How to compact this for?

Question

I have a dictionary of dictionaries. I want to count how many of those dictionaries has element "status" set to "connecting".

This is my working code:

connecting = 0
for x in self.servers:
    if self.servers[x]["status"] == "connecting": connecting += 1

Is there any way of compacting this? I was thinking something like:

connecting = [1 if self.servers[x]["status"] == "closed" else 0 for x in self.servers]

But it just returns a list of 0 and 1, doesn't add that 1 to connecting, which is what I expected.

Answer 1

You can use a generator expression within sum function :

sum(x["status"]=="connecting" for x in self.servers.values()) 

Note that since the result of x["status"]=="connecting" is a boolean value and if it be True python will evaluated it as 1, so in the end it will returns the number of dictionaries that follows your conditions.

Answer 2

You can use sum for this. You have the list comprehension already, so just wrap it in sum.

connecting = sum(1 for x in self.servers if self.servers[x]["status"] == "closed")