Fibonacci前缀码

Question

这件事有点牵扯。是否有更简单的方法在基本python中实现这个前缀代码，我现有的代码在下面，但是任何修改或审查的帮助都是有帮助的。

# Lagged fibonacci numbers the sequence starts: 1,2,3,5,8,11
def fibonacci(n):
    a = 1
    b = 1
    out = []
    for i in range(n):
        out.append(a)
        a,b = a+b,a
    return out

# There has to be a better way to do this, right?
# Go through the list for the first few fibonacci numbers from greatest to
# make a list of fibonacci numbers that add up to each code number (1,2,3...)
# The code is then the indexes of those fibonacci numbers, with leading zeroes
# removed, reversed, with an additional "1" at the end.
def makeCodebook(decode=False):
    F = fibonacci(10)
    F.reverse()
    codes = []
    for x in range(1,27):
        while x != 0:
            code = []
            for f in F:
                if f <= x:
                    code.append("1")
                    x = x-f
                else:
                    code.append("0")

            while code[0] == "0":

                code.pop(0)
            code.reverse()
            code.append("1")
            codes.append("".join(code))

    # Matchup codes with letters
    D = {}
    alpha = "ETAOINSRHLDCUMFPGWYBVKXJQZ"
    if decode == False:
        for a,b in zip(alpha,codes):
            D[a] = b

    if decode == True:
        for a,b in zip(alpha,codes):
            D[b] = a
    return D

def prefixCode(text,decode=False):

    # Build the codebook
    D = makeCodebook(decode=decode)

    # When encoding we simply translate each letter to its code
    if decode == False:
        out = []
        for letter in text:
            out.append(D[letter])
        return "".join(out)

Answer 1

您的fibonacci不需要变量a和b；out已经在跟踪它们。

def fibonacci(n):
    out = [1,2]
    for i in range(n-2):
        out.append(out[i]+out[i+1])
    return out

您应该同时构建编码和编码代码本，这样您就不必运行相同的函数两次才能得到这两种代码。

您可以在找到代码时构建代码本：

def makeCodebook():
    alpha = "ETAOINSRHLDCUMFPGWYBVKXJQZ"
    F = fibonacci(10)
    F.reverse()
    D = {}
    E = {}
    for x,letter in enumerate(alpha,1):
        while x != 0:
            code = []
            for f in F:
                if f <= x:
                    code.append("1")
                    x = x-f
                else:
                    code.append("0")

            while code[0] == "0":

                code.pop(0)
            code.reverse()
            code.append("1")
        D[code] = letter
        E[letter] = code
    return D,E

注意，这消除了保留codes列表的需要。

那你就可以

def prefixCode(text,decode=False):
     D, E = makeCodebook(decode=decode)
     if not decode:
          return "".join([E[letter] for letter in text])