emasatsugu python bootcamp

exercises-hello/hello.py

@@ -9,3 +9,5 @@
 #
 # TODO: write your code below
 
+
+print "hello world";

exercises-hello/script.py

@@ -0,0 +1,3 @@
+#!/usr/bin/env python
+print "this is a python script!"
+

exercises-more/exercises.py

@@ -2,48 +2,68 @@
 # 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 num in lst:
+        sum += num
+    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
+    return len(set(lst))
 
 # 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 []
+    newlst = []
+    for x in reversed(lst):
+        newlst.append(x)
+    return newlst;
+
 
 # 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 []
+    newlst = sorted(lst)
+    return reverse_list(newlst)
 
 # 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
+    news = s;
+    vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
+    for letter in s:
+        if letter in vowels:
+            news = news.replace(letter, "")
+    return news
+
 
 # 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
+    longest = ""
+    if len(lst) == 0:
+        return None;
+    for word in lst:
+        if len(word) > len(longest):
+            longest = word
+    return longest;
 
 # PROB 9
 # Return a dictionary, mapping each word to the number of times the word

exercises-spellchecker/dictionary.py

@@ -16,22 +16,28 @@ 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();
+    word_file = open(dictionary_name, "rb")
+    for word in word_file:
+        word = word.strip()
+        words.add(word) 
+    word_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(list(dictionary))
 
 def unload(dictionary):
     """
     Removes everything from the English `dictionary`.
     """
-    pass
+    dictionary.clear();