Limiting user input to a range in Python

2020-01-29 12:03发布

In the code below you'll see it asking for a 'shift' value. My problem is that I want to limit the input to 1 through 26.

    For char in sentence:
            if char in validLetters or char in space: #checks for
                newString += char                     #useable characters
        shift = input("Please enter your shift (1 - 26) : ")#choose a shift
        resulta = []
        for ch in newString:
            x = ord(ch)      #determines placement in ASCII code
            x = x+shift      #applies the shift from the Cipher
            resulta.append(chr(x if 97 <= x <= 122 else 96+x%122) if ch != \
            ' ' else ch) # This line finds the character by its ASCII code

How do I do this easily?

标签: python range
5条回答
聊天终结者
2楼-- · 2020-01-29 12:23
while True:
     result = raw_input("Enter 1-26:")
     if result.isdigit() and 1 <= int(result) <= 26:
         break;
     print "Error Invalid Input"

#result is now between 1 and 26 (inclusive)
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Rolldiameter
3楼-- · 2020-01-29 12:27

Try something like this

acceptable_values = list(range(1, 27))
if shift in acceptable_values:
    #continue with program
else:
    #return error and repeat input

Could put in while loop but you should limit user inputs so it doesn't become infinite

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何必那么认真
4楼-- · 2020-01-29 12:31

Use an if-condition:

if 1 <= int(shift) <= 26:
   #code
else:
   #wrong input

Or a while loop with the if-condition:

shift = input("Please enter your shift (1 - 26) : ")
while True:
   if 1 <= int(shift) <= 26:
      #code
      #break or return at the end
   shift = input("Try Again, Please enter your shift (1 - 26) : ")  
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爷、活的狠高调
5楼-- · 2020-01-29 12:37

Use a while loop to keep asking them for input until you receive something you consider valid:

shift = 0
while 1 > shift or 26 < shift:
    try:
        # Swap raw_input for input in Python 3.x
        shift = int(raw_input("Please enter your shift (1 - 26) : "))
    except ValueError:
        # Remember, print is a function in 3.x
        print "That wasn't an integer :("

You'll also want to have a try-except block around the int() call, in case you get a ValueError (if they type a for example).

Note if you use Python 2.x, you'll want to use raw_input() instead of input(). The latter will attempt to interpret the input as Python code - that can potentially be very bad.

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We Are One
6楼-- · 2020-01-29 12:43

Another implementation:

shift = 0
while not int(shift) in range(1,27):
    shift = input("Please enter your shift (1 - 26) : ")#choose a shift
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