How to Query model where name contains any word in python list?

Question

Aim to Achieve:

I want all objects where name attribute contains any word from the list.

I have:

list = ['word1','word2','word3']
ob_list = data.objects.filter( // What to write here ?  )
// or any other way to get the objects where any word in list is contained, in 
// the na-me attribute of data.

For example:

if name="this is word2": Then object with such a name should be returned since word2 is in the list.

Please help!

Solution 1

You could use Q objects to constuct a query like this:

from django.db.models import Q

ob_list = data.objects.filter(reduce(lambda x, y: x | y, [Q(name__contains=word) for word in list]))

Edit:

reduce(lambda x, y: x | y, [Q(name__contains=word) for word in list]))

is a fancy way to write

Q(name__contains=list[0]) | Q(name__contains=list[1]) | ... | Q(name__contains=list[-1])

You could also use an explicit for loop to construct the Q object.

Solution 2

ob_list = data.objects.filter(name__in=my_list)

And BTW, avoid using the variable name "list" (Or any other python standard keyword), lest you get into some weird bugs later.

Update: (I guess your question was updated too, because when I wrote the answer, I didn't see the part where you wrote you need a contains match and not an exact match)

You can do that using the regex search too, avoiding many Q expressions (which end up using that many where "and" clauses in the SQL, possibly dampening the performance), as follows:

data.objects.filter(name__regex=r'(word1|word2|word3)')

Solution 3

obj_list = [obj for obj in data.objects.all() if any(name in obj.name for name in list)]

Edit: Just re-read your question. Don't know if you can do that with filter but you can do it with a list comprehension or generator expression.

Solution 4

For anyone comparing Arrays, you could use Django's Overlap filter to achieve this.

From the docs:

Returns objects where the data shares any results with the values passed. Uses the SQL operator &&.

So, you would simply write:

ob_list = data.objects.filter(name__overlap=my_list)

View more solutions

Comments

[Yugal Jindle]

Name should not be in the list, but any word in the list that is contained in the name.

[agf]

I already edited my answer -- it now does what you want (in the idiomatic Python way, non-Django specific).

[Yugal Jindle]

can you explaing.. reduce and lambda...? Your solution is working fine.

[agf]

I'm not the one who posted this, but lambda is just a way of defining a function right there instead of defining it elsewhere, and reduce repeatedly applies an operation repeatedly on a list: reduce(lambda x, y: x+y, [1, 2, 3, 4, 5]) does ((((1+2)+3)+4)+5). The reduction by bitwise OR over the list of Qs is the same as the any over the list in my answer.

[Ismail Badawi]

@agf The difference is here the OR is performed at the database level instead of fetching all the records and filtering them in python.

[agf]

Really? Django is able to translate the reduce into SQL? I think that part is happening in Python inside the ORM. I still wouldn't be suprised if it's faster though. The in is happening in the database instead of Python, but the OR?

[Ismail Badawi]

@agf Sorry, by the OR I meant the actual filtering (i.e. the WHERE clause has name LIKE ... OR name LIKE ... and so on). The reduce only constructs the Q object, which doesn't hit the database.

[Ryder Brooks]

for my purposes the regex approach was much faster than using Q expressions. And made for a much more readable query. Thank you Lakshman Prasad

[primoz]

Note: If you're using Python > 3, you must import reduce with from functools import reduce

[tr0yspradling]

This only works with PostgreSQL.