Remove number from list if the first digit and length are the same

Question

Let's say I have a list of [100, 210, 250, 300, 405, 430, 500, 1850, 1875, 2120, 2150] I want to remove any numbers that start with the same digit and have the same length. The result should be: [100, 210, 300, 405, 500, 1850, 2120]

What I have so far is:

for i in installed_tx_calc:
    if (len(str(i)) == 3) and (str(i)[:1] == str(i-1)[:1]):
        installed_tx_calc.remove(i)
    elif str(i)[:2] == str(i-1)[:2]:
        installed_tx_calc.remove(i)

I have a list of [862, 1930, 2496] and my code outputs [1930].

I couldn't find anything when searching, but I feel like I'm missing something obvious.

Thank you for your time.

Answer 1

You can create the new list with a list comprehension, using itertools.groupby:

from itertools import groupby

numbers =  [100, 210, 250, 300, 405, 430, 500, 1850, 1875, 2120, 2150]

out = [next(group) for key, group in groupby(numbers, key=lambda n: (str(n)[0], len(str(n))))]

print(out)
# [100, 210, 300, 405, 500, 1850, 2120]

We group using the tuple (first digit, length of number), and keep the first number of each group, which we get with next(group).

Answer 2

Create a new set that will keep unique entries. Then, you may filter according to that set:

unique = set()
mylist = [100, 210, 250, 300, 405, 430, 500, 1850, 1875, 2120, 2150]
newlist = []

for num in mylist:
    num_str = str(num)
    length = len(num_str)
    first_digit = num_str[0]
    if (length, first_digit) in unique:
        continue
    newlist.append(num)
    unique.add((length, first_digit))

>>> newlist
[100, 210, 300, 405, 500, 1850, 2120]