Python 3.3.2 - Creating a List of the Length of Words -

i have string of words punctuation, let's example...

string = 'did quick brown fox *really* jump on fence?' 

i have filtered out punctuation, now:

'did quick brown fox jump on fence' 

and have split list.

list = string.split() 

now, list, need count length of each word list, length of list being longest word. setting out of list follows:

lengthlist = [1_letter_words, 2_letter_words, 3_letter_words, ...] 

so, string, be:

lengthlist = [0, 0, 4, 2, 3, 1] 

unfortunately, having trouble doing this. can please provide assistance?

thank you.

i didn't want harangue (at all, not) without giving proper answer, skip ahead if don't care coding practices.

don't use variable names list , string because - in case of list - that's name of type you're making. in fact, that's how make empty instance of type you're making:

something=list()       # empty list! 

this make confusing reference list[2] or along lines. didn't hit errors, sake of readability, try come meaningful variable names.

okay, i'm done rant, code you're looking is

st='did quick brown fox jump on fence'.split() c=[len(i) in st] # gives [3, 3, 5, 5, 3, 6, 4, 4, 3, 5] counts=[0]*max(c)      # gives [0, 0, 0, 0, 0, 0] in range(len(c)):   counts[c[i]-1]+=1    # adds 1 each index of c[i] (we subtract 1 because of 0-based indices) print(counts)          # gives answer: [0, 0, 4, 2, 3, 1] 

i made of these steps way more advanced challenge you're presenting kind of discourage using in assignment, if happens goal. of tools used in solution @ least a little further ahead of you're working with, if you're learning python reward of understanding code hope illuminating , maybe thinking of radically cool stuff can concisely python. said, let's walk through it:

i'm going assume st assignment clear enough don't need discuss it, note split right there when assign it. i'm being lazy , in 2 steps, isn't meat of problem let's move on.

c=[len(i) in st] 

just means "for each element, we'll call i, in st, give me len(i) in list, , make list c". might seem daunting, list comprehensions not bad, , can see save quite bit of time in coding. pretty modest implementation of it, really.

counts=[0]*max(c) 

says make list 0s in each space, , make repeat many times max of c. take longest word, in case 6-letter word 'really', , make list 6 elements long. ensure have list spaces every length word encounter.

for in range(len(c)):   counts[c[i]-1]+=1 

oh boy, we're cooking. see we're iterating through list c, each item through lengths of corresponding words:

• the first element 3, corresponding did.
• the second element 3, corresponding the.
• ...
• the last element 5, corresponding fence.

so that's c[i] about, counts[c[i]-1]? counts going add 1 every length find, it'll add 1 bin when has word 3 characters long. c[i] give 3 on first element, since lists 0-indexed (lists start @ 0 , goes there), need compensate - hence -1. see counts[c[i]-1] , makes little more sense, right?

counts[c[i]-1] # means counts[3-1] means go find bin corresponding counts[2]  # ---v   1 [0,0,0,0,0] 

and +=1 means "add 1 whatever there already".

python happily iterate through , give answer.