python 2.7 - Can't get counter to compare first letter of word against A-Z alphabet -


what i'm trying compare first letter of each word against lower case a-z alphabet , print (similar word_frequency) how many times word starts each letter of alphabet (such this)

a = 0 b = 2, c = 0, d = 2 ------------ y = 1, z = 0

but unable find way through counter of yet or found worked me (beginner). idea had along lines of 

    w in word_count:         l_freq = []         l_freq.append(w[0])

and comparing counter against it? not sure how compare against whole alphabet rather letters in string?

also, there way print frequency cleaner? without counter , brackets showing up?

from collections import counter  def function():     string = "this string written in python."     word_count = string.split()      char_count = 0      char in string:         if char != " " , char != ".":             char_count += 1      word_freq = counter(word_count)      print "word count:  " + str(len(word_count))     print "average length of word: " + str(char_count / len(word_count))     print ""     print "word frequency: "     print word_freq

let's have test data this:

in [1]: test_data = '''    ...: hello    ...: hi    ...: goodbye    ...: bye    ...: bye!    ...: lol    ...: lmao    ...: rofl    ...: '''

to count first letter of each word, you'd this:

counter = counter(w[0] w in test_data.split())

which you're doing. expected output think want, need associate each lowercase letter count, (import string if haven't already):

for letter in string.ascii_lowercase:     print letter, '=', counter[letter]

which test data showed gives you:

a = 0 b = 1 c = 0 d = 0 e = 0 f = 0 g = 1 h = 2 = 0 j = 0 k = 0 l = 2 m = 0 n = 0 o = 0 p = 0 q = 0 r = 1 s = 0 t = 0 u = 0 v = 0 w = 0 x = 0 y = 0 z = 0

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