Did you know that you can apply len python function on your classes? Let's say you have a Lessons class and when a use use len() function on your class it should return number of lessons inside of your class. For our sample let's say we have the below class:

class Lessons(object):
     def __init__(self, lessons=0):
         self.lessons = lessons

     def insert(self, data):
         # let's assume we have save data in database
         self.lessons += 1

     def __len__(self):
         return self.lessons

This is a simple class that sets a fake lesson in your class and insert a new lesson.

std_lessons = Lessons(1)
 print len(std_lessons) # output: 1

 std_lessons.insert({})
 print len(std_lessons) # output: 2

As you can see above we have used len on our class itself. In case you don't implemet __len__ on your class and use len() on it, there will be an AttributeError:

len(std_no_len)
 Traceback (most recent call last):
   File "<input>", line 1, in <module>
 AttributeError: Lessons instance has no attribute '__len__'

__len__ is built-in class and can be implemented in different scenarios. There are tons of built-in functions that you can master and use to improve code quality and prove that you are professional.

#python #class #builtin #len #__len__

One of the benefits of Django`'s `QuerySet is that you can apply filters on it, slice it and move it around your classes without actually hitting the database until you do so.

One of the ways that executes the query is using iteration:

for c in Course.objects.all():
     print(c.title)

NOTE: the first time you iterate over the query, it hits the database and get the result NOT in each iteration.

In the below ways your query will be evaluated (executed):
- repr
- len
- list (e.g.: `list(Entry.objects.all())`)
- bool

#python #django #querySet #query_set #evaluation

# return masked string
def maskify(cc):
  return len(cc[:-4]) * '#' + cc[-4:]

#python #maskify #solution

from itertools import groupby

def duplicate_count(a):
    return len([1 for k, i in groupby(sorted(a.lower())) if len(list(i)) > 1])

#python #solution

from math import pow

def narcissistic(value):
    digits = map(int, list(str(value)))
    power = len(digits)
    final_res = 0
    for digit in digits:
        final_res += pow(digit, power)
    return int(value) == int(final_res)

#python #solution

def narcissistic(value):
    return value == sum(int(x) ** len(str(value)) for x in str(value))

#python #solution

Is there a way to create ObjectID from an INT in MongoDB?

import bson

def object_id_from_int(n):
    s = str(n)
    s = '0' * (24 - len(s)) + s
    return bson.ObjectId(s)

def int_from_object_id(obj):
    return int(str(obj))

n = 12345
obj = object_id_from_int(n)
n = int_from_object_id(obj)
print(repr(obj))  # ObjectId('000000000000000000012345')
print(n)  # 12345

#mongodb #objectid #pymongo #python #bson #int