Hacks by Ruddra

Some Useful Tools/Function for Django

This post is depricated as its based on Django 1.6 mostly.

I am going to share some useful Django tools/functions which are very useful(were for me atleast) to get things done.

Return any model class and its properties

This method will return any model class if you have the name of the class.

from django.db import models

def get_model_description(model_name=None, return_property_list=True):
    for item in models.get_models(include_auto_created=True):
        if item.__name__ == model_name:
            if return_property_list is True:
                return item.get_trigger_properties()
                return item
    return []

For usage, let us think of an example. Let us think, we have a class name ‘X’, we will get the class instance using it like this:

from usefultools import get_model_descriptor

model_x = get_model_descriptor(model_name='X')   #will get class
model_x_objects = get_model_descriptor(model_name='X').objects.all() #will get all the objects of this class

And for its property:

from usefultools import get_model_descriptor

model_x = get_model_descriptor(model_name='X', return_property_list=True)   #will get a list of properties like ['a_property','b_property']

Distance calculator

If you input latitude and longitude of two places, this function will return the distance in between them. Got help from here: http://code.activestate.com/recipes/576779-calculating-distance-between-two-geographic-points/

import math

def distance_calculator(lat1, long1, lat2, long2):

    lat1, long1, lat2, long2 = float(lat1), float(long1), float(lat2), float(long2)

    degrees_to_radians = math.pi/180.0

    phi1 = (90.0 - lat1)*degrees_to_radians
    phi2 = (90.0 - lat2)*degrees_to_radians

    theta1 = long1*degrees_to_radians
    theta2 = long2*degrees_to_radians

    cos = (math.sin(phi1)*math.sin(phi2)*math.cos(theta1 - theta2) +
    arc = math.acos( cos )
    distance = arc*6378.1

    return distance

It will return the distance in KM.

Dynamic relational operations

Suppose we have a sentence like: '5 is greater than 9' and check if its true. We could use eval to dynamically converty string to python but its highly not recommended. So I tried like this:

def calculate_relational_operation(lhs, rhs, operator):
    get_type = type(lhs).__name__
    if get_type == 'str':
        rhs = str(rhs)
    elif get_type == 'float':
        rhs = float(rhs)
    elif get_type == 'int':
        rhs = int(rhs)

    if operator == "==":
        if lhs == rhs:
            return True
        return False
    elif operator == "!=":
        if lhs != rhs:
                return True
        return False
    elif operator == ">":
        if lhs > rhs:
                return True
        return False
    elif operator == "<":
        if lhs < rhs:
                return True
        return False
    elif operator == ">=":
        if lhs >= rhs:
            return True
        return False
    elif operator == "<=":
        if lhs == rhs:
                return True
        return False
    elif operator == "Is":
        if lhs is rhs:
            return True
        return False
    return False

It will return True or False depending on the statement/input.

Get week list

It will return all the weeks list from last 1 year (extendable).

from isoweek import Week

def generate_week():
    max_week = datetime.datetime.combine(Week.thisweek().thursday(), datetime.time(0,0))
    min_week = max_week - datetime.timedelta(days=365)
    _weeks = list()
    while True:
        _weeks.append('Week'+str(max_week.isocalendar()[1])+ ' ' +str(max_week.isocalendar()[0])))
        max_week -= datetime.timedelta(days=7)
        if max_week <= min_week:

    return _weeks

#Output>> ['Week2 2015', 'Week1 2015', 'Week52 2014' ....]

Get month list

It will return last 12 month’s year and month number. Constructed using this SO answer: http://stackoverflow.com/a/6576603/2696165

x = 12
now = time.localtime()
print([time.localtime(time.mktime((now.tm_year, now.tm_mon - n, 1, 0, 0, 0, 0, 0, 0)))[:2] for n in range(x)])

#Output>> [(2015, 2), (2015, 1), (2014, 12), (2014, 11), (2014, 10), (2014, 9), (2014, 8), (2014, 7), (2014, 6), (2014, 5), (2014, 4), (2014, 3)]

Thats it. Thanks for reading.

Topics: Django , Python

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