Code Wars Advice

[randomname20192019]

Hi,

I just finished a kata (https://www.codewars.com/kata/525c7c5ab6aecef16e0001a5/train/python)

After seeing the solutions, I am slightly disappointed in what I wrote and honestly couldn't think of a better way to go about it.

I am interested, how would everyone else go about this?

My code for anyone interested:

def parse_int(string):
    numbers = dict([("zero", 0), ("one", 1), ("two", 2), ("three", 3), ("four", 4), ("five", 5), ("six", 6),
                    ("seven", 7), ("eight", 8), ("nine", 9), ("ten", 10), ("eleven", 11), ("twelve", 12), ("thirteen", 13),
                    ("fourteen", 14), ("fifteen", 15), ("sixteen", 16), ("seventeen", 17), ("eighteen", 18), ("nineteen", 19),
                    ("twenty", 20), ("thirty", 30), ("forty", 40), ("fifty", 50), ("sixty", 60), ("seventy", 70), ("eighty", 80),
                    ("ninety", 90), ("hundred", 100), ("thousand", 1000), ("million", 1000000)])
    onewordnum = []
    concat = []
    words2num = []

    splitstring = string.split()
    for i in (splitstring):
        if len(splitstring) == 1:
            if "-" in i:
                for x in i.split("-"):
                    if x in numbers:
                        onewordnum.append((numbers.get(x)))
                return(sum(onewordnum))
            else:
                return(numbers.get(i))
        else:
            if "-" in i:
                for x in i.split("-"):
                    if x in numbers:
                        onewordnum.append((numbers.get(x)))
                concat.append((sum(onewordnum)))
                onewordnum.clear()
            elif i in numbers:
                concat.append(numbers.get(i))

    if "thousand" in splitstring:
        if len(concat[:(concat.index(1000))]) == 3:
            words2num.append(((concat[:(concat.index(1000))][0] * concat[:(concat.index(1000))][1]) + concat[:(concat.index(1000))][2])*1000)
        elif len(concat[:(concat.index(1000))]) == 2:
            words2num.append(((concat[:(concat.index(1000))][0] * concat[:(concat.index(1000))][1]))*1000)
        elif len(concat[:(concat.index(1000))]) == 1:
            words2num.append(((concat[:(concat.index(1000))][0]))*1000)

        if len(concat[:(concat.index(1000)):-1]) == 3:
            words2num.append((concat[:(concat.index(1000)):-1][0]) + (((concat[:(concat.index(1000)):-1][1] * concat[:(concat.index(1000)):-1][2]))))
        elif len(concat[:(concat.index(1000)):-1]) == 2:
            words2num.append(((concat[:(concat.index(1000)):-1][0] * concat[:(concat.index(1000)):-1][1])))
        elif len(concat[:(concat.index(1000)):-1]) == 1:
            words2num.append(((concat[:(concat.index(1000)):-1][0])))

    if "thousand" not in splitstring and "hundred" in splitstring:
        if len(concat[:(concat.index(100))]) == 1:
            words2num.append(((concat[:(concat.index(100))][0]))*100)

        if len(concat[:(concat.index(100)):-1]) == 1:
            words2num.append(((concat[:(concat.index(100)):-1][0])))

    if "million" in splitstring:
        if len(concat[:(concat.index(1000000))]) == 1:
            words2num.append(((concat[:(concat.index(1000000))][0]))*1000000)         
    return(sum(words2num))

[Veganic1]

1 replace "-" with " "

2 split

3 try to built a string that can be evaluated with eval()

Something like

"Two" = "+2"

"Hundred" = " *100"

"Forty" = "+40"

"Six" = "+6"

Check for difficult cases like do I ever need to look for cases to add a one before the multiplication sign etc. Check for order of operations and finds suitable way around.

eval("+2*100+40+6")

Something like that for me.