Question

本学期开始上Python课程，我遇到了当前任务的麻烦。通过现在在课堂上完成的例子和练习，找不到任何东西。

Q1）已修复

aN = raw_input("Enter a number with decimals")
bN = raw_input("Enter a binary number, 2 to 4 digits in length")

if (bN[-1] == 1):
    print "The binary number was odd so your number will contain 2 decimals"
    print "The number is now",aN[0:4],

elif (bN[-1] == 0):
    print "The binary number was even so your number will contain 1 decimal"
    print "The number is now",bN[0:3]

我希望它能够打印出两个语句，每个结果一个。如果输入的二进制数以“1”结尾，则会以2位小数形式输出aN，如果输入的二进制数以“0”结尾，则会以1位小数吐出aN。

运行时，在显示用户为bN

输入的值后，它不会执行任何操作

Q2）有没有更好的方法在小数点后找到数字？切片仅在数字＆lt; 10。

编辑）对那个指出弦乐的人来说，我完全忘记了:(

aN = raw_input("Enter a number with decimals")
bN = raw_input("Enter a binary number, 2 to 4 digits in length")

if (float(bN[-1]) == 1):
    print "The binary number was odd so your number will contain 2 decimals"
    print "The number is now",aN[0:4],

elif (float(bN[-1]) == 0):
    print "The binary number was even so your number will contain 1 decimal"
    print "The number is now",aN[0:3]

如果有人能回答第二个问题，那就太棒了。

Answer 1

raw_input()将输入作为字符串。在if条件下，您正在使用整数来编写字符串。让你的if像这样，

if float(bN[-1]) == 1:
    print "The binary number was odd so your number will contain 2 decimals"
    print "The number is now %.2f" % float(aN)

elif float(bN[-1]) == 0:
    print "The binary number was even so your number will contain 1 decimal"
    print "The number is now", bN[0:3]

如果要在小数点后打印2位数，可以按照这种方法进行操作。

Answer 2

首先，raw_input()将输入视为str。

因此，aN[-1]与int的任何比较都会产生False。即您的代码将始终运行else块。

此外，您需要首先找到小数点的索引，然后计算是否显示一个或两个位置（或处理没有小数点的情况）

aN = raw_input("Enter a number with decimals")
bN = raw_input("Enter a binary number, 2 to 4 digits in length")

if (bN[-1] == '1'):
    print "The binary number was odd so your number will contain 2 decimals"
    if aN.find('.') == -1:
        print "The number is now ",aN
    else:
        print "The number is now ", aN[:aN.find('.')+3]

elif (bN[-1] == '0'):
    print "The binary number was even so your number will contain 1 decimal"
    if aN.find('.') == -1:
        print "The number is now ",aN
    else:
        print "The number is now ", aN[:aN.find('.')+2]

Answer 3

假设aN总是至少有1个小数，这应该是有效的。您可以使用find()找到“。”小数点。

aN = raw_input("Enter a number with decimals")
bN = raw_input("Enter a binary number, 2 to 4 digits in length")

if (bN[-1] == "1"):
    print "The binary number was odd so your number will contain 2 decimals"
    print "The number is now", aN[:aN.find(".")+3]

elif (bN[-1] == "0"):
    print "The binary number was even so your number will contain 1 decimal"
    print "The number is now", aN[:aN.find(".")+2]

示例：

>>> aN
'1.12345'
>>> bN
'11'
>>> if (bN[-1] == "1"):
...     print "The binary number was odd so your number will contain 2 decimals"
...     print "The number is now", aN[:aN.find(".")+3]
... elif (bN[-1] == "0"):
...     print "The binary number was even so your number will contain 1 decimal"
...     print "The number is now", aN[:aN.find(".")+2]
...
The binary number was odd so your number will contain 2 decimals
The number is now 1.12

字符串切片用于？

3 个答案: