How to check if a user input is a float

Question

I'm doing Learn Python the Hard Way exercise 35. Below is the original code, and we're asked to change it so it can accept numbers that don't have just 0 and 1 in them.

def gold_room():
    print "This room is full of gold. How much do you take?"

    next = raw_input("> ")

    if "0" in next or "1" in next:
        how_much = int(next)

    else:
        dead("Man, learn to type a number.")

    if how_much < 50:
        print "Nice, you're not greedy, you win!"
        exit(0)

    else:
        dead("You greedy bastard!")

This is my solution, which runs fine and recognizes float values:

def gold_room():
    print "This room is full of gold. What percent of it do you take?"

    next = raw_input("> ")

    try:
        how_much = float(next)
    except ValueError:
        print "Man, learn to type a number."
        gold_room()

    if how_much <= 50:
        print "Nice, you're not greedy, you win!"
        exit(0)

    else:
        dead("You greedy bastard!")

Searching through similar questions, I found some answers that helped me write another solution, shown in the below code. The problem is, using isdigit() doesn't let the user put in a float value. So if the user said they want to take 50.5%, it would tell them to learn how to type a number. It works otherwise for integers. How can I get around this?

def gold_room():
    print "This room is full of gold. What percent of it do you take?"

    next = raw_input("> ")

while True:
    if next.isdigit():
        how_much = float(next)

        if how_much <= 50:
            print "Nice, you're not greedy, you win!"
            exit(0)

        else:
            dead("You greedy bastard!")

    else: 
        print "Man, learn to type a number."
        gold_room()

Answer 1

isinstance(next, (float, int)) will do the trick simply if next is already converted from a string. It isn't in this case. As such you would have to use re to do the conversion if you want to avoid using try..except.

I would recommend using the try..except block that you had before instead of a if..else block, but putting more of the code inside, as shown below.

def gold_room():
    while True:
        print "This room is full of gold. What percent of it do you take?"
        try:
            how_much = float(raw_input("> "))

            if how_much <= 50:
                print "Nice, you're not greedy, you win!"
                exit(0)

            else:
                dead("You greedy bastard!")

        except ValueError: 
            print "Man, learn to type a number."

This will try to cast it as a float and if it fails will raise a ValueError that will be caught. To learn more, see the Python Tutorial on it.