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()

Do not use next as a variable name. That is a function in Python. — dawg
– dawg, Commented Apr 25, 2014 at 22:20

Chris Hagmann · Accepted Answer · 2014-05-02 18:02:42Z

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.

I tried this code and it worked well. Thank you. It is a good solution, but since he hasn't yet introduced try/except statements, I think the answer provided by Slick is more along the lines of what the author was looking for.
Oops, sorry. I got mixed up looking at something else. Yes, then both your codes are appropriate and efficient.

Sheng · Accepted Answer · 2014-04-25 22:13:04Z

RE would be a good choice

>>> re.match("^\d+.\d+$","10") >>> re.match("^\d+.\d+$","1.00001") <_sre.SRE_Match object at 0x0000000002C56370>

If the raw input is a float number, it will return an object. Otherwise, it will return None. In case you need to recognize the int, you could:

>>> re.match("^[1-9]\d*$","10") <_sre.SRE_Match object at 0x0000000002C56308>

Julio · Accepted Answer · 2014-04-25 22:13:09Z

You can use a regex to validate the format:

r'^[\d]{2}\.[\d]+$'

You can found the documentation here: https://docs.python.org/2/library/re.html

Slick · Accepted Answer · 2014-04-27 17:41:05Z

The problem I have with your approach is that you're going down a path of "Look Before You Leap" instead of the more Pythonic "Easier to Ask Forgiveness than Permission" path. I think your original solution is better than trying to validate the input in this way.

Here is how I would write it.

GREEDY_LIMIT = 50 def gold_room(): print("This room is full of gold. What percent of it do you take?") try: how_much = float(raw_input("> ")) except ValueError: print("Man, learn to type a number.") gold_room() return if how_much <= GREEDY_LIMIT: print "Nice, you're not greedy, you win!" exit(0) else: dead("You greedy bastard!")

Your code is simple and efficient, and it worked well. Thank you. I like this answer best so far, because it doesn't use any concepts that I haven't learned yet. He has not introduced try/except statements yet, so I feel this answer would be along the lines of what he expects from us.
@pez If you find an answer useful, it is good habit to express it by upvoting (any number of answers), or finally even accepting an answer (only one).
@ Jan Vlcinsky Thanks for the advice. I can't upvote anything yet because I'm a new user, and that requires a reputation of 15. I'm thinking of which answer to accept since multiple ones are valid.
This is wrong. When you call gold_room() at the end of the except clause, you are starting a recursive stack of calls that will cause unexpected behavior if this is used and the user inputs a non float value.

donald · Accepted Answer · 2017-02-27 07:24:17Z

use below python based regex checking for float strings

import re a=re.match('((\d+[\.]\d*$)|(\.)\d+$)' , '2.3') a=re.match('((\d+[\.]\d*$)|(\.)\d+$)' , '2.') a=re.match('((\d+[\.]\d*$)|(\.)\d+$)' , '.3') a=re.match('((\d+[\.]\d*$)|(\.)\d+$)' , '2.3sd') a=re.match('((\d+[\.]\d*$)|(\.)\d+$)' , '2.3')

output: !None, !None, !None, None , !None then use this output to do conversions.

why downvote . it best solution among all existing posted solution ? COuld you reason then just downvote!

emanuel cheng · Accepted Answer · 2019-05-16 14:41:48Z

If you don't want to use try/except, below is my answer :

def gold_room(): print "This room is full of gold. How much do you take?" choice = input("> ") if choice.isdigit(): how_much = int(choice) elif "." in choice: choice_dot = choice choice_dot_remove = choice_dot.replace(".","") if choice_dot_remove.isdigit(): how_much = float(choice) 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!")

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