It works! :D

exercises-hello/hello.py

@@ -9,3 +9,4 @@
 #
 # 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

@@ -2,62 +2,86 @@
 # 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
+    words = s.split()
+    return len(words)
 
 # PROB 2
 # Return the sum of all the numbers in lst. If lst is empty, return 0.
 def sum_list(lst):
-    return 0
+    if len(lst) == 0:
+	return len(lst)
+    else:
+	return sum(lst)
 
 # 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
+    newLst = list(set(lst))
+    return len(newLst)
 
 # 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 []
+    lst.reverse()
+    return lst
 
 # 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 []
+    lst.sort()
+    lst.reverse()
+    return lst
 
 # 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):
+    for letter in s:
+	if letter in ['a', 'A', 'e', 'E', 'i', 'I', 'o', 'O', 'u', 'U']:
+	    s = s.replace(letter, '')
     return s
 
 # 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 len(lst) == 0:
+	return None
+    else:
+	word = max(set(lst), key=len)
+	return word
 
 # 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 {}
+    words = list(set(lst))
+    d = {}
+    for word in words:
+	freq = lst.count(word)
+	d[word] = freq
+    return d
 
 # 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 len(lst) == 0:
+	return None
+    else:
+	word = max(set(lst), key=lst.count)
+	return (word, lst.count(word))
 
 # PROB 11
 # Compares the two lists and finds all the positions that are mismatched in the list.
@@ -70,5 +94,5 @@ def find_mismatch(lst1, lst2):
 # 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 []
+    return list(set(word_list) - set(vocab_list))
 

exercises-spellchecker/dictionary.py

@@ -15,23 +15,28 @@ def load(dictionary_name):
 
     Each line in the file contains exactly one word.
     """
-    # TODO: remove the pass line and write your own code
-    pass
+    words = set()
+    words_file = open(dictionary_name, "rb")
+    for word in words_file:
+        word = word.strip()
+        words.add(word)
+    words_file.close()
+    return words
 
 def check(dictionary, word):
     """
     Returns True if `word` is in the English `dictionary`.
     """
-    pass
+    return word in dictionary
 
 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.clear()