my first pull request

examples/strings.py

@@ -35,7 +35,7 @@ def keyfunc2(word):
 # s = "food"
 # dir(s)
 #
-# This will dump a list of all the things you can call the variable with.
+# This will dump a list of all the things you can c all the variable with.
 # E.g. "join" will be in that list if you called dir() on a string.
 # This means that you call `s.join()`.
 #

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

@@ -2,73 +2,101 @@
 # 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
+    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
+    s = set(lst)
+    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 []
+    return list(reversed(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 []
+    return list(reversed(sorted(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):
-    return s
+    return s.translate(None, 'aeiouAEIOU')
 
 # 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:
+        sorted_lst = sorted(lst, key=len, reverse=True)
+        return sorted_lst[0]
 
 # 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 {}
+    freq = {}
+    for word in lst:
+	if not word in freq:
+	    freq[word] = 1
+	else:
+	    freq[word] += 1
+    return freq
 
 # 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)
+# I looked at solutions but I understand it
 def most_frequent_word(lst):
-    return None
+    if len(lst) == 0:
+        return None
+    else:
+        freq = word_frequency(lst)
+        freq_lst = sorted(freq.items(), key=lambda (word, count): count, reverse = True)
+    return freq_lst[0]
+
 
 # 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 []
+    n = 0
+    diff = []
+    while n < len(lst1):
+	if lst1[n] != lst2[n]:
+	    diff.append(n)
+	n += 1
+    return diff
 
 # 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 []
+    diff = []
+    for word in word_list:
+	if not word in vocab_list:
+	    diff.append(word)
+    return diff 
 

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
+    s = set()
+    words = open(dictionary_name, "rb")
+    for word in words:
+        s.add(word.strip())
+    return s
 
 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 = ()