python学习——如何判断输入是数字

笨办法学python第35节

该节主要是讲分支与函数，主要遇到的问题是python中如何判断输入是数字。

首先原代码如下：

from sys import exit

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

    next = raw_input("> ")
    if "" in next or "" 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!")

def bear_room():
    print "There is a bear here."
    print "The bear has a bunch of honey."
    print "The fat bear is in front of another door."
    print "How are you going to move the bear?"
    bear_moved = False

    while True:
        next = raw_input("> ")

        if next == "take honey":
            dead("The bear looks at you then slaps your face off.")
        elif next == "taunt bear" and not bear_moved:
            print "The bear has moved from the door. You can go through it now."
            bear_moved = True
        elif next == "taunt bear" and bear_moved:
            dead("The bear gets pissed off and chews your leg off.")
        elif next == "open door" and bear_moved:
            gold_room()
        else:
            print "I got no idea what that means."

def cthulhu_room():
    print "Here you see the great evil Cthulhu."
    print "He, it, whatever stares at you and you go insane."
    print "Do you flee for your life or eat your head?"

    next = raw_input("> ")
    if "flee" in next:
        start()
    elif "head" in next:
        dead("Well that was tasty!")
    else:
        cthulhu_room()

def dead(why):
    print why, "Good job!"
    exit(0)

def start():
    print "You are in a dark room."
    print "There is a door to your right and left."
    print "Which one do you take?"

    next = raw_input("> ")

    if next == "left":
        bear_room()
    elif next == "right":
        cthulhu_room()
    else:
        dead("You stumble around the room until you starve.")

start()

将改代码的流程图画出来思路就很清晰（略）（哈哈哈哈想起来小时候看答案，答案“略”，sad）

其中，gold_room函数的一个判断语句“if "0" in next or "1" in next：”，这句话就只能使得输入的数字中有1或0的才可以进行how_much的判断，那么python中有没有一种方法可以直接判断输入是否是数字，有哒，就是可以用语句“if  next.isdigit():”判断，将前一句换成这一句之后再运行，可以得到运行结果如下（其中我把raw_input("> ")换成了raw_input("please input a number: ")）：

注：

补充：isdigit()方法检查字符串是否只包含数字(全由数字组成)。也就是说“isdigit（）本身就是处理字符串的函数，他不分辨你到底是数字还是字符串”

除了判断raw_input()输入的是否是数字，还可以判断是否是字符串，如下

"if next.isdigit():" 都是数字
"if next.isalnum():" 都是数字或者字母
"if next.isalpha():" 都是字母
"if next.islower():" 都是小写
"if next.isupper():" 都是大写
"if next.istitle():" 都是首字母大写，像标题
"if next.isspace():" 都是空白字符、\t、\n、\r

（其他的判断还没有试，有用到的话回来找）