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aalyn
2 months ago
8

Implement the function couple, which takes in two lists and returns a list that contains lists with i-th elements of two sequenc

es coupled together. You can assume the lengths of two sequences are the same. Try using a list comprehension.
Computers and Technology
1 answer:
zubka84 [1K]2 months ago
7 0

Answer:

couple.py

def couple(s1,s2):

  newlist = []

  for i in range(len(s1)):

      newlist.append([s1[i],s2[i]])

  return newlist

s1=[1,2,3]

s2=[4,5,6]

print(couple(s1,s2))

enum.py

def couple(s1,s2):

  newlist = []

  for i in range(len(s1)):

      newlist.append([s1[i],s2[i]])

  return newlist

def enumerate(s,start=0):

  number_Array=[ i for i in range(start,start+len(s))]

  return couple(number_Array,s)

s=[6,1,'a']

print(enumerate(s))

print(enumerate('five',5))

Explanation:

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Suppose a linked list of 20 nodes. The middle node has a data –250. Write the pseudocode to replace the middle node of the linke
oksian1 [950]
The data for the middle node shows as –250.... Please outline the pseudocode to substitute the middle node in the linked list. Assume that the head pointer is designated as Head_ptr and the entry data for the new node is defined as Entry.
4 0
2 months ago
Read 2 more answers
Su now wants to highlight several inventors in her presentation, so she decides to use the underline color option.
Harlamova29_29 [1022]

Answer:

1. The home tab

2. Font group

3. The small arrow at the bottom right of the font command group.

4. Under "Font" heading (then "All text" subheading)

Explanation:

This explanation will guide Su in completing her task.

1. Navigate to the home tab;

2. From the home tab, Su will access a variety of command groups, but for her needs, she requires the Font command group.

3. A tiny arrow can be found at the bottom left of the font command group section; clicking it will unveil the font dialogue.

4. Within the font dialogue, there are two headings. The "font" heading and the "character spacing" heading.

For her purposes, she must focus on the "Font" heading, which includes various subheadings. To reach the underline color option, she needs to select the "All text" subheading.

Refer to the attached image for guidance.

7 0
2 months ago
Describe a strategy for avoiding nested conditionals. Give your own example of a nested conditional that can be modified to beco
maria [1035]

Answer:

To prevent nested conditionals, one can utilize logical expressions like the AND operator.

A recommended approach is creating an interface class with a method designed for a shared function. This design approach is known as the strategy design pattern. The conditional statements can be consolidated into this method. Subsequently, each class can implement this interface and invoke that shared method as needed by constructing instances of subclasses and calling the common method for those objects. This illustrates polymorphism.

Explanation:

Nested conditionals occur when if or else if statements are placed within another condition. For instance:

if( condition1) {

//runs when condition1 is true

  if(condition2) {

//runs when both condition1 and condition2 are true

  }  else if(condition3) {

 //when condition1 is true and condition3 is also true

} else {

 //condition1 is true but neither condition2 nor conditions3 are satisfied

}  }

Excessive nested conditionals can complicate the program, rendering it hard to read or comprehend, particularly if there's improper indentation. Debugging also becomes challenging when there are numerous nested statements.

Thus, several strategies can be adopted to eliminate nested conditionals, such as utilizing a switch statement.

For instance, I’ll present an example of the strategies discussed earlier:

Logical Expressions Usage:

A method to avoid nested conditionals is by employing logical expressions with logical operators like the AND operator. The previous nested conditionals can be reframed as:

if(condition1 && condition2){ //only runs when both condition1 and condition2 are true

} else if(condition1 && condition3) {

executes only if both condition1 and condition3 are true

} else if(condition1 ){

//condition1 is satisfied but neither condtion2 nor condtion3 are true  }

This can be further simplified to one condition as:

if(!condition3){

// when  condition1 and condition2 are both satisfied

}

else

// condition3 is met

Now, consider a simple instance dealing with whether to attend school based on certain conditions.

if (temperature< 40)

{

   if (busArrived=="yes")

   {

       if (!sick)

       {

           if (homework=="done")

           {

               printf("Go to school.");

           }

       }                    

   }

}

Here, nested conditionals are evident. This can be restructured into a solitary conditional using the AND logical operator.

if ((temperature <40) && (busArrived=="yes") &&

(!sick) && (homework=="done"))

{    cout<<"Eligible for promotion."; }

The alternate method is utilizing an interface. For example, you can

abstract class Shape{

//declare a method common to all sub classes

  abstract public int area();

// same method that varies by formula of area for different shapes

}

class Triangle extends Shape{

  public int area() {

     // implementation of area code for Triangle

return (width * height / 2);

  }

}

class Rectangle extends Shape{

  public int area() {

     // implementation of area code for Rectangle

    return (width * height)

  }

}

Now, you can readily create Rectangle or Triangle instances and invoke area() for any of those objects, resulting in execution of the appropriate version accordingly.

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2 months ago
Given an 10-bit two's complement binary number, what is the decimal value of the largest negative integer that can be represente
Natasha_Volkova [1026]
The largest negative integer is -512.
7 0
1 month ago
python Consider this data sequence: "3 11 5 5 5 2 4 6 6 7 3 -8". Any value that is the same as the immediately preceding value i
oksian1 [950]

int currentNumber,previousNumber = -1, countDuplicates = 0;

do {

cin >> currentNumber;

if ( previousNumber == -1) {

previousNumber = currentNumber;

}else {

if ( previousNumber == currentNumber )

countDuplicates++;

else

previousNumber = currentNumber;

}

} while(currentNumber > 0 );

cout << countDuplicates;

7 0
2 months ago
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