How to write dictionary comprehension for the below problem-CodePudding

Can I do the followign problem using Dict comprehensions

Given this list: a_list= [ 22, 55, 77, 234, 765, 1234] Produce a code snippet that produces this output: a_dict = {2:[22, 55 77], 3:[234, 765], 4:[1234]} The keys in the dictionary are the length of the digits in a_list.

My solution:

    tmp_list = []
    key = len(str(i))
    if key in a_dict:
        tmp_list = a_dict[key]
        tmp_list.append(i)
        a_dict[key] = tmp_list
    else:
        tmp_list.append(i)
        a_dict[key] = tmp_list ```

CodePudding user response：

Using groupby, sort and then group by length:

a_list = [22, 55, 77, 234, 765, 1234]
print({k: list(map(int, g)) for k, g in groupby(sorted(map(str, a_list), key=lambda el: len(str(el))), key=len)})

Output:

{2: [22, 55, 77], 3: [234, 765], 4: [1234]}

CodePudding user response：

It ain't pretty, but it works.

{
    key: [i[key] for i in [{len(str(num)): num} for num in a_list] if key in i]
    for key in [len(str(num)) for num in a_list]
}

>>> {2: [22, 55, 77], 3: [234, 765], 4: [1234]}

Be warned, this approach is not readable or efficient in terms of looping.

CodePudding user response：

Using collections.defaultdict is going to be more efficient than a comprehension in this case. Also, math.log10 is more efficient than len(str(digit)):

res = defaultdict(list)
for digit in digits:
    res[int(log10(digit)) 1].append(digit)

{2: [22, 55, 77], 3: [234, 765], 4: [1234]}