Coding examples of python, Python Programming

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Following are some  attempts at defining a function isSubset,  which  takes  two  arguments, a and  b, and  returns True if a is a subset  of b, assuming that  a and  b are represented as lists of elements.

Here is one answer to the problem which  is in the Pythonic functional style.

 

def isSubset(a, b):

return reduce(operator.and_, [x in b for x in a])

 

This is short  and  direct.   The expression x in b tests  to see if the  item  x is in the  structure b.  So, [x in b for x in a] is a list of Booleans,  one for each components in a, indicating whether that component is in b. Finally,  we replace that  list using  the and operator11 , which  will have  the value True if all of the elements of the list are False, and True otherwise.

An alternative is to do it recursively:

 

def isSubset(a, b):

if a == []:

return True

else:

return a[0] in b and isSubset(a[1:], b)

 

The base case of the recursion is that a is the empty list; in that type, it's define that a is a subset  of b. Otherwise, we may give  our answer in terms  of isSubset,  but asking  a simpler question.  So, we call that a is a subset  of b if the ?rst element of a is a member of b and the set of rest of the elements of a is a subset  of b.

 

We can go even farther in the recursive solution for compressing:

def isSubset(a, b):

return a == None or a[0] in b and isSubset(a[1:], b)

 

Here, we are taking  advantage of the fact that in Python (and most other languages), the or Boolean operator has the "early  out"  charters tics. That is, if we are calculating e1 or e2, we start  by evaluating e1, and  if it is True,  then  we know  the  output  of the  expression has  to be True,  and  therefore we give without calculating e2.  So, or can perform as a kind  of conditional operator. Some person would search this program too abstruse (shorter isn't always better),  and  some  would take  it very beautiful.

 

Here is another good solution, this time in the imperative style:

 

def isSubset(a, b):

for x in a:

if not x in b:

return False return True

 

It works  by going through the component in a, in order.  If any of those components is not in b, then we can immediately quit and give False. Otherwise, if we have taken through all of the components in a, and each of them  has been in b, then a really is a subset  of b and we may give True.

Here is another imperative program:

 

def isSubset(a, b): result = True for x in a:

result = result and x in b return result

 

This procedure starts  by initializing a result  to True,  which  is the  check component for and (it plays  the same  role as 0 for add).  Then  it tends all the way  through the structure a, and  ands the last result  with  x in b. This calculation is as the computation done  in the functional version, but rather than  preparing the entire  list of Booleans  and  then  replacing, we inter leave the individual membership tests with combining them  into a result.

 

Posted Date: 8/9/2012 7:25:45 AM | Location : United States







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