Calculating the Letter Frequency in Python -

i need define function slice string according character, sum indices, divide number of times character occurs in string , divide length of text.

here's have far:

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def ave_index(char):   passage = "string"   if char in passage:     word = passage.split(char)     words = len(word)     number = passage.count(char)     answer = word / number / len(passage)     return(answer)    elif char not in passage:     return false 

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so far, answers i've gotten when running have been quite off mark

edit: passage given use string - 'call me ishmael. years ago - never mind how long precisely - having little or no money in purse, , nothing particular interest me on shore, thought sail little , see watery part of world. way have of driving off spleen , regulating circulation. whenever find myself growing grim mouth; whenever damp, drizzly november in soul; whenever find myself involuntarily pausing before coffin warehouses, , bringing rear of every funeral meet; , whenever hypos such upper hand of me, requires strong moral principle prevent me deliberately stepping street, , methodically knocking people's hats off - then, account high time sea can. substitute pistol , ball. philosophical flourish cato throws himself upon sword; quietly take ship. there nothing surprising in this. if knew it, men in degree, time or other, cherish same feelings towards ocean me.'

when char = 's' answer should 0.5809489252885479

the problem how counting letters. take string hello world , trying count how many l there are. know there 3 l, if split:

>>> s.split('l') ['he', '', 'o wor', 'd'] 

this result in count of 4. further, have position of each instance of character in string.

the enumerate built-in helps out here:

>>> s = 'hello world' >>> c = 'l'  # letter looking >>> results = [k k,v in enumerate(s) if v == c] >>> results [2, 3, 9] 

now have total number of occurrences len(results), , positions in string letter occurs.

the final "trick" problem make sure divide float, in order proper result.

working against sample text (stored in s):

>>> c = 's' >>> results = [k k,v in enumerate(s) if v == c] >>> results_sum = sum(results) >>> (results_sum / len(results)) / float(len(s)) 0.5804132973944295