roman numeral generator using for loop

Question

I am trying to generate dictionary values of roman numerals, but I am really stuck in designing my for loop.
In my problem, you have an input called values. For each distinct letter in the input, it corresponds to a roman numeral. In the example below, A=1,B=5,C=10,D=50,E=100,F=500,G=1000 etc.
Thus,the letters of the input are just like the roman numerals, I,V,X,L,C,D.
I want to create a function where you input a string of any length and each distinct letter corresponds to a roman numeral which is put into a dictionary

values='ABCDEFGHI'
dict={}
for i,letter in enumerate(values):
    if i%2==1:
        dict[letter]=5*10**(i)
    else:
        dict[letter]=1*10**i
print(dict)

I have come up with a pattern, but don’t know how to implement that into a for loop:
This link shows a table with the pattern I found

I am having problems because at each certain iteration you need to subtract more and more from the power and I dunno what that looks like.

Asked By: Robert Selangor

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Source

Answer 1

Thanks to the commenter, I think there’s no need to use subtraction in the power, just note that the power increases more slowly, at half the rate of i, so this works:

romans={}
for i,letter in enumerate(values):
  power=int(0.5*(i))
  print(f"{i=}")
  print(f"{power=}")
  if i%2==1:
    romans[letter]=5*10**(power)
  else:
    romans[letter]=1*10**(power)

>>> romans
{'A': 1, 'B': 5, 'C': 10, 'D': 50, 'E': 100, 'F': 500, 'G': 1000, 'H': 5000, 'I': 10000}

Answered By: mjsqu

Answer 2

I actually came up with this solution which works as well haha! Thank you mjsqu for your alternative solution too
here is what i did

def dictcreator(value):
    dict={}
    n=0
    for i,letter in enumerate(reversed(value)):
        if i%2==1:
            dict[letter]=5*10**(i-n)
        else:
            dict[letter]=1*10**(i-n)
            n=n+1
    return dict

Answered By: Robert Selangor

roman numeral generator using for loop

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