>>> print( X ) [0,1,2,3] def mystery(l):??

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1 REVIEW EXERCISES Problem 4 What is the output of the following script? Disclaimer. I cannot guarantee that the exercises of the final exam are going to look like this. Also, for the most part, the exercises are organized to fit tight in each page. They are not ordered by difficulty nor by topic. Problem 1 What is the output of the following instructions? print (7 > 10)... print (4 < 16)... print (4 == 4)... print (4 <= 4)... print (4!= 4)... x = 17.0 / 2 % 2 * 3**3 print (x)... Problem 2 What is the output of the following script? num1 = 5 if num1 >= 91: num2 = 3 if num1 < 6: num2 = 4 num2 = 2 x = num2 * num1 + 1 print (x,x%7) list1 = [1,2,3] list1 = list1 * 2 print (list1) list3 = [2,4,6,8,10,12,14,16,18,20] print (list3[0:1]) print (list3[5:7]) Problem 5 Replace the unknowns in the following script (each unknown consists, at most, of a single line): def mystery(l): if len(l)==0: return -1?? return??? >>> X=list(range(5)) >>> print( mystery(x) ) 0 >>> print( X ) [0,1,2,3]? Problem 3 Fill in the blanks of the following function. You may need only one line here. def swapper(l): This function mutates a list by swapping its first and last elements. Parameters: L: a list, of len(l)>=2. Returns: swapper does not return anything.... = =... 1 Problem 6 Replace the unknowns in the following script (each unknown consists, at most, of a single line): def mystery(l):?? >>> X=list(range(5,0,-1)) >>> print( mystery(x) ) None >>> print( X ) [1, 2, 3, 4, 5] >>> X=[(1,0), (0,2), (-10,10), (10,10)] >>> mystery( X ) >>> print( X ) [(-10,10), (0,2), (1,2), (10,10)]

2 REVIEW EXERCISES 2 Problem 7 Replace the unknowns in the following script (each unknown consists, at most, of a single line): class Empty: pass def putsome(x): def printhello():????? >>> A = Empty() >>> putsome(a) >>> A.f() Hello!? Problem 8 s1 = spamandeggs x = s1.find( and ) print (x)... print (s1[0:1],s1[5:7]) s = one\ttwo\tthree print (s)... print (len(s))... Problem 9 Fill in the blanks so function foof does as expected. def foof(l): Removes the elements of a list if they are strictly smaller than the previous one. Input: L, a list; foof may mutate L (remove elems). Returns: Nothing....=[L[i] for i in range(...) if...] >>> L = [1,3,2,4,3,5] >>> foof(l) >>> L [1, 3, 4, 5] >>> L = [1,-1,1,-1,1] >>> foof(l) [1, 1, 1, 1] >>> L = [ a, c, b, d ] >>> foof(l) [ a, b, d ] Fill in the unknown (??) Printed!: class empty: pass x = empty() y = empty()?? if y is x.z: print("printed!") Problem 10 so that the following script prints... Problem 11 Replace the unknowns in the following script (each unknown consists, at most, of a single line): class Kount: def init (self): self.c = 1 def tellme(self): c = self.c self.c += 1 return c class Negnt(??): def tellme(self): c = self.c??? return c >>> A = Kount() >>> B = Negnt() >>> print( A.tellMe(), B.tellMe() ) 1 1 >>> print( A.tellMe(), B.tellMe() ) 2-1 >>> print( A.tellMe(), B.tellMe() ) 3 1 >>> print( A.tellMe(), B.tellMe() ) 4-1? Problem 12 list2 = ["B","C","A"] list2.extend(["x","y"]) list2.reverse() list2.append("s") list2.sort() list2.reverse() print (list2)

3 REVIEW EXERCISES 3 Problem 13 def f(s): """Parameter s is a string.""" d = {} for c in s: if c in d.keys(): d[c] += 1 d[c] = 1 x = None for k in d.keys(): if x == None: x = d[k] y = k elif d[k] > x: x = d[k] y = k return y For the function defined above, answer the following questions. (1) What is the value of f( )?... (2) What is the value of f( )?... (3) What does f do? Specify its documentation below: Problem 15 Consider the program of creating the type (class) dictionary from scratch. Complete in the following code so the class below becomes functional. class D: # one very simple name! def init (self): self.mem = [] # This will store the (key,value)s def additem( self, key, value ): This function will add pair (key,value) to the internal list MEM. Each key, however, has only one value associated to it. if self.hasitem(key): # To ensure key is unique self.delitem(key) # add the pair self.mem.append( (...,...) ) def hasitem( self, key ): This function checks whether there is a value stored associated to key. return key in [...] def getitem( self, key ): Returns the value associated to key. Does not check if key exists in MEM. return [...][0] def delitem( self, key ): Deletes the pair (key,value) from MEM. Does not check if key exists in MEM. # There are three lines below, but one line is # actually enough! del self.mem[i] Problem 14 Consider the following function definition: def f(l): result = [] for e in L: if type(e)!= list: result.append(e) return f(e) return result Answer the following questions: (1) What does f([ a, b, c ]) do? (2) What does f( hello word ) do? (3) What does f(f( hello word )) do? (4) What does f([f]) return? def length( self ): Returns the number of entries in MEM.... An example implementation of the module above produced the following real output using the console (I stored D in D.py): >>> from D import D >>> A=D() >>> A.addItem(1,2) >>> A.addItem(0,5) >>> A.addItem(3,2) >>> A <D.D instance at 0x7f12bf38e5a8> >>> A.MEM [(1, 2), (0, 5), (3, 2)] >>> A.addItem(3,3) >>> A.MEM [(1, 2), (0, 5), (3, 3)] >>> A.delItem(0) >>> A.MEM [(1, 2), (3, 3)]

