I have 3 tables flotation has many lightresidues has many compositions.
I'm wanting to make a view for each flotation which can access a list of light residues and a list of associated compositions. There are only ever 2-3 light residues to each flotation and the same for compositions, so manageable.
I can get a flotation record and its referenced light residues, but I'm having trouble passing the lightresidue_id to get the compositions. [n.b. I know lightresidue.id is the Django way of doing things, but I opt for this way]
The view code is as follows, I've hard coded for lightresidue.lightresidue_id = 17 which works, but how do I substitute this for the lightresidue.lightresidue_id = composition.lightresidue_id.
def botanyoverview(request, flotation_id):
flotation = get_object_or_404(Flotation, pk=flotation_id)
lightresidue = LightResidue.objects.filter(flotation_id__flotation_id=flotation_id)
# composition = Composition.objects.filter(lightresidue.lightresidue_id)
composition = Composition.objects.filter(lightresidue_id=17)
return render(request, 'dashboard/botanyoverview.html',
{
'flotation':flotation,
'lightresidue':lightresidue,
'composition':composition,
})
You could do this by chaining the compositions of all the lightresidue using itertools.
from itertools import chain
def botanyoverview(request, flotation_id):
flotation = get_object_or_404(Flotation, pk=flotation_id)
lightresidue = LightResidue.objects.filter(flotation_id__flotation_id=flotation_id)
queryset = []
for i in lightresidue:
queryset += Composition.objects.filter(lightresidue_id = i.lightresidue_id)
composition = chain.from_iterable(queryset)
return render(request, 'dashboard/botanyoverview.html',
{
'flotation':flotation,
'lightresidue':lightresidue,
'composition':composition,
})
the lightresidue may contain one or more objects and so here i am getting all the compositions associated with each lightresidue sepearately and combining them using itertools.
Related
i am new to django , would u experts pls kindly help me..
so i have two modules, and in CBV's get_context_data they both return total sum, now I want to add both the totals and display it on my HTML page, I honestly tried many things but I am always getting an error.
here is my views.py
def get_context_data(self, **kwargs):
# Call the base implementation first to get a context
context = super().get_context_data(**kwargs)
today = datetime.datetime.now()
# Order.objects.filter(created_at__year=today.year, created_at__month=today.month)
context['expense1'] = Tracker.objects.filter(user=self.request.user)
context['Total1'] =(Body.objects.filter(user=self.request.user, pub_date__year=today.year, pub_date__month=today.month).aggregate(total=Sum(F('price') * F('quantity'))),
Sport.objects.filter(user=self.request.user, pub_date__year=today.year, pub_date__month=today.month).aggregate(total=Sum(F('price') * F('quantity'))))
return context
so what I want is , total of body + total of sports , which is being asgn to the context "total1" ,and then I want to show it in HTML
my HTML file
this is how I am displaying the total,
Total: {{Total1}}
Your oneliner code is a bit confusing but I think I get where you made the mistake - the aggregate returns a dict with the key in the argument as the return value. See below:
body_total = Body.objects.filter(
user=self.request.user, pub_date__year=today.year, pub_date__month=today.month
).aggregate(total=Sum(F('price') * F('quantity')))['total']
sport_total = Sport.objects.filter(
user=self.request.user, pub_date__year=today.year, pub_date__month=today.month
).aggregate(total=Sum(F('price') * F('quantity')))['total']
# Note: this is a tuple, not a sum of the two, so check what you need :)
context['Total1'] = (body_total, sport_total)
# For total sum of two totals, just add them of course.
# It seems like this is what you need from the post, and not the
# line above.
context['Total1'] = body_total + sport_total
PS. the code is not tested. I'm a bit unsure if the F expressions work fine (probably yes). Let me know if there are any issues.
In our Django project, there is a view which creates multiple objects (from 5 to even 100). The problem is that creating phase takes a very long time.
Don't know why is that so but I suppose that it could be because on n objects, there are n database lookups and commits.
For example 24 objects takes 67 seconds.
I want to speed up this process.
There are two things I think may be worth to consider:
To create these objects in one query so only one commit is executed.
Create a ThreadPool and create these objects parallel.
This is a part of the view which causes problems (We use Postgres on localhost so connection is not a problem)
#require_POST
#login_required
def heu_import(request):
...
...
product = Product.objects.create(user=request.user,name=name,manufacturer=manufacturer,category=category)
product.groups.add(*groups)
occurences = []
counter = len(urls_xpaths)
print 'Occurences creating'
start = datetime.now()
eur_currency = Currency.objects.get(shortcut='eur')
for url_xpath in urls_xpaths:
counter-=1
print counter
url = url_xpath[1]
xpath = url_xpath[0]
occ = Occurence.objects.create(product=product,url=url,xpath=xpath,active=True if xpath else False,currency=eur_currency)
occurences.append(occ)
print 'End'
print datetime.now()-start
...
return render(request,'main_app/dashboard/new-product.html',context)
Output:
Occurences creating
24
.
.
.
0
End
0:01:07.727000
EDIT:
I tried to put the for loop into the with transaction.atomic(): block but it seems to help only a bit (47 seconds instead of 67).
