12

README.md

@@ -57,6 +57,7 @@
 * __11-6 homophone (sound check)__
 * 12-1 letter frequecy
 * __12-2 anagram search and sort__
+* 12-3 metathesis pairs
 
 
 
@@ -87,3 +88,9 @@ I was reviewing all the materials from school recently but figured out it could
 
 ### _31/12/2016_
 The last day of this year, but I am still stuck with a lot of exercises in the book. I don't think I am able to finish in this year. So good luck to next year. 皆さんあけおめ!
+
+### _11/01/2017_
+Arrr, back to work.
+
+### _11/01/2017_
+Low effiency :/ I did 2 different versions for exercise 12-1, 12-2.

TP reading notes.md

@@ -83,10 +83,15 @@
 # Chapter 11 Dictionaries
 
 * key and values -> key-value pair (aka item)
+* associative memories
+* string and numbers can always be keys!!!
+* keys are immutable
+* Tuples can be used as keys if they contain only strings, numbers, or tuples; if a tuple contains any mutable object either directly or indirectly, it cannot be used as a key.
+* list CANNOT be keys!!!!
 
+# Chapter 12 Tuples
 
-
-
+* The key 
 
 
 

ch12_ex.py

@@ -0,0 +1,9 @@
+def sumall(*args):
+# use of gather opeartion
+# on page 142
+	sum_result = 0
+	for i in range(len(args)):
+		sum_result = sum_result + args[i]
+	return sum_result
+
+print sumall(1,2,3,4)

ex12_1_v2.py

@@ -0,0 +1,45 @@
+#version 2 for exercise 1 in chapter 12
+
+def map_dict(s):
+	'''
+	Returns a dictionary mapping each letter with corresponding frequency
+	
+	s: string
+	
+	'''
+	d = {}
+	for i in s:
+		d[i] = d.get(i, 0) + 1
+	return d
+
+def most_frequent(s):
+	'''
+	Sorts the letters in reverse order of frequence
+	
+	s: string
+	
+	Retruns a list of letters
+	'''
+	d = map_dict(s)
+
+	l = []
+
+	for x, f in d.items():
+		l.append((f,x))
+
+	l.sort(reverse = True)
+
+	result = []
+
+	for f, x in l:
+		result.append(x)
+
+	return result
+
+
+
+if __name__ == '__main__':
+	# emma.txt is provided from author's code
+	with open('emma.txt', 'r') as fin:
+		string = fin.read()
+		print most_frequent(string)

ex12_2.py

@@ -1,13 +1,13 @@
-# http://stackoverflow.com/questions/8286554/using-python-find-anagrams-for-a-list-of-words
-
-#hughdbrown did an amazing job and I tried to understand his code and found this one is the better so far.
+'''http://stackoverflow.com/questions/8286554/using-python-find-anagrams-for-a-list-of-words
 
+# hughdbrown did an amazing job and I tried to understand his code and found this one is the better so far.
+'''
 from collections import defaultdict
 
 def load_words(filename = "words.txt"):
 	with open(filename) as f:
 		for word in f:
-			yield word.rstrip()
+			yield word.rstrip() # return a generator
 
 def all_anagram(l):
 	'''Reads a list and return a set of anagram words
@@ -16,12 +16,12 @@ def all_anagram(l):
 	
 	Returns: set
 	'''
-	d = defaultdict(list)
+	d = defaultdict(list) # avoid KeyError in dict()
 	for word in l:
-		key = "".join(sorted(word))
-		d[key].append(word)
+		signature = "".join(sorted(word)) #sorted: leave the original word untouched
+		d[signature].append(word)
 	return d
-	
+
 def print_anagram(l):
 	d = all_anagram(l)
 	for key, anagram in d.iteritems():
@@ -39,10 +39,10 @@ def print_anagram_in_order(l):
 
 	for x in new_list:
 		print x
-
+
+
+
 if __name__ == '__main__':
 	l = load_words()
-	# print_anagram(l)
+	#print_anagram(l)
 	print_anagram_in_order(l)
-
-

ex12_2_v2.py

@@ -0,0 +1,44 @@
+from collections import defaultdict
+
+def load_words(filename = "words.txt"):
+	with open(filename) as f:
+		for word in f:
+			yield word.rstrip() # return a generator
+
+def all_anagram(s):
+	'''
+	s: string
+	'''
+	d = defaultdict(list)
+	for i in s:
+		signature = ''.join(sorted(i)) # siganature is a after-sort string of chars in the word
+		d[signature].append(i)
+
+	return d
+
+def print_anagram_in_order(s):
+	d = all_anagram(s)
+	l = []
+	for signature, words in d.items():
+		if len(words) > 1:
+			l.append((len(words),words))
+	l.sort(reverse=True)
+	for x in l:
+		print x
+
+def select_bingo(s, n=8):
+	d = all_anagram(s)
+	l = {}
+	for signature, words in d.iteritems():
+		if len(words) > 1:
+			if len(signature) == n:
+				l[signature] = words 
+	res = []
+	for k, v in l.iteritems():
+		res.append((len(v),v))
+
+	for x in sorted(res, reverse=False):
+		print x
+
+#print_anagram_in_order(load_words())
+select_bingo(load_words())

ex12_3.py

@@ -0,0 +1,48 @@
+# mathesis pair 
+# my idea: to start the pair of word need to be anagram (consist of all the same letter)
+# secondly only 1 char's position can be changed. 
+
+
+'''
+According to author's answer it is easier to find the distance between a pair of word!
+'''
+from collections import defaultdict
+
+def load_words(filename = "words.txt"):
+	with open(filename) as fin:
+		for line in fin:
+			yield line.rstrip()
+
+def all_anagram(s):
+	'''
+	s: string
+	'''
+	d = defaultdict(list)
+	for i in s:
+		signature = ''.join(sorted(i)) # siganature is a after-sort string of chars in the word
+		d[signature].append(i)
+
+	for k, v in d.items():
+		if len(v) == 1:
+			del d[k]
+
+	return d
+
+def pair_distance(a, b):
+	# credit to http://www.greenteapress.com/thinkpython/code/metathesis.py
+	assert len(a) == len(b) # if not equal raise AssertionError
+	ctr = 0
+	for c1, c2 in zip(a,b): # zip(a,b) returns a tuple for each corresponding char pair eg. "apple" & "paple" returns [('a', 'p'), ('p', 'a'), ('p', 'p'), ('l', 'l'), ('e', 'e')]
+		if c1 != c2:
+			ctr += 1
+	return ctr
+
+def find_metathesis(s):
+	d = all_anagram(s)
+	for v in d.itervalues():
+		for a in v:
+			for b in v:
+				if a < b and pair_distance(a,b) == 2:
+					print a, b
+
+find_metathesis(load_words())

ex12_4.py

@@ -0,0 +1,5 @@
+def child_of_word(word):
+	'''
+	Returns a list of words by removing one char
+	'''
+