from difflib import Differ, SequenceMatcher
[docs]def get_max_len(wildcards):
'''
find the max length from wildcards.
'''
mlen = 0
for item in wildcards.split('*'):
length = len(item)
if length > mlen:
mlen = length
# FIXME 直接生成*.*.*.*， 或者，compare的顺序相反？但是仍然不能避免。
[docs]def get_wildcards(str1, str2, min_length=0):
'''
获取2个字符串的通配符字符串,
length，2个*之间的字符串的最小长度，默认为0。
如果小于这个长度，那么会变成*；如果min_length=1，*a* -> *
'''
if not str1 or not str2:
return None
num1 = str1.count('.')
num2 = str2.count('.')
if num2 < num1:
num1 = num2
diff = Differ().compare(str1, str2)
diff = Differ().compare(str2, str1)
# print('-' * 100)
# print(str1, str2)
# print(num1, num2)
wildcards = ''
for item in list(diff):
if '-' in item or '+' in item:
if not wildcards.endswith('*'):
wildcards = wildcards + '*'
else:
wildcards = wildcards + item.strip()
if not wildcards:
return wildcards
result = ''
if min_length > 0:
if wildcards[0] == '*':
result = '*'
is_first = True
for item in wildcards.split('*'):
if is_first:
is_first = False
if len(item) < min_length and not result.endswith('*'):
result = result + '*'
elif not result.endswith('*'):
result = result + '*'
if len(item) > min_length:
result = result + item
elif '.' in item:
result = result + '.'
else:
result = wildcards
return result
[docs]def get_wildcards_in_list(str_list, min_length=0):
'''
获取一个通配字符串，可以通配符该列表里面所有的字符串。
'''
wildcards = str_list[0]
str_list.remove(wildcards)
for item in str_list:
wildcards = get_wildcards(wildcards, item, min_length)
return wildcards
[docs]def get_best_wildcard_from_list(str1, str_list, min_length=0):
'''
从列表str_list中，找出一个与str最相似的通配字符串。
'''
best_radio = 0.0
best_str = ''
for sss in str_list:
radio = get_ratio(str1, sss)
if best_radio < radio:
best_radio = radio
best_str = sss
return get_wildcards(str1, best_str, min_length)
[docs]def get_ratio(str1, str2, weight=3):
len1 = len(str1)
len2 = len(str2)
if len1 < weight or len2 < weight:
return 0
# print(int(len1 / len2 + len2 / len1))
if int(len1 / len2 + len2 / len1) >= weight:
return 0
return SequenceMatcher(None, str1, str2).ratio()
[docs]def gen_wildcard_str(str1, str2, min_length=0):
'''
get commom opcode
'''
result = ''
fcp = find_common_opcodes(str1, str2)
for key, value in fcp[0]:
if key and len(value) > min_length:
result = result + value + '*'
result = result.replace('**', '*')
if result.endswith('*'):
return result[:-1]
return result
[docs]def longest_common_subopcode(s1, s2):
'''
如果是2个普通串还好，但是，如果里面包含*，这种符号，那就完蛋了
'''
m = [[0] * (1 + len(s2)) for i in range(1 + len(s1))]
longest, x_longest = 0, 0
for x in range(1, 1 + len(s1)):
for y in range(1, 1 + len(s2)):
if s1[x - 1] == s2[y - 1]:
m[x][y] = m[x - 1][y - 1] + 1
if m[x][y] > longest:
longest = m[x][y]
x_longest = x
else:
m[x][y] = 0
if '*' not in s1:
end = x_longest & ~1
longest = longest - (x_longest - end)
start = end - (longest & ~1)
subopcode = s1[start:end]
else:
start = x_longest - longest
end = x_longest
subopcode = s1[start:end]
if subopcode.startswith('*') and len(subopcode) % 2 == 0:
subopcode = subopcode[0:-1]
return subopcode
[docs]def find_common_opcodes(s1, s2): # used recursively
if s1 == '' or s2 == '':
return [], []
com = longest_common_subopcode(s1, s2)
if len(com) < 2:
return ([(0, s1)], [(0, s2)])
s1_bef, _, s1_aft = s1.partition(com)
s2_bef, _, s2_aft = s2.partition(com)
before = find_common_opcodes(s1_bef, s2_bef)
after = find_common_opcodes(s1_aft, s2_aft)
return (before[0] + [(1, com)] + after[0],
before[1] + [(1, com)] + after[1])
[docs]def longest_common_substring(s1, s2):
m = [[0] * (1 + len(s2)) for i in range(1 + len(s1))]
longest, x_longest = 0, 0
for x in range(1, 1 + len(s1)):
for y in range(1, 1 + len(s2)):
if s1[x - 1] == s2[y - 1]:
m[x][y] = m[x - 1][y - 1] + 1
if m[x][y] > longest:
longest = m[x][y]
x_longest = x
else:
m[x][y] = 0
return s1[x_longest - longest:x_longest]
[docs]def find_common_patterns(s1, s2): # used recursively
if s1 == '' or s2 == '':
return [], []
com = longest_common_substring(s1, s2)
if len(com) < 2:
return ([(0, s1)], [(0, s2)])
s1_bef, _, s1_aft = s1.partition(com)
s2_bef, _, s2_aft = s2.partition(com)
before = find_common_patterns(s1_bef, s2_bef)
after = find_common_patterns(s1_aft, s2_aft)
return (before[0] + [(1, com)] + after[0],
before[1] + [(1, com)] + after[1])
# 将opcode 分成opcode数组，然后，再diff，也许效果会好一些，只要有差异就不接受？
if __name__ == '__main__':
print('hello')
a = '*o*.*.*.*m*'
c = '*.*o*.*.*m'
b = 'mb.cvf.jt.jge.pt.oirg'
d = 'com.android.dyad.MyApplication'
d2 = 'com.system.activity.MyApplication'
e = 'com.wjbl.mio.efj.ycgh.rywy.prl'
print(get_wildcards(a, b))
print(gen_wildcard_str(d, d2))
print(get_wildcards(c, b))
print(get_wildcards(d, e, 1))