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()
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.
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*$","10")
<_sre.SRE_Match object at 0x0000000002C56308>
You can use a regex to validate the format:
r'^[\d]{2}\.[\d]+$'
You can found the documentation here: https://docs.python.org/2/library/re.html
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!")
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.