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python - 算术编码和解码算法 Python

转载 作者:行者123 更新时间:2023-12-01 04:42:28 26 4
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我正在研究算术编码和解码算法的自适应实现,并且我已经用Python实现了它,但是对于某些字符串我得到了正确的答案,但对于其他字符串我得到了正确的答案。

当程序第一次启动时,会提供一个参数来决定符号概率更改的频率。例如,如果参数为10,则在发送/接收10个符号之后,概率表根据迄今为止发送/接收的所有符号而改变。因此,域分配也发生了变化。最初,我的均匀分布 [a-z] 的概率为 1/26。

它不适用于“heloworldheloworld”和许多其他情况。

此外,我已经了解了下溢问题,但如何解决该问题。

import sys
import random
import string


def encode(encode_str, N):
count = dict.fromkeys(string.ascii_lowercase, 1) # probability table
cdf_range = dict.fromkeys(string.ascii_lowercase, 0)
pdf = dict.fromkeys(string.ascii_lowercase, 0)

low = 0
high = float(1)/float(26)

for key, value in sorted(cdf_range.iteritems()):
cdf_range[key] = [low, high]
low = high
high += float(1)/float(26)

for key, value in sorted(pdf.iteritems()):
pdf[key] = float(1)/float(26)

# for key, value in sorted(cdf_range.iteritems()):
# print key, value

# for key, value in sorted(pdf.iteritems()):
# print key, value

i = 26

lower_bound = 0 # upper bound
upper_bound = 1 # lower bound

u = 0

# go thru every symbol in the string
for sym in encode_str:
i += 1
u += 1
count[sym] += 1

curr_range = upper_bound - lower_bound # current range
upper_bound = lower_bound + (curr_range * cdf_range[sym][1]) # upper_bound
lower_bound = lower_bound + (curr_range * cdf_range[sym][0]) # lower bound

# update cdf_range after N symbols have been read
if (u == N):
u = 0

for key, value in sorted(pdf.iteritems()):
pdf[key] = float(count[key])/float(i)

low = 0
for key, value in sorted(cdf_range.iteritems()):
high = pdf[key] + low
cdf_range[key] = [low, high]
low = high

return lower_bound

def decode(encoded, strlen, every):
decoded_str = ""

count = dict.fromkeys(string.ascii_lowercase, 1) # probability table
cdf_range = dict.fromkeys(string.ascii_lowercase, 0)
pdf = dict.fromkeys(string.ascii_lowercase, 0)

low = 0
high = float(1)/float(26)

for key, value in sorted(cdf_range.iteritems()):
cdf_range[key] = [low, high]
low = high
high += float(1)/float(26)

for key, value in sorted(pdf.iteritems()):
pdf[key] = float(1)/float(26)


lower_bound = 0 # upper bound
upper_bound = 1 # lower bound

k = 0

while (strlen != len(decoded_str)):
for key, value in sorted(pdf.iteritems()):

curr_range = upper_bound - lower_bound # current range
upper_cand = lower_bound + (curr_range * cdf_range[key][1]) # upper_bound
lower_cand = lower_bound + (curr_range * cdf_range[key][0]) # lower bound

if (lower_cand <= encoded < upper_cand):
k += 1
decoded_str += key

if (strlen == len(decoded_str)):
break

upper_bound = upper_cand
lower_bound = lower_cand

count[key] += 1

if (k == every):
k = 0
for key, value in sorted(pdf.iteritems()):
pdf[key] = float(count[key])/float(26+len(decoded_str))

low = 0
for key, value in sorted(cdf_range.iteritems()):
high = pdf[key] + low
cdf_range[key] = [low, high]
low = high

print decoded_str

def main():
count = 10
encode_str = "yyyyuuuuyyyy"
strlen = len(encode_str)
every = 3
encoded = encode(encode_str, every)
decoded = decode(encoded, strlen, every)

if __name__ == '__main__':
main()

最佳答案

发生这种情况是因为 Python float 具有 53 位精度。您无法对很长的字符串进行编码。

您可能想使用decimal而不是 floats 来获得任意精度

import sys
import random
import string

import decimal
from decimal import Decimal

decimal.getcontext().prec=100

def encode(encode_str, N):
count = dict.fromkeys(string.ascii_lowercase, 1) # probability table
cdf_range = dict.fromkeys(string.ascii_lowercase, 0)
pdf = dict.fromkeys(string.ascii_lowercase, 0)

low = 0
high = Decimal(1)/Decimal(26)

for key, value in sorted(cdf_range.iteritems()):
cdf_range[key] = [low, high]
low = high
high += Decimal(1)/Decimal(26)

for key, value in sorted(pdf.iteritems()):
pdf[key] = Decimal(1)/Decimal(26)

# for key, value in sorted(cdf_range.iteritems()):
# print key, value

# for key, value in sorted(pdf.iteritems()):
# print key, value

i = 26

lower_bound = 0 # upper bound
upper_bound = 1 # lower bound

u = 0

# go thru every symbol in the string
for sym in encode_str:
i += 1
u += 1
count[sym] += 1

curr_range = upper_bound - lower_bound # current range
upper_bound = lower_bound + (curr_range * cdf_range[sym][1]) # upper_bound
lower_bound = lower_bound + (curr_range * cdf_range[sym][0]) # lower bound

# update cdf_range after N symbols have been read
if (u == N):
u = 0

for key, value in sorted(pdf.iteritems()):
pdf[key] = Decimal(count[key])/Decimal(i)

low = 0
for key, value in sorted(cdf_range.iteritems()):
high = pdf[key] + low
cdf_range[key] = [low, high]
low = high

return lower_bound

def decode(encoded, strlen, every):
decoded_str = ""

count = dict.fromkeys(string.ascii_lowercase, 1) # probability table
cdf_range = dict.fromkeys(string.ascii_lowercase, 0)
pdf = dict.fromkeys(string.ascii_lowercase, 0)

low = 0
high = Decimal(1)/Decimal(26)

for key, value in sorted(cdf_range.iteritems()):
cdf_range[key] = [low, high]
low = high
high += Decimal(1)/Decimal(26)

for key, value in sorted(pdf.iteritems()):
pdf[key] = Decimal(1)/Decimal(26)


lower_bound = 0 # upper bound
upper_bound = 1 # lower bound

k = 0

while (strlen != len(decoded_str)):
for key, value in sorted(pdf.iteritems()):

curr_range = upper_bound - lower_bound # current range
upper_cand = lower_bound + (curr_range * cdf_range[key][1]) # upper_bound
lower_cand = lower_bound + (curr_range * cdf_range[key][0]) # lower bound

if (lower_cand <= encoded < upper_cand):
k += 1
decoded_str += key

if (strlen == len(decoded_str)):
break

upper_bound = upper_cand
lower_bound = lower_cand

count[key] += 1

if (k == every):
k = 0
for key, value in sorted(pdf.iteritems()):
pdf[key] = Decimal(count[key])/Decimal(26+len(decoded_str))

low = 0
for key, value in sorted(cdf_range.iteritems()):
high = pdf[key] + low
cdf_range[key] = [low, high]
low = high

print decoded_str

def main():
count = 10
encode_str = "heloworldheloworld"
strlen = len(encode_str)
every = 3
encoded = encode(encode_str, every)
decoded = decode(encoded, strlen, every)

if __name__ == '__main__':
main()

关于python - 算术编码和解码算法 Python,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30278704/

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