EDIT2:
I'm not sure but it seems that SQL queries are not a problem:
Please use bulk_create for inserting multiple objects.
occurences = []
for url_xpath in urls_xpaths:
counter-=1
print counter
url = url_xpath[1]
xpath = url_xpath[0]
occurances.append(Occurence(product=product,url=url,xpath=xpath,active=True if xpath else False,currency=eur_currency))
Occurence.objects.bulk_create(occurences)
A lot of websites will display:
"1.8K pages" instead of "1,830 pages"
or
"43.2M pages" instead of "43,200,123 pages"
Is there a way to do this in Django?
For example, the following code will generate the quantified amount of objects in the queryset (i.e. 3,123):
Books.objects.all().count()
Is there a way to add a custom count filter to return "3.1K pages" instead of "3,123 pages?
Thank you in advance!
First off, I wouldn't do anything that alters the way the ORM portion of Django works. There are two places this could be done, if you are only planning on using it in one place - do it on the frontend. With that said, there are many ways to achieve this result. Just to spout off a few ideas, you could write a property on your model that calls count then converts that to something a little more human readable for the back end. If you want to do it on the frontend you might want to find a JavaScript lib that could do the conversion.
I will edit this later from my computer and add an example of the property.
Edit: To answer your comment, the easier one to implement depends on your skills in python vs in JavaScript. I prefer python so I would probably do it in there somewhere on the model.
Edit2: I have wrote an example to show you how I would do a classmethod on a base model or on the model that you need these numbers on. I found a python package called humanize and I took its function that converts these to readable and modified it a bit to allow for thousands and took out some of the super large number conversion.
def readable_number(value, short=False):
# Modified from the package `humanize` on pypy.
powers = [10 ** x for x in (3, 6, 9, 12, 15, 18)]
human_powers = ('thousand', 'million', 'billion', 'trillion', 'quadrillion')
human_powers_short = ('K', 'M', 'B', 'T', 'QD')
try:
value = int(value)
except (TypeError, ValueError):
return value
if value < powers[0]:
return str(value)
for ordinal, power in enumerate(powers[1:], 1):
if value < power:
chopped = value / float(powers[ordinal - 1])
chopped = format(chopped, '.1f')
if not short:
return '{} {}'.format(chopped, human_powers[ordinal - 1])
return '{}{}'.format(chopped, human_powers_short[ordinal - 1])
class MyModel(models.Model):
#classmethod
def readable_count(cls, short=True):
count = cls.objects.all().count()
return readable_number(count, short=short)
print(readable_number(62220, True)) # Returns '62.2K'
print(readable_number(6555500)) # Returns '6.6 million'
I would stick that readable_number in some sort of utils and just import it in your models file. Once you have that, you can just stick that string wherever you would like on your frontend.
You would use MyModel.readable_count() to get that value. If you want it under MyModel.objects.readable_count() you will need to make a custom object manager for your model, but that is a bit more advanced.
I'm working with the program Autodesk Maya.
I've made a naming convention script that will name each item in a certain convention accordingly. However I have it list every time in the scene, then check if the chosen name matches any current name in the scene, and then I have it rename it and recheck once more through the scene if there is a duplicate.
However, when i run the code, it can take as long as 30 seconds to a minute or more to run through it all. At first I had no idea what was making my code run slow, as it worked fine in a relatively low scene amount. But then when i put print statements in the check scene code, i saw that it was taking a long time to check through all the items in the scene, and check for duplicates.
The ls() command provides a unicode list of all the items in the scene. These items can be relatively large, up to a thousand or more if the scene has even a moderate amount of items, a normal scene would be several times larger than the testing scene i have at the moment (which has about 794 items in this list).
Is this supposed to take this long? Is the method i'm using to compare things inefficient? I'm not sure what to do here, the code is taking an excessive amount of time, i'm also wondering if it could be anything else in the code, but this seems like it might be it.
Here is some code below.
class Name(object):
"""A naming convention class that runs passed arguments through user
dictionary, and returns formatted string of users input naming convention.
"""
def __init__(self, user_conv):
self.user_conv = user_conv
# an example of a user convention is '${prefix}_${name}_${side}_${objtype}'
#staticmethod
def abbrev_lib(word):
# a dictionary of abbreviated words is here, takes in a string
# and returns an abbreviated string, if not found return given string
#staticmethod
def check_scene(name):
"""Checks entire scene for same name. If duplicate exists,
Keyword Arguments:
name -- (string) name of object to be checked
"""
scene = ls()
match = [x for x in scene if isinstance(x, collections.Iterable)
and (name in x)]
if not match:
return name
else:
return ''
def convert(self, prefix, name, side, objtype):
"""Converts given information about object into user specified convention.
