0004_bad.py

#!/usr/bin/env
'''
I HATE REGULAR EXPRESSION :(

So.. the following code is a very  preliminary word processing, 
it won't split abbrevations, like "I'd", nor it won't split words
concatenated with underscore, like "bad_game"
'''
import os
import re
def word_count(file_path):
	word_dict = {}
	with open(file_path) as txt:
		for line in txt:
			words = re.findall(r"\w+", line.strip()) 
			for word in words:
				word = word.lower()
				# dict.get() saves trouble of if/else block for checking
				# key existence. 
				# some_dict.get(some_key, default_value) will return 
				# default value is key is not in dict, while returns 
				# actual value if key is present!
				word_dict[word] = word_dict.get(word,0) + 1
	return word_dict

result = word_count(os.path.join(os.path.dirname(os.path.abspath(__file__)), "sampletext.txt"))

###################################################################################

'''
share a gain :)
"How to "sort" a dict by its key/value"

Obviously, a dict can not be sorted, since it is orderless in nature.
However, what we can do is to sort a representation(a list of tuples or a list of keys) of dict. 
Here is How:
http://stackoverflow.com/questions/613183/sort-a-python-dictionary-by-value
'''

import operator
d = {"what": 3, "I": 19, "the":30}
# to get a list of tuples with sorted items by the value
sorted_d_by_value = sorted(d.items(), key=operator.itemgetter(1))
# to get a list of tuples with sorted key
sorted_d_by_key = sorted(d.items(), key=operator.itemgetter(0))
print sorted_d_by_key