xor solution is n. Most efficient

Good one! You also are good writing without "!=0" because boolean for "0" would be "False" anyway, and all others would be "True!
I used set, but actually this one is shorter, maybe it's more expensive if talking about productivity...
Thank you!

Pleasant, simple task! I like it)

It is a generator btw.

The examples show one with a list of string values, e.g. count_if(["John","Steve","John"], "John") but it looks like the tests only give you numbers

Edit: see my recent solution here which should fail, if it were given the example tests in the description, but instead passes

approximate equality is now used for average_if()

added here

approved

he sad you crate the fun to do this work not use a fun or method from string module

bruh

python new test framework + random tests are required. updated in this fork

Already raised as issue

I'm confused as to why it knows c represents characters in the strings