Tracking current during iteration

Question

I have a list of posts. I need to generate it each one by one. For generation of each post, I need to know about other posts. However, I want to make the current post marked.

def generate(posts):
 print(posts) #faking generation
def marked_active_post(orig_posts, active_index):
 posts_view = orig_posts.copy()
 active_post = orig_posts[active_index]
 active_post = orig_posts[active_index].copy()
 active_post['active?'] = True
 posts_view[active_index] = active_post
 return posts_view
posts = [{'foo':'bar'}, {'aaa':'bbb'}]
for i in range(0, len(posts)):
 generate(marked_active_post(posts, i))

How can I do better?

• Stan Kurilin: Created as a more detailed question from codereview.stackexchange.com/questions/63174/…
• Gareth Rees: The loop variable i tells you which post is current. So why not use that?
• Stan Kurilin: @Gareth - I can not do any logic in my template engine (pystache).

Answer 1

It's too bad your template engine doesn't support conditions. What about this?

def marked_active_post(orig_posts, active_index):
 posts_view = orig_posts.deepcopy()
 posts_view[active_index]['active?'] = True
 return posts_view

That means n+1 copies instead of 2, but no need to optimize it until it proves to be slow.

• Stan Kurilin: I don't like this code either. But can not come up with better approach.

Answer 2

It's a very simple point which I'm sure you already know, but I wanted to say it anyway just in case.

You should wrap the main execution like the following:

if __name__ == '__main__':
 posts = [{'foo':'bar'}, {'aaa':'bbb'}]
 for i in range(0, len(posts)):
 generate(marked_active_post(posts, i))

This allows you to import your script into other scripts without anything actually running.

• Stan Kurilin: Thank you for mention it. It could be useful for readers.