Remove the smallest N values from python dictionary

Question

Suppose that I have the following dictionary:

d = {'q' : 2,
     'b': 22,
     'c': 19,
     'e': 3,
     'd': 9}

What I want to do is to iteratively remove the smallest values (with their keys) from the dictionary. For instance, if I want to remove 2 smallest values, I expect to obtain the following as remaining dictionary:

d = {'b': 22,
     'c': 19,
     'd': 9}

Note that the ordering matters for me in this case. That is, the items in the original dictionary should NOT be reordered after processing.

I tried the below code, but it does not work for an example where dict's length is 19, and I need to remove 18 least values (w/ their associated keys).

def remaining(dict, smallest_n):
    remaining = dict
    s = {}
    for i in range(smallest_n-1):
        m = min(remaining.values())
        for k, v in remaining.items():
            if v != m:
                s[k] = v
        remaining = s.copy()
        s.clear()

    return remaining

For the case I mentioned above, when the smallest_n (which is the number of minimum elements that should be removed) equals to 18, and the input dictionary's size is 19, what I obtain as remaining should have size 1 (19-18). However, it doesn't contain anything after the code lines are being executed.

if ordering matters, you shouldn't be using a regular dictionary. you should use an [`OrderedDict`](https://docs.python.org/3.5/library/collections.html#collections.OrderedDict) — wpercy, Apr 18 '19 at 20:21
also worth mentioning that you probably shouldn't be overwriting `remaining` with the value of `d`... — wpercy, Apr 18 '19 at 20:22
@wpercy thanks! by ordering, I mean that I'm not allowed to modify the initial ordering of the main dictionary. Just updated! — inverted_index, Apr 18 '19 at 20:26
normal dictionaries like you used do not have an ordering. They often appear to be ordered, but that is simply a coincidence and not guaranteed. If you need it guaranteed, you use an OrderedDict — Tom Lubenow, Apr 18 '19 at 20:28
Yep! "removing the smallest N values without repositioning the items in the dictionary"? Actually, I'm skeptical what the wrong is with my script? @BradSolomon — inverted_index, Apr 18 '19 at 20:28
@Tom [Dicts in Python3.7+ are ordered](https://stackoverflow.com/a/39980744/4518341), though not the way OP needs. — wjandrea, Apr 18 '19 at 20:33

Tom Lubenow · Accepted Answer · 2019-04-18T20:32:34.053

3

min_key = min(d.keys(), key=lambda k: d[k])
del d[min_key]

This will delete the entry with the minimum value. Do this in a loop if you want to remove multiple.

as a function:

def remove_n_minimums(d, n):
    for _ in range(n):
        min_key = min(d.keys(), key=lambda k: d[k])
        del d[min_key]

edited Apr 18 '19 at 20:32

answered Apr 18 '19 at 20:31

Tom Lubenow

1,109
7
15

score 1 · Answer 2 · answered Apr 18 '19 at 20:38

You could use

def remove(some_dict, smallest_n = 0):
    # nothing to do
    if not smallest_n:
        return some_dict

    keys_tuple = sorted(
        [(v, k) for k,v in some_dict.items()],
        key = lambda x: x[0]
    )
    for x in range(smallest_n):
        key = keys_tuple[x][1]
        del some_dict[key]

    return some_dict

remove(d, 2)
# {'b': 22, 'c': 19, 'd': 9}

We build a list of tuples with the keys and values. The list will be sorted and the keys in question deleted afterwards. The original dict will be changed with this approach.

Remove the smallest N values from python dictionary

2 Answers2