Elso hazi

basic/list1.py

@@ -1,28 +1,14 @@
-#!/usr/bin/python -tt
-# Copyright 2010 Google Inc.
-# Licensed under the Apache License, Version 2.0
-# http://www.apache.org/licenses/LICENSE-2.0
-
-# Google's Python Class
-# http://code.google.com/edu/languages/google-python-class/
-
-# Basic list exercises
-# Fill in the code for the functions below. main() is already set up
-# to call the functions with a few different inputs,
-# printing 'OK' when each function is correct.
-# The starter code for each function includes a 'return'
-# which is just a placeholder for your code.
-# It's ok if you do not complete all the functions, and there
-# are some additional functions to try in list2.py.
-
 # A. match_ends
 # Given a list of strings, return the count of the number of
 # strings where the string length is 2 or more and the first
 # and last chars of the string are the same.
 # Note: python does not have a ++ operator, but += works.
 def match_ends(words):
-    # +++your code here+++
-    return
+    n = 0
+    for word in words:
+        if len(word) >= 2 and word[0] == word[len(word) - 1]:
+            n += 1
+    return n
 
 
 # B. front_x
@@ -33,8 +19,15 @@ def match_ends(words):
 # Hint: this can be done by making 2 lists and sorting each of them
 # before combining them.
 def front_x(words):
-    # +++your code here+++
-    return
+    listx = []
+    listnorm = []
+    for word in words:
+        if word[0] == 'x':
+            listx.append(word)
+        else:
+            listnorm.append(word)
+
+    return sorted(listx) + sorted(listnorm)
 
 
 # C. sort_last
@@ -44,8 +37,17 @@ def front_x(words):
 # [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
 # Hint: use a custom key= function to extract the last element form each tuple.
 def sort_last(tuples):
-    # +++your code here+++
-    return
+    endings = []
+    res = []
+    for e in tuples:
+        endings.append(e[-1])
+    endings.sort()
+    for ending in endings:
+        for tup in tuples:
+            if tup[-1] == ending:
+                res.append(tup)
+    return res
+
 
 
 # Simple provided test() function used in main() to print

basic/list2.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python -tt
+# !/usr/bin/python -tt
 # Copyright 2010 Google Inc.
 # Licensed under the Apache License, Version 2.0
 # http://www.apache.org/licenses/LICENSE-2.0
@@ -13,17 +13,15 @@
 # so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
 # modify the passed in list.
 def remove_adjacent(nums):
-    # +++your code here+++
-    return
+    return list(set(nums))
 
 
 # E. Given two lists sorted in increasing order, create and return a merged
 # list of all the elements in sorted order. You may modify the passed in lists.
 # Ideally, the solution should work in "linear" time, making a single
 # pass of both lists.
 def linear_merge(list1, list2):
-    # +++your code here+++
-    return
+    return sorted(list1 + list2)
 
 
 # Note: the solution above is kind of cute, but unforunately list.pop(0)
@@ -62,4 +60,4 @@ def main():
 
 
 if __name__ == '__main__':
-    main()
+    main()

basic/string1.py

@@ -24,8 +24,10 @@
 # So donuts(5) returns 'Number of donuts: 5'
 # and donuts(23) returns 'Number of donuts: many'
 def donuts(count):
-    # +++your code here+++
-    return
+    if count < 10:
+        return "Number of donuts: " + str(count)
+    else:
+        return "Number of donuts: many"
 
 
 # B. both_ends
@@ -34,22 +36,26 @@ def donuts(count):
 # so 'spring' yields 'spng'. However, if the string length
 # is less than 2, return instead the empty string.
 def both_ends(s):
-    # +++your code here+++
-    return
+    if len(s) >= 2:
+        return s[0:2] + s[-2:]
+    else:
+        return ''
 
 
 # C. fix_start
 # Given a string s, return a string
-# where all occurences of its first char have
+# where all occurrences of its first char have
 # been changed to '*', except do not change
 # the first char itself.
 # e.g. 'babble' yields 'ba**le'
 # Assume that the string is length 1 or more.
 # Hint: s.replace(stra, strb) returns a version of string s
 # where all instances of stra have been replaced by strb.
 def fix_start(s):
-    # +++your code here+++
-    return
+    first = s[0]
+    rest = s[1:]
+    res = rest.replace(first, '*')
+    return first + res
 
 
 # D. MixUp
@@ -60,8 +66,7 @@ def fix_start(s):
 #   'dog', 'dinner' -> 'dig donner'
 # Assume a and b are length 2 or more.
 def mix_up(a, b):
-    # +++your code here+++
-    return
+    return b[0:2] + a[2:] + ' ' + a[0:2] + b[2:]
 
 
 # Provided simple test() function used in main() to print
@@ -108,4 +113,4 @@ def main():
 
 # Standard boilerplate to call the main() function.
 if __name__ == '__main__':
-    main()
+    main()

basic/string2.py

@@ -16,8 +16,13 @@
 # If the string length is less than 3, leave it unchanged.
 # Return the resulting string.
 def verbing(s):
-    # +++your code here+++
-    return
+    if len(s) > 2:
+        if s[-3:] != 'ing':
+            return s + 'ing'
+        else:
+            return s + 'ly'
+    else:
+        return s
 
 
 # E. not_bad
@@ -29,9 +34,11 @@ def verbing(s):
 # So 'This dinner is not that bad!' yields:
 # This dinner is good!
 def not_bad(s):
-    # +++your code here+++
-    return
-
+    not_ind = s.find('not')
+    bad_ind = s.find('bad')
+    if not_ind != -1 and bad_ind != -1 and not_ind < bad_ind:
+        s = s[:not_ind] + 'good' + s[bad_ind + 3:]
+    return s
 
 # F. front_back
 # Consider dividing a string into two halves.
@@ -41,8 +48,13 @@ def not_bad(s):
 # Given 2 strings, a and b, return a string of the form
 #  a-front + b-front + a-back + b-back
 def front_back(a, b):
-    # +++your code here+++
-    return
+    amid = len(a) / 2
+    bmid = len(b) / 2
+    if len(a) % 2 == 1:
+        amid += 1
+    if len(b) % 2 == 1:
+        bmid += 1
+    return a[:int(amid)] + b[:int(bmid)] + a[int(amid):] + b[int(bmid):]
 
 
 # Simple provided test() function used in main() to print
@@ -76,4 +88,4 @@ def main():
 
 
 if __name__ == '__main__':
-    main()
+    main()