4 REVIEW EXERCISES 4 Problem 16 Problem 18 Fill in the blanks of the following script to make it print the results desired. def xpand(word): S = [...] print("-".join(s)) def phoo(x): if type(x)==type(""): print(... ) elif type(x)==type([]): xpand(x) print(...) return... >>> phoo( BINGO ) B-I-N-G-O None >>> phoo(10) 20 >>> phoo(0) 0 >>> phoo(20) 40 >>> phoo([1,2,3,"4"]) Implement the following function according to its definition. def average_word_length (text): text is a str consisting only of words and spaces (no punctuation). text is guaranteed to have at least one word. There can be one or more than one whitespace character between pairs of words. Return the average length of all words in text, as a float. # This can be done in a couple of lines Problem 19 Write a good docstring for the following function. Problem 17 Fill in the blanks of the following function. def distionaree(d): This function creates a tuple with the elements of the dictionary such that keys are immediately followed by their values in the tuple, i.e., if D[x]=y, then (...,x,y,...) in the returned tuple by distionaree(d). No additional order is enforced on the returned tuple. Examples: distionary({1:0,2:2, a :3}) -> (1,0,2,2, a,3) distionary({10:5, ow :()}) - > ( ow,(),10,5) Parameters: D: a dictionary. Returns: a tuple. accum = () for key in??:??? return accum? def enigma(s, c, n): count = 0 result = "" for char in s: if char == c and count < n: result += X count += 1 result += char return result

5 REVIEW EXERCISES 5 Problem 20 L = [[ on, off ], [ go, slow, stop ], [ up ]] print( L[1][2][3] )... print( L[0][0][0] )... L[2:] = L L[2:] = L print( L ) Problem 21 Implement the following function according to the functionality specified in its docstring. def longest_sequence(r): This function counts the number of lines in the longest consecutive sequence of blank lines in open reader r (a file handle, i.e., r was defined using r=open(...) or something alike), or zero if there are no blank lines at all. Parameters: r, a file handle. Returns: An integer specifying the longest consec. sequence of blank lines read from r. Problem 22 D = {} D[ apple ] = 4 D[ apple ] = 8 print( D )... D = {2 : [ d ], 0 : [ c ]} D[0].append( a ) print( D[0] )... Problem 23 Implement the following function according to the functionality specified in its docstring. def swapcases(s): This function swaps the case of each symbol in string s; lowercase symbols become uppercase and uppercase symbols become lowercase. The function returns nothing; rather, it mutates string s. Problem 24 D = {4 : 3, 3 : 2, 2 : 1} for x in D.values(): print(d[x]) D1 = { a : 1, b : 2, c : 3} D2 = { b : 2, c : 3, a : 1} print D1 == D Problem 25 Write a good docstring for the following function. def enigmatic(a, b): i = 0 while b!= "" and i < len(a): if a[i] in b: b = b.replace(a[i], "") i += 1 return b == ""

6 REVIEW EXERCISES 6 Problem 26 def mystery(s): result = for ch in s: if ch in AEIOUaeiou : result += x elif ch.isalpha(): result += _ result += ch return result if name == main : print( mystery( Why ) ) print( mystery( am ) ) print( mystery( I ) ) print( mystery( so tired? ) ) Problem 27 Write a proper documentation for the following function. def enigma(d1, d2): ans = {} for key, value in d1.items(): if key in d2: ans[key] = max(value, d2[key]) ans[key] = value for key, value in d2.items(): if key in d1: ans[key] = max(value, d1[key]) ans[key] = value return ans Problem 28 def bar1(y): x = len(y) - 1 while(x >= 0): if(x % 2 == 0): y[x] = x = x - 1 return # Main x = 5 names =[ Foo, Boo, Goo, Hoo, Voo, Doo ] result = bar1(names) print(x) print(names) print(result) Problem 29 Consider the following function. def whatdoido(x): Parameter x can be a tuple, a list or a string. This function returns True or False after testing some property on x. for i in range( 1, len(x) ): if x[i]==x[i-1]: return True return False Rewrite the function without using loops (for, while). Use recursion instead. def whatdoido(x): Parameter x can be a tuple, a list or a string. This function returns True or False after testing some property on x. if len(x)<=1:

