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TEA [102]
2 years ago
11

The groups_per_user function receives a dictionary, which contains group names with the list of users. Users can belong to multi

ple groups. Fill in the blanks to return a dictionary with the users as keys and a list of their groups as values.
def groups_per_user(group_dictionary):
user_groups = {}
# Go through group_dictionary
for ___:
# Now go through the users in the group
for ___: # Now add the group to the the list of
# groups for this user, creating the entry
# in the dictionary if necessary
return(user_groups)
print(groups_per_user({"local": ["admin", "userA"],
"public": ["admin", "userB"],
"administrator": ["admin"] }))
Computers and Technology
1 answer:
Akimi4 [234]2 years ago
3 0

The groups_per_user function receives a dictionary, which contains group names with the list of users.

Explanation:

The blanks to return a dictionary with the users as keys and a list of their groups as values is shown below :

def groups_per_user(group_dictionary):

   user_groups = {}

   # Go through group_dictionary

   for group,users in group_dictionary.items():

       # Now go through the users in the group

       for user in users:

       # Now add the group to the the list of

         # groups for this user, creating the entry

         # in the dictionary if necessary

         user_groups[user] = user_groups.get(user,[]) + [group]

   return(user_groups)

print(groups_per_user({"local": ["admin", "userA"],

       "public":  ["admin", "userB"],

       "administrator": ["admin"] }))

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2 years ago
Recall the problem of finding the number of inversions. As in the text, we are given a sequence of n numbers a1, . . . , an, whi
Kay [80]

Answer:

The algorithm is very similar to the algorithm of counting inversions. The only change is that here we separate the counting of significant inversions from the merge-sort process.

Algorithm:

Let A = (a1, a2, . . . , an).

Function CountSigInv(A[1...n])

if n = 1 return 0; //base case

Let L := A[1...floor(n/2)]; // Get the first half of A

Let R := A[floor(n/2)+1...n]; // Get the second half of A

//Recurse on L. Return B, the sorted L,

//and x, the number of significant inversions in $L$

Let B, x := CountSigInv(L);

Let C, y := CountSigInv(R); //Do the counting of significant split inversions

Let i := 1;

Let j := 1;

Let z := 0;

// to count the number of significant split inversions while(i <= length(B) and j <= length(C)) if(B[i] > 2*C[j]) z += length(B)-i+1; j += 1; else i += 1;

//the normal merge-sort process i := 1; j := 1;

//the sorted A to be output Let D[1...n] be an array of length n, and every entry is initialized with 0; for k = 1 to n if B[i] < C[j] D[k] = B[i]; i += 1; else D[k] = C[j]; j += 1; return D, (x + y + z);

Runtime Analysis: At each level, both the counting of significant split inversions and the normal merge-sort process take O(n) time, because we take a linear scan in both cases. Also, at each level, we break the problem into two subproblems and the size of each subproblem is n/2. Hence, the recurrence relation is T(n) = 2T(n/2) + O(n). So in total, the time complexity is O(n log n).

Explanation:

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Why can a failure in a database environment be more serious than an error in a nondatabase environment?
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2 years ago
Read 2 more answers
Write a second constructor as indicated. Sample output: User1: Minutes: 0, Messages: 0 User2: Minutes: 1000, Messages: 5000
Mashcka [7]

Complete Question:

Write a second constructor as indicated. Sample output:User1: Minutes: 0, Messages: 0User2: Minutes: 1000, Messages: 5000// ===== Code from file PhonePlan.java =====public class PhonePlan { private int freeMinutes; private int freeMessages; public PhonePlan() { freeMinutes = 0; freeMessages = 0; } // FIXME: Create a second constructor with numMinutes and numMessages parameters. /* Your solution goes here */ public void print() { System.out.println("Minutes: " + freeMinutes + ", Messages: " + freeMessages); return; }}

Answer:

The second constructor is given as:

//This defines the constructor, the name has to be the same as the class //name

PhonePlan(int numOfMinutes, int numberOfMessages) {

this.freeMinutes = numOfMinutes;

this.freeMessages = numberOfMessages

}

Explanation:

The second constructor is defined using java programming language.

  1. The given class has two constructors This is called "Constructor Overloading) which implements polymophism
  2. In the second constructor that we created, we pass in two arguments of type integer numOfMinutes and numberOfMessages.
  3. In the constructor's body we assign these values to the initially declared variables freeMinutes and freeMessages
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Taylor and Rory are hosting a party. They sent out invitations, and each one collected responses into dictionaries, with names o
disa [49]

Answer:

Following are the code to this question:

def combine_guest(guest1, guest2):#defining a method combine_guest that accepts two dictionary

   guest2.update (guest1)#use dictionary guest2 that use update method to update guest2 dictionary

   return guest2#return guest2 dictionary values

Rory_guest= { "Ada":2, "Ben":3, "Dav":1, "Jo":3, "Charry":2, "Terry":1, "bob":4}#defining a dictionary and add value

Taylor_guest = { "Dav":4, "Nan":1, "bobert":2, "Ada":1, "Samantha":3, "Chr":5}#defining a dictionary and add value

print(combine_guest(Rory_guest,Taylor_guest))#calling the combine_guest method

Output:

{'Nan': 1, 'Samantha': 3, 'Ada': 2, 'bob': 4, 'Terry': 1, 'Jo': 3, 'Ben': 3, 'Dav': 1, 'Charry': 2, 'bobert': 2, 'Chr': 5}

Explanation:

In the code a method, "combine_guest" is defined, that accepts two dictionaries "guest1, guest2" inside the method, in which the "guest2" dictionary uses the update method to combine the value of the guest1 dictionary and use a return keyword to return guest2 values.

In the next step, two dictionaries are declared, that holds some values and use a print method to call the "combine_guest" method and prints its return values.  

7 0
1 year ago
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