Need to filter list of dictionaries

Given a prefix, the filtering code comes down to an if-statement and a str.startwith() test:

prefix = '220'
for d in list_of_dict:
    if d['label'].startswith(prefix):
        print(d)

For purposes of a dropdown menu, it may be preferable to group the data in advance using defaultdict() and extracting the prefix with slicing:

from collections import defaultdict

grouped = defaultdict(list)
for d in list_of_dict:
    prefix = d['label'][:3]
    grouped[prefix].append( d['value'])

Now you can lookup results directly:

>>> grouped['220']
['220_2_INTL_PRSTR_ET_619076', '220_4_KAL_T2E_DOLE_344657']

This strategy is also faster than filtering the entire list for every lookup.

It's fairly straightforward to turn your list into a dictionary that has exactly what you need:

list_of_dict = [
    {'label': '220_2_INTL_PRSTR_ET_619076', 'value': '220_2_INTL_PRSTR_ET_619076'},
    {'label': '220_4_KAL_T2E_DOLE_344657', 'value': '220_4_KAL_T2E_DOLE_344657'},
    {'label': '221_1_PB_520_REF_174049', 'value': '221_1_PB_520_REF_174049'},
    {'label': '222_5_KAL_T2E_YT_344991', 'value': '222_5_KAL_T2E_YT_344991'},
    {'label': '223_2_PB_520_REF_174050', 'value': '223_2_PB_520_REF_174050'},
    {'label': '224_3_PB_520_REF_174051', 'value': '224_3_PB_520_REF_174051'}
]

# it's a "one-liner", a dict comprehension split over a few lines for readability:
result = {
    p: [d for d in list_of_dict if d['label'].startswith(p)]
    for p in set(d['label'][:3] for d in list_of_dict)
}

print(result['220'])  # the contents for this prefix
print(result.keys())  # the keys for your first dropdown

Output:

[{'label': '220_2_INTL_PRSTR_ET_619076', 'value': '220_2_INTL_PRSTR_ET_619076'}, {'label': '220_4_KAL_T2E_DOLE_344657', 'value': '220_4_KAL_T2E_DOLE_344657'}]
['222', '220', '224', '223', '221']

Note that the keys are out of order, but sorting is straightforward.

Instead of d['label'][:3] you could consider d['label'].split('_')[0], if the prefixes aren't all 3 characters long, but instead are "everything before the first underscore".

Edit: in the comments, you asked for some additional explanation of the core bit of code:

{
    p: [d for d in list_of_dict if d['label'].startswith(p)]
    for p in set(d['label'][:3] for d in list_of_dict)
}

• Anything of the form {..: .. for .. in ..} is a dictionary comprehension, constructing a dictionary using a very efficient loop.
• Here it's {p: ... for p in set(d['label'][:3] for d in list_of_dict)}. So, p loops over the elements of set(d['label'][:3] for d in list_of_dict) and for every p, a key is added to the dictionary.
• That d['label'][:3] for d in list_of_dict is a generator expression that generates the first three characters ([:3]) of every 'label' value for every dictionary d in your list_of_dict. I.e. ['220', '220', '221', '222', etc.]. And the set() around it reduces it to have only unique elements.
• The value part of the dictionary comprehension is a list comprehension, so a list is construction as a value for each key p. A list comprehension looks like [.. for .. in ..] (with an optional if ..-part to filter the contents)
• The comprehension [d for d in list_of_dict if d['label'].startswith(p)] takes each dictionary d from your list_of_dict, but only keeps it in the resulting list if d['label'].startswith(p) is True (i.e. only if d['label'] starts with p, which is the current 3-letter string being used as a key.

So, it gathers all of the 3-letter prefixes in a set, and then generates a dictionary with those unique prefixes as keys, and a list of all the dictionaries that have 'label' values starting with the matching 3-letter prefix, as their value.

As first step, you can use set() to obtain the first tree characters from labels without duplicates.

Then you can create a function to filter the list using str.startswith:

def get_options(l):
    return {d["label"][:3] for d in l}


def filter_list(l, s):
    return [d for d in l if d["label"].startswith(s)]


print(get_options(list_of_dict))
print(filter_list(list_of_dict, "220"))

Prints:

{"222", "221", "224", "220", "223"}
[
    {
        "label": "220_2_INTL_PRSTR_ET_619076",
        "value": "220_2_INTL_PRSTR_ET_619076",
    },
    {
        "label": "220_4_KAL_T2E_DOLE_344657",
        "value": "220_4_KAL_T2E_DOLE_344657",
    },
]