django orm查询性能

发布于 2025-02-08 12:29:44 字数 1555 浏览 0 评论 0 原文

我试图为我的Django应用找到最好的Django查询。我使用默认的SQLite DB作为后端。我正在使用 timeit 找到查询时间。

>>> import timeit
>>>
>>> setup_arg = 'from .models import AwsConsoleAccess'
>>> stmt1 = """AwsConsoleAccess.objects.filter(request_id='8548a2d5-4bb7-4fa9-add2-a41219dc8785')"""
>>> stmt2 = """AwsConsoleAccess.objects.filter(request_id='8548a2d5-4bb7-4fa9-add2-a41219dc8785').values('status')"""
>>>
>>>
>>> print(timeit.timeit(setup=setup_arg, stmt=stmt1, number=100) * 1000, 'ms')
7.873513997765258 ms
>>> print(timeit.timeit(setup=setup_arg, stmt=stmt2, number=100) * 1000, 'ms')
11.816384001576807 ms

执行SQL查询：


>>> print(AwsConsoleAccess.objects.filter(request_id='8548a2d5-4bb7-4fa9-add2-a41219dc8785').query)
SELECT "aws_console_awsconsoleaccess"."created",.....(all fields) FROM "aws_console_awsconsoleaccess" WHERE "aws_console_awsconsoleaccess"."request_id" = 8548a2d54bb74fa9add2a41219dc8785
>>> print(AwsConsoleAccess.objects.filter(request_id='8548a2d5-4bb7-4fa9-add2-a41219dc8785').values('status').query)
SELECT "aws_console_awsconsoleaccess"."status" FROM "aws_console_awsconsoleaccess" WHERE "aws_console_awsconsoleaccess"."request_id" = 8548a2d54bb74fa9add2a41219dc8785

从理论上从表选择字段应比 SELECT *从表中选择 *，但我的结果没有反映它。

我是否正确进行了测试？
还有其他方法可以测试Django Orm中两个查询之间的性能吗？

原文

I am trying to find best the django query for my Django app. I am using default Sqlite DB as backend. I am using timeit to find time taken for query.

>>> import timeit
>>>
>>> setup_arg = 'from .models import AwsConsoleAccess'
>>> stmt1 = """AwsConsoleAccess.objects.filter(request_id='8548a2d5-4bb7-4fa9-add2-a41219dc8785')"""
>>> stmt2 = """AwsConsoleAccess.objects.filter(request_id='8548a2d5-4bb7-4fa9-add2-a41219dc8785').values('status')"""
>>>
>>>
>>> print(timeit.timeit(setup=setup_arg, stmt=stmt1, number=100) * 1000, 'ms')
7.873513997765258 ms
>>> print(timeit.timeit(setup=setup_arg, stmt=stmt2, number=100) * 1000, 'ms')
11.816384001576807 ms

SQL query executed:


>>> print(AwsConsoleAccess.objects.filter(request_id='8548a2d5-4bb7-4fa9-add2-a41219dc8785').query)
SELECT "aws_console_awsconsoleaccess"."created",.....(all fields) FROM "aws_console_awsconsoleaccess" WHERE "aws_console_awsconsoleaccess"."request_id" = 8548a2d54bb74fa9add2a41219dc8785
>>> print(AwsConsoleAccess.objects.filter(request_id='8548a2d5-4bb7-4fa9-add2-a41219dc8785').values('status').query)
SELECT "aws_console_awsconsoleaccess"."status" FROM "aws_console_awsconsoleaccess" WHERE "aws_console_awsconsoleaccess"."request_id" = 8548a2d54bb74fa9add2a41219dc8785

In theory select field from table should run faster than select * from table but my result are not reflecting it.

Did I do my testing properly?
Is there any other way to test the performance between two queries in Django ORM?

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日记撕了你也走了 2025-02-15 12:29:44

您没有考虑到，即创建QuerySet时，没有对数据库进行实际查询。仅当实际需要查询结果时，查询才会运行。

因此，当您编写 model.objects.filter（...）时，所有的工作都是创建查询。取而代之的是，您应该通过将其转换为列表来强制对QuerySet的评估。这实际上将运行查询并为您提供更准确的结果：

list(Model.objects.filter(...))

You have not considered the fact that QuerySets are lazy, that is when you create a queryset no actual query is made to the database. The query will run only when the results of the query are actually needed.

So when you write Model.objects.filter(...) all that does is create a query. You should instead force the evaluation of the queryset, for example by converting it to a list. This will actually run the query and give you more accurate results: