Question

首先，我使用以下代码从文件中提取了一些文本：

from collections import Counter

def n_gram_opcodes(source, n):
    source = open(source).read()
    OPCODES = set(["add","call","cmp","mov","jnz","jmp","jz","lea","pop","push",
    "retn","sub","test","xor"])

    source_words = source.split()
    opcodes = [w for w in source_words if w in OPCODES]

    return Counter(zip(*[opcodes[i:] for i in range(n)]))

该代码还允许计算文件中某些单词出现的频率。以字典格式存储单词，如下所示：

Counter({('mov', 'mov', 'mov'): 18, ('xor', 'mov', 'mov'): 6, ('mov', 'mov', 'pop'): 3, ('mov', 'mov', 'push'): 3, ('pop', 'mov', 'mov'): 3, ('mov', 'call', 'cmp'): 3, ('push', 'pop', 'mov'): 3, ('mov', 'add', 'mov'): 3, ('call', 'mov', 'call'): 3, ('mov', 'mov', 'xor'): 3, ('cmp', 'mov', 'cmp'): 2, ('pop', 'mov', 'add'): 2, ('mov', 'pop', 'mov'): 2, ('mov', 'cmp', 'sub'): 2, ('mov', 'mov', 'sub'): 2, ('mov', 'mov', 'call'): 2})

通过上面的这个词典，我想取值（出现频率）并在下面的loglikelihood公式中使用。我的问题是如何修改代码，以便它可以从上面的任何字典中获取值，并将其与下面的代码一起使用。最终结果应返回数字并使用matplotlib绘制图形。

import math
# The placeholder value for 0 counts
epsilon = 0.0001
def opcode_llr(opcode, freq_table_before, freq_table_after):

'''
Args:
    opcode: A single opcode mnemonic, e.g., 'mov'

    freq_table_before: The frequency table for opcode trigrams *before*
                       extraction.

    freq_table_after: The frequency table for opcode trigrams *after*
                      extraction.

The keys for both tables are tuples of string. So, each is of the form

    {
        ('mov', 'mov', 'mov'): 5.0,
        ('mov', 'jmp', 'mov'): 7.0,
        ...
    }

'''
    t_b = len(freq_table_before) or epsilon
    t_a = len(freq_table_after) or epsilon

    # Compute the opcode counts when occurring in positions 0, 1, 2
    opcode_counts = [epsilon, epsilon, epsilon]
    for triplet in freq_table_after.keys():
        for i, comp in enumerate(triplet):
            if comp == opcode:
                opcode_counts[i] += 1

    f1 = opcode_counts[0]
    f2 = opcode_counts[1]
    f3 = opcode_counts[2]

    return (f1 + f2 + f3) * math.log(float(t_b) / t_a)

Answer 1

以下是从Counter计算LLR的一般方法。

from collections import Counter
import random
import math

def CntToLLR(cnt):
    n = sum(cnt.values())   # total number of samples
    LLR = {}                # dict to store LLRs (same keys as counter)
    for x,y in cnt.items(): # x is the key, and y the count
        LLR[x] = math.log(y) - math.log(n - y)
    return LLR

# populate a counter with random values
cnt = Counter([random.randrange(10) for x in range(100)])

llrs = CntToLLR(cnt)

# You can convert the dictionary to a list of (key, value)
llrs = list(llrs.iteritems())

使用dict中的值计算对数似然比

1 个答案: