字符串切片是否？

Question

started a class on Python this semester and I'm having troubles with a current assignment. Went through the examples and exercises done in class uptil now and couldn't find anything.

Q1) Fixed

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]

I want it to be able to print out two statements, one for each outcome. If the binary number entered ends with a "1" it would spit out aN with 2 decimals and if the binary number entered ends with a "0" it would spit out aN with 1 decimal.

When it runs, it doesn't do anything after showing the value the user inputs for bN

Q2) Is there a better way to find digits after the decimal? The slicing only works when it's numbers < 10.

edit) Ty to the guy that pointed out the string, I totally forgot that :(

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]

If anyone could answer the second question though, that would be great.

Answer 1

raw_input() takes the input as string. In your if condition you are compering a string with a integer. make your if like this,

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]

If you want to print 2 digits after the decimal point, you can follow this approach.

Answer 2

First, raw_input() takes in the input as a str .

So any comparison of aN[-1] with an int will result in False . ie your code will always run the else block.

Further, you need to first locate the index of the decimal point and then calculate whether to display one or two places (or deal with the case in case there is no decimal point)

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

Assuming aN always have at least 1 decimals, this should be working. You can use find() to locate the "." for decimals.

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]

Example:

>>> 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