Pull request ipod/Roger hey it works!

exercises-hello/hello.py

@@ -9,3 +9,5 @@
 #
 # 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 +2,130 @@
 # 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
+   s = s.split()
+   return len(s)
 
 # PROB 2
 # Return the sum of all the numbers in lst. If lst is empty, return 0.
 def sum_list(lst):
-    return 0
+    n = 0
+    for num in lst:
+        n += num
+    return n
 
 # 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
+    s = set()
+    for a in lst:
+        s.add(a)
+    return len(s)
 
 # 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 []
+    #
+    #for i in len(lst) - 2:
+    #    if list[i] > list[i + 1]
+    lst.sort()
+    return reverse_list(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):
+    s = s.replace("a", "")
+    s = s.replace("A", "")
+    s = s.replace("e", "")
+    s = s.replace("E", "")
+    s = s.replace("i", "")
+    s = s.replace("I", "")
+    s = s.replace("o", "")
+    s = s.replace("O", "")
+    s = s.replace("u", "")
+    s = s.replace("U", "")
     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
+    n = 0
+    g = ""
+    for word in lst:
+        if (len(word) > n):
+            g = word
+            n = len(word)
+    if n == 0:
+        return None
+    else:
+        return g
 
 # 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 {}
+    di = {}
+    for word in lst:
+        if word in di.keys():
+            di[word] += 1
+        else:
+            di[word] = 1
+    return di
 
 # 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
+    di = word_frequency(lst)
+    hold = ""
+    n = 0
+    for key in di:
+        if di[key] > n:
+            hold = key
+            n = di[key]
+    if n == 0:
+        return None
+    else:
+        return (hold, n)
 
 # 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 []
+    num = []
+    a = 0
+    for l in lst1:
+        if l not in lst2[a]:
+            num.append(a)
+        a += 1
+    return num
 
 # 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 []
+    lstr = []
+    for word in word_list:
+        if not word in vocab_list:
+            lstr.append(word)
+    return lstr
 

exercises-spellchecker/dictionary.py

@@ -16,22 +16,25 @@ def load(dictionary_name):
     Each line in the file contains exactly one word.
     """
     # TODO: remove the pass line and write your own code
-    pass
-
+    f = open(dictionary_name, "r")
+    words = set()
+    for line in f:
+        stripper = line.strip() #takes away new lines
+        words.add(stripper)
+    f.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()