Question

Python

Ask a user for their address. Based on their address, tell them what days they should bring an umbrella (>=50% chance of rain they should bring an umbrella).

Output example
Enter address: 15 Lisa Ct. Sewell NJ 08080

04/13/18 - Friday - Sun

04/14/18 - Saturday - Sun

04/15/18 - Sunday - Umbrella

04/16/18 - Monday - Sun

04/17/18 - Tuesday - Umbrella

04/18/18 - Wednesday - Sun

04/19/18 - Thursday - Umbrella

04/20/18 - Friday - Sun

Weather for another address? (y/n)

import requestsresponse = requests.get('https://www.google.com')
if response.status_code == 200:
    print('yay!')
else:
    print('oops!')

Use this website to obtain a lat/lon based on an address use:

https://my.locationiq.com/dashboard#geocoding

Use this website to obtain a forecast use:

https://darksky.net/dev

Example of request for location:

https://locationiq.org/v1/search.php?key=50ed712f145e62&q=15+Lisa+Ct.+Sewell%2C+NJ+08080&format=json

Starting point:

import requests
import pprintaddress = '15 Lisa Ct. Sewell NJ 08080'
response = requests.get('https://locationiq.org/v1/search.php?key=50ed712f145e62&q=' + address + '&format=json')
json = response.json()
lat = json[0]['lat']
lon = json[0]['lon']print(lat)
print(lon)

from datetime import datetime
print(datetime.fromtimestamp(1509993277))

Answer 1

I have provided screenshots of the code at the end because indentation in text is sometimes not retained in the answers, which can cause a lot of confusion especially for python.

Feel free to explore the API and use other different information from the response.
Finally, since a key for the darksky api wasn't provided I used a personal one for testing the code. I cannot share that key here for obvious reasons.
You can easily get a new key for yourself though and get 1000 requests per day for free.
This is required for the program to run

Code in Python:

import requests
import pprint
from datetime import datetime

repeat = True
while repeat:
repeat = False
address = input("\nEnter address: ")

#Get latitude and longitude of address by issuing a GET request to https://my.locationiq.com
#The key has already been provided in the question
response1 = requests.get('https://locationiq.org/v1/search.php?key=50ed712f145e62&q=' + address + '&format=json')

#Convert the response into a dictionary format so that we can work with it
json1 = response1.json()

#In this particular API response, the latitude and longitude are stored in 'lat' and 'lon' keys
lat = json1[0]['lat']
lon = json1[0]['lon']

#Now using these lat and lon values we will send a request to another API - the actual weather forecasting API
#Sign up for a darksky account and you will get a free key you can use for up to 1000 requests per day
#Use that key instead, it's really easy :)
#Simply plug in that key in here and the program is good to run
darksky_key = "GET ANOTHER KEY"

#Use the key to get a response
#This is the standard format for request as given in the Dark Sky docs:
#https://api.darksky.net/forecast/[key]/[latitude],[longitude]
response2 = requests.get("https://api.darksky.net/forecast/"+darksky_key+"/"+lat+","+lon);
json2 = response2.json()

#The response contains lots of information that we could use but here we only need the daily forecast
#By reading the docs, we find that we can access daily forecast data by:
#json2['daily']['data'][n]['precipProbability']
#where n is an integer representing the day: 0 is today, 1 is tomorrow and so on

#The timestamps for each day are also provided in the response so it makes sense to read them too
#This will help later to print the dates of the week
#List to store rain chances of the week
rain_chances = []
#List to store the timestamps of the days of the week
time_stamps = []

#Reading 8 values in total - current day and next 7 days
for i in range(0, 8):
rain_chances.append(json2['daily']['data'][i]['precipProbability'])
time_stamps.append(datetime.fromtimestamp(json2['daily']['data'][i]['time']))

#We safely have the rain chances for the week!
#We display the datetime objects in the specific mm/dd/yy format, the weekday is found by strftime("%A")
for i in range(0, 8):
#Since the probablities are measured as a number between 0 and 1, being greater than 0.5 means >= 50% chance
if rain_chances[i] >= 0.5:
message = "Umbrella"
else:
message = "Sun"
print (time_stamps[i].strftime("%m/%d/%y") + " - " + time_stamps[i].strftime("%A") + " - " + message)

#Ask for repeat
inp = input ("\nWeather for another address? (y/n): ")
inp = inp.lower()
if inp == 'y':
repeat = True

Sample Output:

Code Screenshot: