Converting a dict to a list + ID

Question

I wrote this code:

PROJECTS_LIST = [
 project if not project.update({"project_id": project_id}) else None for project_id, project in PROJECTS.items()
]

where PROJECTS is a dict. The goal is to convert a dict like {"project123": {"a": "b"}} to [{"project_id": "project123", "a": "b"}]

I worry that this isn't the best approach.

Answer 1

project if not project.update({"project_id": project_id}) else None

The above section is rather odd. I would prefer to one of the following solutions depending on Python version:

• Python 3.9+: use the dictionary merging operator |:

  project | {"project_id": project_id}
  

• Before Python 3.9: The, in my opinion, nicest alternate was to use dictionary unpacking:

  {**project, "project_id": project_id}
  

I'd recommend having a newline before the for, but otherwise the code looks fine.

PROJECTS_LIST = [
 project | {"project_id": project_id}
 for project_id, project in PROJECTS.items()
]

Answer 2

I agree with @Peilonrayz that a "constructive" functional approach such as using the | operator is the way to go for a list comprehension. Comprehensions shouldn't really be about side effects -- that's not how human minds view them, independent of what the machine does.

But let's accept that .update() shall be used, and it will be used by a comprehension. Then instead of the cryptic

... = [
 project if not project.update({"project_id": project_id}) else None
 for project_id, project in PROJECTS.items()
]

I feel this is far more clear:

... = [
 project.update({"project_id": project_id}) or project
 for project_id, project in PROJECTS.items()
]

We have an "A or B" disjunct, where we evaluate A for side effects, and then since it is guaranteed to return None the short-circuiting or operator moves along and returns the (updated) project value.

The original formulation had an awkward else clause in order to satisfy syntax rules.

---

If we're evaluating for side effects, a list comprehension is not a natural way to accomplish that.

Prefer:

for project_id, project in PROJECTS.items():
 project.update({"project_id": project_id})

and subsequently use a comprehension to read out the results.

Answer 3

I worry

A little worry is a good thing because it spurs you into action, like posting a question here on Code Review :)

Approaches to alleviate your worry:

• Add documentation to your code (comments/docstrings) to more clearly specify the PROJECTS input dict
• Add documentation to more clearly specify the PROJECTS_LIST output
• Add tests to check a variety of inputs
• Make sure the tests hit the else clause

I assume the PROJECTS input you showed is a minimal example. Make sure to test more realistic input, with multiple projects and multiple key/values within a project (if that is supported).

Here is another possible input. I split the long line because long lines are hard to understand and maintain.

PROJECTS = {
 "project123": {"a": "b"},
 "project45" : {"c": "d"},
 "project6" : {"e": "f", "this" : "that"}
}
PROJECTS_LIST = [
 project if not project.update({"project_id": project_id}) else None
 for project_id, project in PROJECTS.items()
]
print(PROJECTS_LIST)