# Some Useful Tools/Function for Django

September 17, 2015 · 3 Min Read · In Django, PythonThis post is deprecated 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()
else:
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) +
math.cos(phi1)*math.cos(phi2))
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:
break
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 Stack Overflow answer.

```
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.

**Last updated**: July 2, 2020

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