Question

我有一些这样的数据，其中第二个字段是第一个字段的概率，因此“0：0.017”表示存在0.017的概率。所有概率的总和为1.

我的问题是：如何根据概率“范围线”，以便找到每个角色的下限和上限？所以0将是[0,0.017），[0.017,0.022]，依此类推。

我正在尝试实现算术编码。

(0: 0.017,
1: 0.022,
2: 0.033,
3: 0.033,
4: 0.029,
5: 0.028,
6: 0.035,
7: 0.032,
8: 0.028,
9: 0.027,
a: 0.019,
b: 0.022,
c: 0.029,
d: 0.03,
e: 0.028,
f: 0.035,
g: 0.026,
h: 0.037,
i: 0.029,
j: 0.025,
k: 0.025,
l: 0.037,
m: 0.025,
n: 0.023,
o: 0.026,
p: 0.035,
q: 0.033,
r: 0.031,
s: 0.023,
t: 0.022,
u: 0.038,
v: 0.022,
w: 0.016,
x: 0.026,
y: 0.021,
z: 0.033,)

编辑*

nvm我想通了，只是弄乱了愚蠢的数学......感谢所有的输入！

Answer 1

将您的数据转换为python是一项练习：

>>> corpus = [('0', 0.017), ('1', 0.022), ('2', 0.033), ('3', 0.033), ('4', 0.029),
...           ('5', 0.028), ('6', 0.035), ('7', 0.032), ('8', 0.028), ('9', 0.027),
...           ('a', 0.019), ('b', 0.022), ('c', 0.029), ('d', 0.030), ('e', 0.028),
...           ('f', 0.035), ('g', 0.026), ('h', 0.037), ('i', 0.029), ('j', 0.025),
...           ('k', 0.025), ('l', 0.037), ('m', 0.025), ('n', 0.023), ('o', 0.026),
...           ('p', 0.035), ('q', 0.033), ('r', 0.031), ('s', 0.023), ('t', 0.022),
...           ('u', 0.038), ('v', 0.022), ('w', 0.016), ('x', 0.026), ('y', 0.021),
...           ('z', 0.033)]

创建累积总和：

>>> distribution = []
>>> total = 0.0
>>> for letter, frequency in corpus:
...     distribution.append(total)
...     total += frequency
...

实际上使用这种数据是bisect模块的基础。

>>> import bisect, random
>>> def random_letter():
...     value = random.random()
...     index = bisect.bisect(distribution, value) - 1
...     return corpus[index][0]
... 
>>> [random_letter() for n in range(10)]  # doctest: +SKIP
['d', '6', 'p', 'c', '8', 'f', '7', 'm', 'z', '7']

Answer 2

# The data is input as '1: 0.022,' format
def process_data(line):
    # for returning the new string that is cleaned up
    result_line = ''
    for character in line:
        # check if it is either a number or a letter
        if character.isdigit() or character.isalpha():
            result_line += character
        # we want the decimal point
        elif character == '.':
            result_line += character
        # else we replace it with space ' '
        else:
            result_line += ' '
    return result_line

my_list = []  

with open('input.txt') as file:
    for lines in file:
        processed_line = process_data(lines)
        # temp_list has ['letter', 'frequency']
        temp_list = (processed_line.split())
        value = temp_list[0]
        # Require to cast it to a float, since it is a string
        frequency = float(temp_list[1])
        my_list.append([value, frequency])

print(my_list)

从这一点开始，您可以弄清楚如何处理您的价值观。我记录了代码（授予了处理输入文件的一种非常简单的天真方式）。但my_list现在干净且格式正确，string（值）和float（频率）。希望这有帮助。

从上面输出代码：

[['0', 0.017], ['1', 0.022], ['2', 0.033], ['3', 0.033], 
['4', 0.029], ['5', 0.028], ['6', 0.035], ['7', 0.032], 
['8', 0.028], ['9', 0.027], ['a', 0.019], ['b', 0.022], 
['c', 0.029], ['d', 0.03], ['e', 0.028], ['f', 0.035], 
['g', 0.026], ['h', 0.037], ['i', 0.029], ['j', 0.025], 
['k', 0.025], ['l', 0.037], ['m', 0.025], ['n', 0.023], 
['o', 0.026], ['p', 0.035], ['q', 0.033], ['r', 0.031], 
['s', 0.023], ['t', 0.022], ['u', 0.038], ['v', 0.022], 
['w', 0.016], ['x', 0.026], ['y', 0.021], ['z', 0.033]]

然后......

# Took a page out of TokenMacGuy, credit to him
distribution = []
distribution.append(0.00)  
total = 0.0 # Create a float here

for entry in my_list:
    distribution.append(entry[1])
    total += frequency
    total = round(total, 3) # Rounding to 2 decimal points

distribution.append(1.00) # Missing the 1.00 value
print(distribution) # Print to check

输出在这里：

[0.0, 0.017, 0.022, 0.033, 0.033, 0.029, 0.028, 0.035, 0.032, 
0.028, 0.027, 0.019, 0.022, 0.029, 0.03, 0.028, 0.035, 0.026, 
0.037, 0.029, 0.025, 0.025, 0.037, 0.025, 0.023, 0.026, 0.035, 
0.033, 0.031, 0.023, 0.022, 0.038, 0.022, 0.016, 0.026, 0.021, 
0.033, 1.0]

最后，输出最终结果：没有什么特别之处，我使用pattern和format让它们看起来更漂亮。并且它几乎是根据ninjagecko的方法来计算它的。由于计算没有显示，我确实必须将0.00和1.00填入分布。非常直接的实施在后我们弄清楚如何做概率。

pattern = '{0}: [{1:1.3f}, {2:1.3f})'
count = 1 # a counter to keep track of the index

pre_p = distribution[0] 
p = distribution[1]

# Here we will print it out at the end in the format you said in the question
for entry in my_list:
    print(pattern.format(entry[0], pre_p, p))
    pre_p += distribution[count]
    p += distribution[count+1]
    count = count + 1

输出：

0: [0.000, 0.017)
1: [0.017, 0.039)
2: [0.039, 0.072)
3: [0.072, 0.105)
4: [0.105, 0.134)
5: [0.134, 0.162)
6: [0.162, 0.197)
7: [0.197, 0.229)
8: [0.229, 0.257)
9: [0.257, 0.284)
a: [0.284, 0.303)
b: [0.303, 0.325)
c: [0.325, 0.354)
d: [0.354, 0.384)
e: [0.384, 0.412)
f: [0.412, 0.447)
g: [0.447, 0.473)
h: [0.473, 0.510)
i: [0.510, 0.539)
j: [0.539, 0.564)
k: [0.564, 0.589)
l: [0.589, 0.626)
m: [0.626, 0.651)
n: [0.651, 0.674)
o: [0.674, 0.700)
p: [0.700, 0.735)
q: [0.735, 0.768)
r: [0.768, 0.799)
s: [0.799, 0.822)
t: [0.822, 0.844)
u: [0.844, 0.882)
v: [0.882, 0.904)
w: [0.904, 0.920)
x: [0.920, 0.946)
y: [0.946, 0.967)
z: [0.967, 1.000)

完整来源位于：http://codepad.org/a6YkHhed

Answer 3

创建字典，键是您的字符，值是定义下限和上限的对。

prev_p = 0
bounds = {}
for line in open(a_file):
   character, p = parse_the_line(line)
   bounds[character] = (prev_p, p)
   prev_p = p

在数字线上设置范围的好方法是什么？

3 个答案: