Question

我希望使这个代码更加优雅，使用循环将用户输入放入列表中，并将列表作为浮点列表，而不必在列表中将每个参数定义为浮动时自行使用他们或打印他们...... 我是python 3.x或python的新手，这是我迄今为止写过的第一个代码，所以请原谅我！＆＃39;

Name = "john"
Place = "Colorado"
print (("Hello %s What's up? \nare you coming to the party tonight in %s\n 
if not at least try to make simple calculator:") % (Name, Place))

print ("you will input 2 numbers now and i will devide them for you:")
calc =list(range(2))
calc[0] = (input("number 1:"))
calc[1] = (input("number 2:"))
print (float(calc[0])/float(calc[1]))

Answer 1

既然你说你不熟悉Python，我建议你自己尝试一些方法。这将是一个很好的学习经历。我不打算直接在这里给你答案，因为那会破坏目的。我会提供关于从哪里开始的建议。

旁注：你使用Python3很棒。 Python3.6支持f-strings。这意味着您可以使用print函数替换该行，如下所示。

print(f"Hello {Name} What's up? "
"\nare you coming to the party tonight in {Place}"
"\n if not at least try to make simple calculator:")

好的，您应该按顺序查看以下内容：

for loops
列表理解
命名元组
功能
ZeroDivisionError

Answer 2

你在寻找这样的东西：

values=[float(i) for i in input().split()]
print(float(values[0])/float(values[1]))

输出：

1 2
0.5

Answer 3

通过在构造你的2个数字的列表理解中使用为你输入的函数：

def inputFloat(text):
    inp = input(text)   # input as text
    try:                # exception hadling for convert text to float
        f = float(inp)  # if someone inputs "Hallo" instead of a number float("Hallo") will
                        # throw an exception - this try: ... except: handles the exception
                        # by wrinting a message and calling inputFloat(text) again until a
                        # valid input was inputted which is then returned to the list comp
        return f        # we got a float, return it
    except:
        print("not a number!") # doofus maximus user ;) let him try again
        return inputFloat(text) # recurse to get it again

其余部分来自您的代码，更改的是列表comp以处理消息和输入创建：

Name = "john"
Place = "Colorado"
print (("Hello %s What's up? \nare you coming to the party tonight in %s\n"+
        " if not at least try to make simple calculator:") % (Name, Place))

print ("you will input 2 numbers now and i will devide them for you:")

# this list comprehension constructs a float list with a single message for ech input
calc = [inputFloat("number " + str(x+1)+":") for x in range(2)] 
if (calc[1]):  # 0 and empty list/sets/dicts/etc are considered False by default
    print (float(calc[0])/float(calc[1]))
else:
    print ("Unable to divide through 0")

输出：

"
Hello john What's up?
are you coming to the party tonight in Colorado
 if not at least try to make simple calculator:
you will input 2 numbers now and i will devide them for you:
number 1:23456dsfg
not a number!
number 1:hrfgb
not a number!
number 1:3245
number 2:sfdhgsrth
not a number!
number 2:dfg
not a number!
number 2:5
649.0
"

链接：

用户

3 个答案: