Write a function 

`titleToNumber(title) or title_to_number(title) or titleToNb title ...`

(depending on the language)

that given a column title as it appears in an Excel sheet, returns its corresponding column number. All column titles will be uppercase.

Examples:
```
titleTonumber('A') === 1
titleTonumber('Z') === 26
titleTonumber('AA') === 27
```

Difficulty: introductory

Starter code:

def title_to_number(title):
	

Solutions:
def title_to_number(title):
    ret = 0
    for i in title:
        ret = ret*26 + ord(i) - 64
    return ret

def title_to_number(title):
    return sum((ord(c) - 64) * 26**i for i, c in enumerate(title[::-1]))

def title_to_number(title):
    return sum((ord(char) - 64) * 26 ** i for i, char in enumerate(title[::-1]))

def title_to_number(title):
    #your code here
    offset = ord('A') - 1
    radix = 26
    
    val = 0
    for i in title:
        val = radix * val + (ord(i) - offset)
        
    return val

from string import ascii_uppercase
def title_to_number(t):
  return 0 if len(t) == 0 else (ascii_uppercase.index(t[0]) + 1) * 26 ** (len(t) - 1) + title_to_number(t[1:])

from string import ascii_uppercase
from functools import reduce

val = {c:i+1 for i,c in enumerate(ascii_uppercase)}
add = lambda x,y: 26*x + y

def title_to_number(title):
    return reduce(add, map(val.get, title))

from string import ascii_uppercase as letters


def title_to_number(title):
    transfer = list(title)[::-1]
    result = 0
    for i, char in enumerate(transfer):
        result += (letters.find(char)+1) * 26**i
    return result

title_to_number = lambda a: 0 if a == '' else 1 + ord(a[-1]) - ord('A') + 26 * title_to_number(a[:-1])

from string import ascii_uppercase as a

def title_to_number(title):
    return sum(
        (a.index(x) + 1) * 26 ** i
        for i, x in enumerate(title[::-1])
    )