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Is there a way to linking two data values, simpler than Dictionary?

By : Marcea Baker
Date : August 01 2020, 01:00 AM
help you fix your problem You can simplify Dictionary hasSpell into HashSet (you want only Key, not Value from the initial Dictionary):
code :
  public enum Spell { 


  HashSet<Spell> abilities = new HashSet<Spell>() { 


  // If the hero can resurrect...
  if (abilities.Contains(Spell.Resurrect)) {
    // ...let him raises from the dead

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Is there a simpler way to reorder data by the values of a column?

By : user3280421
Date : March 29 2020, 07:55 AM
hop of those help? I'm not into the smiting business for well-formed questions. And I thought the code was readable and sensible. If you wanted to tighten it up a bit you can drop the paste() operation by using "[[" and creating the index inside "[":
code :
ReorderDataByColumn2 <- function(x, column) {
    return(x[ order( x[[column]]), ])
 ReorderDataByColumn2 <- function(x, column, desc=FALSE) {
      x[ do.call( order, x[ , column, drop=FALSE ]  ), ]
      ) }
ReorderDataByColumn2 <- function(x, column) {
    if(column %in% names(x)){return(x[ order( x[[column]]), ]) 
     }else{ cat("Column ", column, "not in dataframe ", deparse(substitute(x))) }

Simpler way to traverse complex dictionary in python?

By : user3233371
Date : March 29 2020, 07:55 AM
I wish this helpful for you You ask "any obvious inefficiencies in my code" -- the answer is yes, specifically where you're looping over dictionaries (thus getting all their keys sequentially, which is O(N), i.e, takes a time proportional to the number of keys in the dictionary) rather than just using them as dictionaries (which takes time O(1), i.e, constant time -- fast too).
So for example where you have
code :
for key2 in key:
    if key2 == "values":
       ...use key.get(key2)...
    if key2 == "name":
       ...use key.get(key2)...
if 'values' in key:
   ...use key['values']...
if 'name' in key:
   ...use key['name']...
values = key.get('values')
if values is not None:
    ...use values...
name = key.get('name')
if name is not None:
    ...use name...

How to use dictionary for simpler invocation of attributes

By : Ron Oh
Date : March 29 2020, 07:55 AM
around this issue Just define the details of the classes Student and GraduateStudent as attributes. Each instance can have a unique id by defining a class attribute studentid:
code :
class Student(object):   
    studentid = -1 
    def __init__(self, name, dob):
        Student.studentid += 1
        self.id = Student.studentid
        self.details = dict()
        self.name = name
        # Date of birth more useful than age :)
        self.dob = dob
        self.address = ""

    def add_address(self, street, pin):
        self.address = Address(street, pin)

    def get_address(self):
        return self.address

    def __str__(self):
        return "{}: Name {}, Address: {}".format(self.id, self.name, self.address)

class GraduateStudent(Student):
    def __init__(self, name, dob, course):
        super(GraduateStudent, self).__init__(name, dob)
        self.course = course
    def __str__(self):
        return "{}: Name: {}, Course: {}, Address: {}".format(self.id, self.name, self.course, self.address)
class Directory(object):
    def __init__(self):
        self.student_dict = dict()

    def add_student(self, student):
        if student.id in self.student_dict:
            print "Student %d already exists" % student.id
        self.student_dict[student.id] = student
        return student.id

    def get_student(self, id):
            return self.student_dict[id]
        except KeyError:
            print "<No Student record found: %d>" % id
            return None

    def add_address(self, id, street, pin):
        student = self.get_student(id)
        if student is not None:
            student.add_address(street, pin)

    def get_address(self, id):
        student = self.get_student(id)
        if student is not None:
            return student.get_address()

    def print_students(self):
        for student in self.student_dict.values():
            print student
>>> d = Directory()
>>> id1 = d.add_student(GraduateStudent('Swad', 1999, 'Computer Science'))
>>> id2 = d.add_student(GraduateStudent('Vish', 1998, 'MECH.'))
>>> id3 = d.add_student(GraduateStudent('Vino', 1998,  'MECH.'))
>>> d.add_address(id1, "11 vvk street", 600015)
>>> d.add_address(id2, "22 vvk street", 600015)
>>> d.add_address(id3, "33 vvk street", 600015)
>>> d.print_students()
0: Name: Swad, Course: Computer Science, Address: Address: 11 vvk street 600015
1: Name: Vish, Course: MECH., Address: Address: 22 vvk street 600015
2: Name: Vino, Course: MECH., Address: Address: 33 vvk street 600015

linking between tables in abap data dictionary

By : J.Doe
Date : March 29 2020, 07:55 AM
Any of those help They are equal, because they have the same domain.
Go to se11 -> lips -> vgbel -> column data elemnt vgbel If you are doubleklicking, you see, that the domain is vbeln

is there any simpler coding in python for encoding and decoding using dictionary data structure

By : bdk0172
Date : March 29 2020, 07:55 AM
this one helps. You can use str.translate in conjunction with str.maketrans:
code :
qwerty_encrypt = {'a': 'q', 'b': 'w', 'c': 'e', 'd': 'r', 'e': 't', 'f': 'y',
                  'g': 'u', 'h': 'i', 'i': 'o', 'j': 'p', 'k': 'a', 'l': 's',
                  'm': 'd', 'n': 'f', 'o': 'g', 'p': 'h', 'q': 'j', 'r': 'k',
                  's': 'l', 't': 'z', 'u': 'x', 'v': 'c', 'w': 'v', 'x': 'b',
                  'y': 'n', 'z': 'm', '1': '9', '2': '8', '3': '7', '4': '6',
                  '5': '5', '6': '4', '7': '3', '8': '2', '9': '1', '0': '0'}

qwerty_decrypt = {value: key for key, value in qwerty_encrypt.items()}
assert len(qwerty_decrypt) == len(qwerty_encrypt)

table_encrypt = str.maketrans(qwerty_encrypt)
table_decrypt = str.maketrans(qwerty_decrypt)

msg = 'Hello Stack Overflow'.lower()

print('After encryption:', msg.translate(table_encrypt))
print('After decryption:',
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