Keyword Arguments:
prefix -- what is prefixed before the name
name -- name of the object or node
side -- what side the object is on, example 'left' or 'right'
obj_type -- the type of the object, example 'joint' or 'multiplyDivide'
"""
prefix = self.abbrev_lib(prefix)
name = self.abbrev_lib(name)
side = ''.join([self.abbrev_lib(x) for x in side])
objtype = self.abbrev_lib(objtype)
i = 02
checked = ''
subs = {'prefix': prefix, 'name': name, 'side':
side, 'objtype': objtype}
while self.checked == '':
newname = Template (self.user_conv.lower())
newname = newname.safe_substitute(**subs)
newname = newname.strip('_')
newname = newname.replace('__', '_')
checked = self.check_scene(newname)
if checked == '' and i < 100:
subs['objtype'] = '%s%s' %(objtype, i)
i+=1
else:
break
return checked
are you running this many times? You are potentially trolling a list of several hundred or a few thousand items for each iteration inside while self.checked =='', which would be a likely culprit. FWIW prints are also very slow in Maya, especially if you're printing a long list - so doing that many times will definitely be slow no matter what.
I'd try a couple of things to speed this up:
limit your searches to one type at a time - why troll through hundreds of random nodes if you only care about MultiplyDivide right now?
Use a set or a dictionary to search, rather than a list - sets and dictionaries use hashsets and are faster for lookups
If you're worried about maintining a naming convetion, definitely design it to be resistant to Maya's default behavior which is to append numeric suffixes to keep names unique. Any naming convention which doesn't support this will be a pain in the butt for all time, because you can't prevent Maya from doing this in the ordinary course of business. On the other hand if you use that for differntiating instances you don't need to do any uniquification at all - just use rename() on the object and capture the result. The weakness there is that Maya won't rename for global uniqueness, only local - so if you want to make unique node name for things that are not siblings you have to do it yourself.
Here's some cheapie code for finding unique node names:
def get_unique_scene_names (*nodeTypes):
if not nodeTypes:
nodeTypes = ('transform',)
results = {}
for longname in cmds.ls(type = nodeTypes, l=True):
shortname = longname.rpartition("|")[-1]
if not shortname in results:
results[shortname] = set()
results[shortname].add(longname)
return results
def add_unique_name(item, node_dict):
shortname = item.rpartition("|")[-1]
if shortname in node_dict:
node_dict[shortname].add(item)
else:
node_dict[shortname] = set([item])
def remove_unique_name(item, node_dict):
shortname = item.rpartition("|")[-1]
existing = node_dict.get(shortname, [])
if item in existing:
existing.remove(item)
def apply_convention(node, new_name, node_dict):
if not new_name in node_dict:
renamed_item = cmds.ls(cmds.rename(node, new_name), l=True)[0]
remove_unique_name(node, node_dict)
add_unique_name ( renamed_item, node_dict)
return renamed_item
else:
for n in range(99999):
possible_name = new_name + str(n + 1)
if not possible_name in node_dict:
renamed_item = cmds.ls(cmds.rename(node, possible_name), l=True)[0]
add_unique_name(renamed_item, node_dict)
return renamed_item
raise RuntimeError, "Too many duplicate names"
To use it on a particular node type, you just supply the right would-be name when calling apply_convention(). This would rename all the joints in the scene (naively!) to 'jnt_X' while keeping the suffixes unique. You'd do something smarter than that, like your original code did - this just makes sure that leaves are unique:
joint_names= get_unique_scene_names('joint')
existing = cmds.ls( type='joint', l = True)
existing .sort()
existing .reverse()
# do this to make sure it works from leaves backwards!
for item in existing :
apply_convention(item, 'jnt_', joint_names)
# check the uniqueness constraint by looking for how many items share a short name in the dict:
for d in joint_names:
print d, len (joint_names[d])
But, like i said, plan for those damn numeric suffixes, maya makes them all the time without asking for permission so you can't fight em :(
Instead of running ls for each and every name, you should run it once and store that result into a set (an unordered list - slightly faster). Then check against that when you run check_scene
def check_scene(self, name):
"""Checks entire scene for same name. If duplicate exists,
Keyword Arguments:
name -- (string) name of object to be checked
"""
if not hasattr(self, 'scene'):
self.scene = set(ls())
if name not in self.scene:
return name
else:
return ''
I want to concatenate two queryset obtained from two different models and i can do it using itertools like this:
ci = ContributorImage.objects.all()
pf = Portfolio.objects.all()
cpf = itertools.chain(ci,pf)
But the real fix is paginating results.If i pass a iterator(cpf, or our concatenated queryset) to Paginator function, p = Paginator(cpf, 10), it works as well but fails at retrieving first page page1 = p.page(1) with an error which says:
TypeError: object of type 'itertools.chain' has no len()
What can i do in case like this ?
The itertools.chain() will return a generator. The Paginator class needs an object implementing __len__ (generators, of course do not support it since the size of the collection is not known).
Your problem could be resolved in a number of ways (including using list to evaluate the generator as you mention) however I recommending taking a look at the QuerySetChain mentioned in this answer:
https://stackoverflow.com/a/432666/119071
I think it fits exactly to your problem. Also take a look at the comments of that answer - they are really enlightening :)
I know it's too late, but because I encountered this error, I would answer to this question.
you should return a list of objects:
ci = ContributorImage.objects.all()
pf = Portfolio.objects.all()
cpf = itertools.chain(ci,pf)
cpf_list = list(cpf)