7 REVIEW EXERCISES 7 Problem 30 Problem 31 Consider the linked list problem. We will implement a list-like class but using a recursive structure in it. Note: this class definition makes intensive use of recursion. class LL: def init (self): self.next=none self.value=none def append(self,value): This method inserts "appends" value into the linked list. if self.value == None: self.value =... elif self.next == None: self.next = LL() self.next.insert( value ) self.next.insert( value ) def has(self,value): Returns true if "value" is stored in the linked list, and returns false otherwise. if self.value == value:... if self.next!= None: return self.next... # If the above were unsuccessful... def get(self,i): Returns the value stored in the i-th element of the linked list. If i is negative or is after the last index of the linked list, we return None. if i < 0: if i == 0: if i > 0 and self.next == None: # Getting here means that: # i > 0 and self.next!= None... def set(self,i,value): Replaces the i-th element of the linked list with value. If i is negative or after the last index of the linked list, we do nothing. if i == 0: self.value... if i > 0 and self.next!= None:... def len(self): Returns the length of the linked list. if self.value == None: if self.next == None:... Consider the following function. def whatdoido(x): Parameter x can be a tuple, a list or a string. This function returns a list the reverse of x. For example, whatdoido([1, 2, 3]) is [3, 2, 1], and whatdoido( hello ) is [ o, l, l, e, h ]. R = [] for i in range(len(x)): R = [ x[i] ] + R return R Rewrite the function without using loops (for, while). Use recursion instead. def whatdoido(x): Parameter x can be a tuple, a list or a string. This function returns a list the reverse of x. For example, whatdoido([1, 2, 3]) is [3, 2, 1], and whatdoido( hello ) is [ o, l, l, e, h ]. if len(x)<=0: Problem 32 Create a function that takes a list L and creates a new function f, such that f(0) = L[0], f(1) = L[1], f(-1) = None, f(len(l))=none, etc. Part of the code is given; you just need to complete it. def list2func( L ): if L == []: # if the list is empty, we return the trivial # function, which always returns None def emptycase(n): return... return emptycase # now, we are not in the trivial case next_f = list2func( L[...] ) def nonemptycase(n): if n == 0: # return the current value return L[...] if n < 0: # this is "f(-1)" return... if n > 0: return next_f( n-1 ) return nonemptycase # we give the ball to the next guy >>> f = list2func( [4,1,2,3] ) >>> f(-1), f(0), f(1), f(2), f(3), f(4) (None, 4, 1, 2, 3, None)

8 REVIEW EXERCISES 8 Problem 33 Problem 35 def foo(*x,**y): print(x) print(y) foo( hello world ) foo( hello, world,0) foo() foo(150,x=150) foo({}) Problem 34 Consider the following function. def whatdoido(f,iters = 1000): This function computes the "fixed" point of function "f". A fixed point is a number such that f(z)=z. To avoid issues with arithmetic precision, we require that z-f(z) <threshold to accept z as an answer. Our starting point is z=0. Also, since the search could go on forever, parameter "iters" tells when to stop the search, if we have not found a fixed point. threshold = 10.0**-3 z = 0 while abs( z - f(z) ) < threshold and iters > 0: z = f(z) iters-= 1 return z Rewrite the function without using loops (for, while). Use recursion instead. def whatdoido(...): This function computes the "fixed" point of function "f". A fixed point is a number such that f(z)=z. To avoid issues with arithmetic precision, we require that z-f(z) <threshold to accept z as an answer. Our starting point is z=0. Also, since the search could go on forever, parameter "iters" tells when to stop the search, if we have not found a fixed point File wordos.txt contains a bunch of words. Write a script that counts how many words and how many different words are there in wordos.txt, and then prints those numbers. The file may be multiline, but there is no punctiation. (Hint: make good use of str.split.) f = open(...) for word in ListOfWords: print( Number of words:,...) print( Number of different words:,...) Problem 36 Let us define an -list using recursion. An -list is a normal Python list whose elements are either: (1) the empty list [ ], or (2) another -list. For example, the following is an -list: [ [ [ [ ] ] ], [ ], [ [ ], [ ] ] ] For this problem, answer the following questions: (1) Write a function that checks whether a list is an -list. The function must return True or False. (2) Write a function that returns the height (number of levels) of a given -list. For example, [ ] has height 1, [ [ ] ] has height 2, [ [ ], [ ] ] also has height 2, [ [ [ ] ] ] has height 3, etc. (3) Write a function that counts the number of -lists contained in a -list, including the list itself. For example, [ ] has one -list, [ [ ], [ ] ] has three -lists, etc.