Mean Reputation Change

Question

Challenge

Given a the ID for a PPCG user who has existed for at least a month, output the average amount of reputation they gain per month. You can do this by dividing their reputation by the time since their account was created (which you can find on the profile). This should be returned floored; that is, truncate all of the decimals. (We don't care about the small stuff).

Input

A single integer, which is the user's ID

Output

A single integer, representing the floored rep change per month.

Test Cases

input -> output
42649 -> 127 (me)
12012 -> 3475 (Dennis)
8478  -> 3015 (Martin Ender)

Answer 1

JS, ES6 243 235 168 166 165 164 bytes

i=>fetch(`//api.stackexchange.com/users/${i}?site=codegolf`).then(x=>x.json()).then(x=>alert(~~((y=x.items[0]).reputation/((new Date/1e3-y.creation_date)/2592e3))))

Based on the Python answer.

Run this in the console:

(i=>fetch(`//api.stackexchange.com/users/${i}?site=codegolf`).then(x=>x.json()).then(x=>alert(~~((y=x.items[0]).reputation/((new Date/1e3-y.creation_date)/2592e3)))))(+prompt("Enter a userId"))

-4 thanks to @insert_name_here

Old answer:

z=>fetch("//codegolf.stackexchange.com/u/"+z).then(x=>x.text()).then(x=>alert((/([\d,]+) <span class="label-uppercase">repu/.exec(x)[1].replace(/,/g,"")/(((y=/Member for <.+>((\d+) years?, )?(\d+) month/.exec(x))[2]||0)*12+ +y[3]||0))))

Improvements welcome! Just fetches the profile page and parses it with regular expressions!

Won't work as a stack snippet, but you can paste this in your console:

(z=>fetch("//codegolf.stackexchange.com/u/"+z).then(x=>x.text()).then(x=>alert(Math.floor(/([\d,]+) <span class="label-uppercase">repu/.exec(x)[1].replace(/,/g,"")/(((y=/Member for <.+>((\d+) years?, )?(\d+) month/.exec(x))[2]||0)*12+ +y[3]||0)))))(+prompt("Enter a userId"))

Answer 2

Python 3 2, 206 219 201 193 bytes

import requests,time
a=requests.get('http://api.stackexchange.com/users/'+str(input())+'?site=codegolf').json()['items'][0]
print a['reputation']/((int(time.time())-a['creation_date'])/2592000)

I am estimating the months as the OP suggested, but actually print(int(b/((d-c)/2592000))) is a more accurate solution than print(int(b/int((d-c)/2592000))) and also saves a few bytes.