如何将字符填充添加到字符串？

Question

好的，这很难解释，但我想在句子中添加填充，以便该句子中的字符是 n 的倍数。

但是这个数字“4”必须更改为 3 和 5，所以它也必须工作..

谁知道我在说什么？？以及怎么做？

Answer 1

def pad(yourString,blockLength):
    return yourString + ("X" * (blockLength - (len(yourString) % blockLength)))

是你的填充函数。用于结束填充。 如果您需要中心填充使用：

def centerPad(yourString,blockLength):
    return ("X" * ((blockLength - (len(yourString) % blockLength))/2)) + yourString + ("X" * ((blockLength - (len(yourString) % blockLength))/2))

如果您想实现块密码，则需要更仔细地查看其余代码。

Answer 2

>>> import math
>>> def encrypt(string, length):
        inverse_string = string.replace(' ','X')[::-1]

        center_width = int(math.ceil(len(inverse_string)/float(length)) * length) # Calculate nearest multiple of length rounded up

        inverse_string = inverse_string.center(center_width,'X')

        create_blocks = ' '.join(inverse_string[i:i+length] for i in xrange(0,len(inverse_string),length))
        return create_blocks

>>> encrypt('THE PRICE OF FREEDOM IS ETERNAL VIGILENCE', 4)
'XECN ELIG IVXL ANRE TEXS IXMO DEER FXFO XECI RPXE HTXX'

Answer 3

我希望下面的自我注释代码可以帮助您掌握这个概念。你只需要做一些数学运算就可以得到两端的填充字符

一些概念

1. 需要额外字符填充 = len(string) % block_length
2. Total_Pad_Characters = block_length - len(string) % block_length
3. 前面的填充字符 = Total_Pad_Characters/2
4. 填充字符末尾 = Total_Pad_Characters - Total_Pad_Characters/2

这里是代码

>>> def encrypt(st,length):
    #Reversed the String and replace all Spaces with 'X'
    st = st[::-1].replace(' ','X')
    #Find no of characters to be padded.
    padlength = (length - len(st)%length) % length
    #Pad the Characters at either end
    st = 'X'*(padlength/2)+st+'X'*(padlength-padlength/2)
    #Split it with size length and then join with a single space
    return ' '.join(st[i:i+length] for i in xrange(0,len(st),length))

>>> encrypt('THE PRICE OF FREEDOM IS ETERNAL VIGILENCE', 4) #Your Example
'XECN ELIG IVXL ANRE TEXS IXMO DEER FXFO XECI RPXE HTXX'
>>> encrypt('THE PRICE', 5) # One Extra Character at end for Odd Numbers
'ECIRP XEHTX'
>>> encrypt('THE PRIC', 5) # 1 Pad Characters at either end
'XCIRP XEHTX'
>>> encrypt('THE PRI', 5) # 1 Pad Characters at either end and one Extra for being Odd
'XIRPX EHTXX'
>>> encrypt('THE PR', 5) # 2 Pad Characters at either end
'XXRPX EHTXX'
>>> encrypt('THE P', 5) # No Pad characters required
'PXEHT'
>>> encrypt('THE PRICE OF FREEDOM IS ETERNAL VIGILENCE', 5) #Ashwini's Example
'XXECN ELIGI VXLAN RETEX SIXMO DEERF XFOXE CIRPX EHTXX'
>>>