How can I remove an dictionary element in list?

Question

There are some dictionary in list:

a = [{1: 2}, {1: 3}, {2: 5}, {2: 3}]

All of the keys and values of this dictionary are int type, some dicts have the same key, I want to remain those dictionaries who has the same key and the value is bigger than others, how can I do it? example: I want to obtain this list:

[{1: 3}, {2: 5}]

is it guaranteed that each dictionary only has a single key/value pair? — BrownieInMotion, Nov 05 '21 at 08:02

Dani Mesejo · Answer 1 · 2021-11-06T08:57:46.493

2

Simple approach using collections.defaultdict:

from collections import defaultdict

a = [{1: 2}, {1: 3}, {2: 5}, {2: 3}]

temp = defaultdict(lambda: float("-inf"))
for item in a:
    [(i, v)] = item.items()
    temp[i] = max(v, temp[i])

res = [{i: v} for i, v in temp.items()]
print(res)

Output

[{1: 3}, {2: 5}]

In the code the expression float("-inf") acts an unbounded lower value for comparison, this means that:

float("-inf") < x

for any value of x, unless x is float("nan") or float("-inf") itself. Read more in here.

After each first call setting temp[i] for a given i, the value is going to be the first appearing v.

edited Nov 06 '21 at 08:57

answered Nov 05 '21 at 08:57

Dani Mesejo

61,499
6
49
76

I'm new to python, how to understand 'float("-inf))' ? – BraveRock Nov 06 '21 at 01:30
@BraveRock it creates a float that represents infinity, it acts a kind of marker that is less than any number, it will be overriten in the first iteration. – Dani Mesejo Nov 06 '21 at 08:47

BrownieInMotion · Accepted Answer · 2021-11-05T08:33:23.697

Assuming that each dictionary has only one key/value pair, here is a possible solution. Essentially, create a new dictionary to keep track of the maximum values for each key. Then, turn this into the final list.

def remove_duplicates(dicts):
    merged = {}
    for d in dicts:
        key, value = list(d.items())[0]
        if key not in merged:
            merged[key] = value
        else:
            merged[key] = max(
                merged[key], value
            )
    return [
        {key: value}
        for key, value in merged.items()
    ]

score 1 · Answer 3 · edited May 05 '22 at 07:07

1

a = [{1: 2}, {1: 3}, {2: 5}, {2: 3}]

output = {}

for d in a:
    k = list(d.keys())[0]
    v = list(d.values())[0]
    if k not in output.keys():
        output[k] = v
    else:
        if output[k] < v:
            output[k] = v

output = [{k:v} for k, v in output.items()]

print(output)

edited May 05 '22 at 07:07

Dharman

30,962
25
85
135

answered Nov 05 '21 at 08:23

Rish

804
8
15

Christian · Answer 4 · 2021-11-05T08:26:04.283

0

from itertools import groupby
a = [{1: 2}, {1: 3}, {2: 5}, {2: 3}]
[{k, max(y[1] for y in g)} # build up the result by recreating the list of maps by taking the keys in the group and map them to the biggest value in the group
 for k, g
 in groupby((sorted( # group everything together, so that those touples with the same key (=index 0) are in the same group
     ((k, v) for m in a for k, v in m.items())  # transform the list of maps to a generator of tuples
     , key=lambda x: x[0])), key=lambda x: x[0])]

So here I first transform the list of maps to a list of tuples (which makes much more sense, but that is just me), group them together by the key, and afterwards create new maps for each key, by taking the biggest value inside each group.

edited Nov 05 '21 at 08:26

answered Nov 05 '21 at 08:21

Christian

1,437
2
8
14

3

While correct, I would not want to encounter this "line" in the wild. – timgeb Nov 05 '21 at 08:25
1

Me neither, but everything is cooler in one line :D (kidding) – Christian Nov 05 '21 at 08:26
This is also inefficient, sorting to get the maximum is a bad idea – Dani Mesejo Nov 05 '21 at 08:32

score 0 · Answer 5 · answered Nov 05 '21 at 08:32

Same logic as BrownieInMotion's approach, but with cleaner code using list comprehension for the result.

a = [{1: 2}, {1: 3}, {2: 5}, {2: 3}]

temp = {}
for item in a:
    i, v = list(*item.items())

    if i not in temp:
        temp[i] = v
    else:
        temp[i] = max(v, temp[i])

print([{i: v} for i, v in temp.items()])

The unpacking usage list(*item.items()) restricts to dictionaries of single key. If you want to remove the restriction, you can go ahead and use list(item.items())[0].

How can I remove an dictionary element in list?

5 Answers5