in the [site guidelines](/help/how-to-ask), rather than leaving [comments](/help/privileges/comment). For one thing, a question should be understandable without reading comments. For another, SO is a Q&A site, not a forum, and comments aren't intended (nor are they well suited) for discussions. — outis, Sep 30 '22 at 18:51

score 2 · Answer 1 · answered Sep 16 '22 at 08:19

You could refigure your data into a dict of lists, then you can use code similar to what you would use for a list of values:

dd = { k : [d[k] for d in l] for k in l[0] }
indexes = { k : sorted(dd[k].index(x) for x in set(dd[k])) for k in dd }

Output:

{'id': [0, 1, 3], 'score': [0], 'remarks': [0, 1, 2, 4]}

blhsing · Answer 2 · 2022-09-16T08:32:04.927

Since dict keys are inherently unique, you can use a dict to keep track of the first index of each unique value by storing the values as keys of a dict and setting the index as the default value of a key with dict.setdefault:

for key in 'id', 'score', 'remarks':
    unique = {}
    for i, d in enumerate(l):
        unique.setdefault(d[key], i)
    print(key, list(unique.values()))

This outputs:

id [0, 1, 3]
score [0]
remarks [0, 1, 2, 4]

Demo: https://replit.com/@blhsing/PalatableAnxiousMetric

score 1 · Answer 3 · answered Sep 16 '22 at 09:11

With functools.reduce:

l=[
    {"id":1, "score":80, "remarks":"B" },
    {"id":2, "score":80, "remarks":"A" },
    {"id":1, "score":80, "remarks":"C" },
    {"id":3, "score":80, "remarks":"B" },
    {"id":1, "score":80, "remarks":"F" },
]

from functools import reduce

result = {}
reduce(lambda x, y: result.update({y[1]['remarks']:y[0]}) \
                     if y[1]['remarks'] not in result else None, \
        enumerate(l), result)

result
# {'B': 0, 'A': 1, 'C': 2, 'F': 4}

unique_items = list(result.keys())
unique_items
# ['B', 'A', 'C', 'F']
unique_index = list(result.values())
unique_index
# [0, 1, 2, 4]

Explanation: the lambda function adds to the dictionary result at each step a list containing index (in l) and id but only at the first occurrence of a given value for remarks.

The dictionary structure for the result makes sense since you're extracting unique values and they can therefore be seen as keys.

How to find indexes of first unique value from a list of dictionary using a key

3 Answers3