q7_lists.py

# Based on materials copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
#

def match_ends(words):

    """
    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.
    >>> match_ends(['aba', 'xyz', 'aa', 'x', 'bbb'])
    3
    >>> match_ends(['', 'x', 'xy', 'xyx', 'xx'])
    2
    >>> match_ends(['aaa', 'be', 'abc', 'hello'])
    1
    """
    count = 0
    for word in words:
      if len(word) >= 2 and word[0] == word[len(word)-1]:
        count+=1
    return count
print match_ends(['aaa', 'be', 'abc', 'hello'])
print match_ends(['', 'x', 'xy', 'xyx', 'xx'])

def comparefunctionx(x,y):
  if x[0] == 'x' and y[0] == 'x':
    if x > y:
      return 1
    elif x < y:
      return -1
    else:
      return 0
  elif x[0] == 'x':
    return -1
  elif y[0]=='x':
    return 1
  else:
    if x>y:
      return 1
    elif x<y:
      return -1
    else:
      return 0
def front_x(words):
    """
    Given a list of strings, return a list with the strings in sorted
    order, except group all the strings that begin with 'x' first.
    e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
         ['xanadu', 'xyz', 'aardvark', 'apple', 'mix'].
    >>> front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa'])
    ['xaa', 'xzz', 'axx', 'bbb', 'ccc']
    >>> front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa'])
    ['xaa', 'xcc', 'aaa', 'bbb', 'ccc']
    >>> front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark'])
    ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
    """
    return sorted(words, cmp=comparefunctionx)

print front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa'])
print front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark'])





def sort_last(tuples):
    """
    Given a list of non-empty tuples, return a list sorted in
    increasing order by the last element in each tuple.
    e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
         [(2, 2), (1, 3), (3, 4, 5), (1, 7)].
    >>> sort_last([(1, 3), (3, 2), (2, 1)])
    [(2, 1), (3, 2), (1, 3)]
    >>> sort_last([(2, 3), (1, 2), (3, 1)])
    [(3, 1), (1, 2), (2, 3)]
    >>> sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)])
    [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
    """
    getkey = lambda item: item[len(item)-1]
    return sorted(tuples,key=getkey)

print sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)])
print sort_last([(2, 3), (1, 2), (3, 1)])
def remove_adjacent(nums):
    """
    Given a list of numbers, return a list where all adjacent equal
    elements have been reduced to a single element, so [1, 2, 2, 3]
    returns [1, 2, 3]. You may create a new list or modify the passed
    in list.
    >>> remove_adjacent([1, 2, 2, 3])
    [1, 2, 3]
    >>> remove_adjacent([2, 2, 3, 3, 3])
    [2, 3]
    >>> remove_adjacent([3, 2, 3, 3, 3])
    [3, 2, 3]
    >>> remove_adjacent([])
    []
    """
    retlist = []
    for x in range(0,len(nums)):
      if x > 0:
        if retlist[-1] == nums[x]:
          pass
        else:
          retlist.append(nums[x])
      else:
        retlist.append(nums[0])
    return retlist
print remove_adjacent([3, 2, 3, 3, 3])
print remove_adjacent([])
print remove_adjacent([2, 2, 3, 3, 3])

def linear_merge(list1, list2):
    """
    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.
    >>> linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc'])
    ['aa', 'bb', 'cc', 'xx', 'zz']
    >>> linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz'])
    ['aa', 'bb', 'cc', 'xx', 'zz']
    >>> linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb'])
    ['aa', 'aa', 'aa', 'bb', 'bb']
    """
    list1.extend(list2)
    return sorted(list1)
print linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc'])
print linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz'])
print linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb'])
#