pull request

exercises-hello/hello.py

@@ -8,4 +8,4 @@
 # Run ./test.sh to make sure your program matches our expected output.
 #
 # TODO: write your code below
-
+print "hello world"

exercises-hello/script.py

@@ -0,0 +1,2 @@
+#!/usr/bin/env python
+print "this is a python script!"

exercises-more/exercises.py

@@ -2,73 +2,107 @@
 # Return the number of words in the string s. Words are separated by spaces.
 # e.g. num_words("abc def") == 2
 def num_words(s):
-    return 0
+    return len(s.split())
 
 # PROB 2
 # Return the sum of all the numbers in lst. If lst is empty, return 0.
 def sum_list(lst):
-    return 0
+    sum = 0
+    for elt in lst:
+	sum += elt
+    return sum
 
 # PROB 3
 # Return True if x is in lst, otherwise return False.
 def appears_in_list(x, lst):
-    return False
+    return x in lst
 
 # PROB 4
 # Return the number of unique strings in lst.
 # e.g. num_unique(["a", "b", "a", "c", "a"]) == 3
 def num_unique(lst):
-    return 0
+    seen = set([])
+    for elt in lst:
+	seen.add(elt)
+    return len(seen)
 
 # PROB 5
 # Return a new list, where the contents of the new list are lst in reverse order.
 # e.g. reverse_list([3, 2, 1]) == [1, 2, 3]
 def reverse_list(lst):
-    return []
+    return lst[::-1]
 
 # PROB 6
 # Return a new list containing the elements of lst in sorted decreasing order.
 # e.g. sort_reverse([5, 7, 6, 8]) == [8, 7, 6, 5]
 def sort_reverse(lst):
-    return []
+    return sorted(lst, key=lambda num : -1 * num)
 
 # PROB 7
 # Return a new string containing the same contents of s, but with all the
 # vowels (upper and lower case) removed. Vowels do not include 'y'
 # e.g. remove_vowels("abcdeABCDE") == "bcdBCD"
 def remove_vowels(s):
-    return s
+    outstr = ''
+    for character in s:
+	if not(character.lower() in ['a','e','i','o','u']):
+		outstr += character
+    return outstr
 
 # PROB 8
 # Return the longest word in the lst. If the lst is empty, return None.
 # e.g. longest_word(["a", "aaaaaa", "aaa", "aaaa"]) == "aaaaaa"
 def longest_word(lst):
-    return None
+    if not lst:
+	return None
+    else:
+    	newlst = sorted(lst, key=lambda word : len(word))
+    	return newlst[-1]
 
 # PROB 9
 # Return a dictionary, mapping each word to the number of times the word
 # appears in lst.
 # e.g. word_frequency(["a", "a", "aaa", "b", "b", "b"]) == {"a": 2, "aaa": 1, "b": 3}
 def word_frequency(lst):
-    return {}
+    dict = {}
+    for word in lst:
+	if word in dict:
+		dict[word]=dict[word]+1
+	else:
+		dict[word]=1
+    return dict
 
 # PROB 10
 # Return the tuple (word, count) for the word that appears the most frequently
 # in the list, and the number of times the word appears. If the list is empty, return None.
 # e.g. most_frequent_word(["a", "a", "aaa", "b", "b", "b"]) == ("b", 3)
 def most_frequent_word(lst):
-    return None
+    if not lst:
+	return None
+    else:
+    	dict = word_frequency(lst)
+    	keys = dict.keys()
+    	sortedkeys = sorted(keys, key=lambda word : dict[word])
+    	return (sortedkeys[-1], dict[sortedkeys[-1]])
 
 # PROB 11
 # Compares the two lists and finds all the positions that are mismatched in the list.
 # Assume that len(lst1) == len(lst2). Return a list containing the indices of all the
 # mismatched positions in the list.
 # e.g. find_mismatch(["a", "b", "c", "d", "e"], ["f", "b", "c", "g", "e"]) == [0, 3]
 def find_mismatch(lst1, lst2):
-    return []
+    mismatches = []
+    for pos in range(len(lst1)):
+	if (lst1[pos]!=lst2[pos]):
+		mismatches.append(pos)
+    return mismatches
 
 # PROB 12
 # Returns the list of words that are in word_list but not in vocab_list.
 def spell_checker(vocab_list, word_list):
-    return []
+    returnme = []
+    for word in word_list:
+	if not(word in vocab_list):
+		returnme.append(word)
+    return returnme
 

exercises-spellchecker/dictionary.py

@@ -16,22 +16,26 @@ def load(dictionary_name):
     Each line in the file contains exactly one word.
     """
     # TODO: remove the pass line and write your own code
-    pass
+    lines = set([line.strip() for line in open(dictionary_name)])
+    return lines
 
 def check(dictionary, word):
     """
     Returns True if `word` is in the English `dictionary`.
     """
-    pass
+    if word in dictionary:
+	return True
+    else:
+	return False
 
 def size(dictionary):
     """
     Returns the number of words in the English `dictionary`.
     """
-    pass
+    return len(dictionary)
 
 def unload(dictionary):
     """
     Removes everything from the English `dictionary`.
     """
-    pass
+    dictionary=set([])