dictionaries = [] dictionaries.append({"First":"Bob", "Last":"Jones"}) dictionaries.append({"First":"Harpreet", "Last":"Kaur"}) dictionaries.append({"First":"Mohamad", "Last":"Argani"}) for i in range(0, len(dictionaries)): # Condition that's print the full name when Condition is True if(dictionaries[i]['First']=='Bob')or(dictionaries[i]['Last']=='Kaur'): print(dictionaries[i]['First']+" "+dictionaries[i]['Last']) ********************************** please modify the code to use a while loop instead of a for loop

dictionaries = [] dictionaries.append({"First":"Bob", "Last":"Jones"}) dictionaries.append({"First":"Harpreet", "Last":"Kaur"}) dictionaries.append({"First":"Mohamad", "Last":"Argani"}) for i in range(0, len(dictionaries)): # Condition that's print the full name when Condition is True if(dictionaries[i]['First']=='Bob')or(dictionaries[i]['Last']=='Kaur'): print(dictionaries[i]['First']+" "+dictionaries[i]['Last']) ********************************** please modify the code to use a while loop instead of a for loop

EBK JAVA PROGRAMMING

9th Edition

ISBN:9781337671385

Author:FARRELL

Publisher:FARRELL

Chapter11: Advanced Inheritance Concepts

Section: Chapter Questions

Problem 8RQ

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dictionaries = []dictionaries.append({"First":"Bob", "Last":"Jones", "Color":"red"})dictionaries.append({"First":"Harpreet", "Last":"Kaur", "Color":"green"})dictionaries.append({"First":"Mohamad", "Last":"Argani", "Color":"blue"})for i in range(0, -1,len(dictionaries)):print(dictionaries[i]['First']) ******************** Please modify the code, so the result will show first, last names and colour separated by a single space
dictionaries = [] dictionaries.append({"First":"Bob", "Last":"Jones"}) dictionaries.append({"First":"Harpreet", "Last":"Kaur"}) dictionaries.append({"First":"Mohamad", "Last":"Argani"}) for i in range(0, len(dictionaries)): print(dictionaries[i]['First']) ************************************************* please modity the above code to "Add an ‘if’ statement(s) to the loop in your solution in Exercise 7 to only show the full name if the first name is “Bob” or if the last name is “Kaur”.
Trace through the following program and show the output. Show your work for partial credit. public class Employee { private static int empID = 1111l; private String name , position; double salary; public Employee(String name) { empID ++; this.name 3 пате; } public Employee(Employee obj) { empID = obj.empĪD; пате %3D оbj.naте; position = obj.position; %3D public void empPosition(String empPosition) {position = empPosition;} public void empSalary(double empSalary) { salary = empSalary;} public String toString() { return name + " "+empID + " "+ position +" $"+salary; public void setName(String empName){ name = empName;} public static void main(String args[]) { Employee empOne = new Employee("James Smith"), empTwo; %3D empOne.empPosition("Senior Software Engineer"); етрOпе.етpSalary(1000); System.out.println(empOne); еmpTwo empTwo.empPosition("CEO"); System.out.println(empOne); System.out.println(empTwo); %3D етpОпе empOne ;
Python Data = { "logs": [ { "logType": “xxxxx” }, { "logType": “43435” }, { "logType": “count” }, { "logType": “4452” }, { "logType": “count” }, { "logType": “count” }, { "logType": “43435” }, { "logType": "count" }, { "logType": "count" } ] } So, please count the total of "logType": “count” My desired output: { "count”: 5 }
class Solution(object): def longestCommonPrefix(self, strs): result ="" for i in strs[0]: for j in strs[1]: for k in strs[2]: if i == j and i == k: result+=i if len(result) >0: return result else: return result IndexError: list index out of range for j in strs[1]: Line 5 in longestCommonPrefix (Solution.py) ret = Solution().longestCommonPrefix(param_1) Line 31 in _driver (Solution.py) _driver() Line 41 in <module> (Solution.py) can someone explain why this is wrong?
class Artist{StringsArtist(const std::string& name="",int age=0):name_(name),age_(age){}std::string name()const {return name_;}void set_name(const std::string& name){name_=name;}int age()const {return age_;}void set_age(int age){age_=age;}virtual std::string perform(){return std::string("");}private:int age_;std::string name_;};class Singer : public Artist{public:Singer():Artist(){};Singer(const std::string& name,int age,int hits);int hits() const{return hits_;}void set_hits(int hit);std::string perform();int operator+(const Singer& rhs);int changeme(std::string name,int age,int hits);Singer combine(const Singer& rhs);private:int hits_=0;};int FindOldestandMostSuccess(std::vector<Singer> list); Task: Implement the function int Singer::changeme(std::string name,int age,int hits) : This function should check if the values passed by name, age and hits are different than those stored. And if this is the case it should change the values. This should be done by…
class Artist{StringsArtist(const std::string& name="",int age=0):name_(name),age_(age){}std::string name()const {return name_;}void set_name(const std::string& name){name_=name;}int age()const {return age_;}void set_age(int age){age_=age;}virtual std::string perform(){return std::string("");}private:int age_;std::string name_;};class Singer : public Artist{public:Singer():Artist(){};Singer(const std::string& name,int age,int hits);int hits() const{return hits_;}void set_hits(int hit);std::string perform();int operator+(const Singer& rhs);int changeme(std::string name,int age,int hits);Singer combine(const Singer& rhs);private:int hits_=0;};int FindOldestandMostSuccess(std::vector<Singer> list); Implement the function Singer Singer::combine(const Singer& rhs):It should create a new Singer object, by calling the constructor with the following values: For name it should pass a combination of be the name of the calling object followed by a '+' and then…
Data Structure & Algorithum java program Do the following: 1) Add a constructor to the class "LList" that creates a list from a given array of objects.2) Add a method "addAll" to the "LList" class that adds an array of items to the end of the list. The header of the method is as follows, where "T" is the generic type of the objects in the list. 3) Write a Test/Driver program that thoroughly tests all the methods in the class "LList".
A class object can encapsulate more than one [answer].
Programming Language: C++ I need the codes for arrayListType.h, main.cpp, myString.cpp, myString.h
Doctor -signature:String -doctorID:int - medicine:Arraylist +Doctor(signature:String,doctorID:int) +PrescribeMedicine():void + salary () +checkRecords():void Medicine Pharmacist -medName:String -startTime:int -dosage :int -endTime:int -date_prescribed:int - medicine:Arraylist +Medicine(medName:String,-dosage :int,date_prescribed:int) +Pharmacist (startTime:int,endTime:int) +checkForConflict():double +confirm_prescription():String +getStartTime():int +getEndTime():int +setStartTime(time:int):void +setEndTime(time1:int).void
public class Runner2{ public static void main (String[] args) { StudentRecord myRecord1=new StudentRecord(); //create one object //set values myRecord1.setName("Sai"); myRecord1.setAddresss("VZM"); myRecord1.setAge(22); myRecord1.setMathGrade(97); myRecord1.setScienceGrade(85.7); myRecord1.setEnglishGrade(88); StudentRecord myRecord2=new StudentRecord(); //create another object //set values myRecord2.setName("Pavan"); myRecord2.setAddresss("VSKP"); myRecord2.setAge(20); myRecord2.setMathGrade(89.5); myRecord2.setScienceGrade(80.6); myRecord2.setEnglishGrade(78.9); }} USE ARRAYS FOR THIS