如何检查用户输入是否是浮动的 [英] How to check if a user input is a float
问题描述
def gold_room（）：
print这个房间充满了黄金。 ？
$ b $ next = raw_input（>）
如果下一个为0或下一个为1：
how_much = int（next）
else：
dead（Man，learn to type a number。）
if how_much< 50：
print不错，你不贪心，你赢了！
exit（0）
else：
dead（You greedy bastard！）
这是我的解决方案，运行良好，可以识别浮点值：
def gold_room（）：
print这个房间里充满了黄金，你拿的百分比是多少？
$ b $ next = raw_input（>）
尝试：
how_much = float（next）
除ValueError：
print人，学会打一个号码。
gold_room（）
如果how_much <= 50：
print很好，你不贪心，你赢了！
exit（0）
else：
dead（You greedy bastard！）
通过类似的问题搜索，我找到了一些答案，帮助我编写了另一个解决方案，如下面的代码所示。问题是，使用isdigit（）不允许用户输入一个浮点值。所以如果用户说要拿50.5％，就会告诉他们学习如何输入一个数字。它适用于整数。
$ b
def gold_room（）：
printThis this this get this around？房间充满了黄金，你拿的是多少？
next = raw_input（>）
而真：
if next.isdigit（）：
how_much = float（next）
如果how_much <= 50：
print很好，你不贪心，你赢了！
exit（0）
else：
dead（You greedy bastard！）
else：
print键入一个数字。
gold_room（）
<$ c $如果 next 已经从一个字符串转换过来，那么c> isinstance（next，（float，int））这不是在这种情况下。因此，如果您想避免使用 try..except 。，则必须使用 re / p>
我建议使用前面的 try..except 块来代替 if..else block，但是把更多的代码放在里面，如下所示：
def gold_room（）：
，True：
print这个房间里充满了黄金，你拿的是多少？
try：
how_much = float（raw_input（>））
如果how_much< = 50：
print不错，你不贪心， 你赢了！
exit（0）
else：
dead（You greedy bastard！）
除了ValueError：
print学会键入一个数字。
这将尝试将其转换为浮点数，如果失败将会引发 ValueError 将被捕获。要了解更多信息，请参阅上面的 Python教程。 p>
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.
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