class User(models.Model):
email = models.EmailField()
name = models.CharField()
How to get email and name of User as a list of tuples? My current solution is this:
result = []
for user in User.objects.all():
result.append((user.email, user.name))
If this is an django ORM queryset (or result of it), you can just use values_list
method instead of values
. That will give exactly what you want.
Assuming queryset
is supposed to look like this, with 'x'
and 'y'
as string keys:
>>> queryset = [{'x':'1', 'y':'a'}, {'x':'2', 'y':'b'}]
>>> result = [(q['x'], q['y']) for q in queryset]
>>> result
[('1', 'a'), ('2', 'b')]
>>> # or if x and y are actually the correct names/vars for the keys
... result = [(q[x], q[y]) for q in queryset]
If you can have multiple keys and you just want certain key values, you can use itemgetter with map passing the keys you want to extract:
from operator import itemgetter
result = list(map(itemgetter("x", "y"), queryset)))
your_tuple = [(x.get('attrA'), x.get('attrB')) for x in queryset.values()]
You can use dict.values()
:
queryset = [{x:'1',y:'a'}, {x:'2',y:'b'}]
result = []
for i in queryset:
result.append(tuple(i.values()))
Or in one line:
result = [tuple(i.values()) for i in queryset]
If you want them in a particular order:
result = [(i[x], i[y]) for i in queryset]
Use list comprehension and dict.values()
>>> queryset = [{'x': '1', 'y': 'a'}, {'x': '2', 'y': 'b'}]
>>> result = [tuple(v.values()) for v in queryset]
>>> result
[('1', 'a'), ('2', 'b')]
UPDATE
as @aneroid reasonably mentioned, since dict
object are not ordered the code snippet could return different order in tuple
So since i don't want to add duplicate solution. There is one option, not so elegant, and maybe with a lack of efficiency, to use OrderedDict
>>> from collections import OrderedDict
>>> queryset = [{'x': '1', 'y': 'a'}, {'x': '2', 'y': 'b'}]
>>> order = ('x', 'y')
>>> result = [tuple(OrderedDict((k, v[k]) for k in myorder).values()) for v in queryset]
>>> result
[('1', 'a'), ('2', 'b')]
But i personally think that @PadraicCunningham 's solution is the most elegant here.
为此使用values_list :
result = User.objects.all().values_list("email", "name")
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