python3，衍射图SequenceMatcher

Question

下面是两个字符串，比较它们之间的差异，并将它们作为同义词以及它们的差异返回，用空格分隔(保持最长的刺的长度)。

代码中的注释区域是应该返回的4个字符串。

from difflib import SequenceMatcher




t1 = 'betty:  backstreetvboysareback"give.jpg"LAlarrygarryhannyhref="ang"_self'

t2 = 'bettyv:  backstreetvboysareback"lifeislike"LAlarrygarryhannyhref="in.php"_self'


#t1 = 'betty :  backstreetvboysareback" i e      "LAlarrygarryhannyhref=" n    "_self'
#t2 = 'betty :  backstreetvboysareback" i e      "LAlarrygarryhannyhref=" n    "_self'

#o1 = '                                g v .jpg                          g           '
#o2 = '     v                          l f islike                        i .php      '



matcher = SequenceMatcher(None, t1, t2)
blocks = matcher.get_matching_blocks()

bla1 = []
bla2 = []

for i in range(len(blocks)):
    if i != len(blocks)-1:
        bla1.append([t1[blocks[i].a + blocks[i].size:blocks[i+1].a], blocks[i].a + blocks[i].size, blocks[i+1].a])
        bla2.append([t2[blocks[i].b + blocks[i].size:blocks[i+1].b], blocks[i].b + blocks[i].size, blocks[i+1].b])



cnt = 0
for i in range(len(bla1)):


    if bla1[i][1] < bla2[i][1]:
        num = bla2[i][1] - bla1[i][1]
        t2 = t2[0:bla2[i][1]] + ' '*num + t2[bla2[i][1]:len(t2)]
        bla2[i][0] = ' '*num + bla2[i][0]
        bla2[i][1] = bla1[i][1]

    if bla2[i][1] < bla1[i][1]:
        num = bla1[i][1] - bla2[i][1]
        t1 = t1[0:bla1[i][1]] + ' '*num + t1[bla1[i][1]:len(t1)]
        bla1[i][0] = ' '*num + bla1[i][0]
        bla1[i][1] = bla2[i][1]

    if bla1[i][2] > bla2[i][2]:
        num = bla1[i][2] - bla2[i][2]
        t2 = t2[0:bla2[i][2]] + ' '*num + t2[bla2[i][2]:len(t2)]
        bla2[i][0] = bla2[i][0] + ' '*num
        bla2[i][2] = bla1[i][2]

    if bla2[i][2] > bla1[i][2]:
        num = bla2[i][2] - bla1[i][2]
        t1 = t1[0:bla1[i][2]] + ' '*num + t1[bla1[i][2]:len(t1)]
        bla1[i][0] = bla1[i][0] + ' '*num
        bla1[i][2] = bla2[i][2]




t11 = []
t11 = t1[0:bla1[0][1]]
t11 += t1[bla1[0][2]:bla1[1][1]]
t11 += t1[bla1[1][2]:bla1[2][1]]
t11 += t1[bla1[2][2]:bla1[3][1]]
t11 += t1[bla1[3][2]:bla1[4][1]]
t11 += t1[bla1[5][2]:bla1[6][1]]
t11 += t1[bla1[6][2]:len(t1)]

t12 = []
t12 = t2[0:bla1[0][1]]
t12 += t2[bla1[0][2]:bla1[1][1]]
t12 += t2[bla1[1][2]:bla1[2][1]]
t12 += t2[bla1[2][2]:bla1[3][1]]
t12 += t2[bla1[3][2]:bla1[4][1]]
t12 += t2[bla1[5][2]:bla1[6][1]]
t12 += t2[bla1[6][2]:len(t2)]

将块排列成一个有组织的格式bla1，bla2，其中每个差异都存储为一个字符串，其起始位置和结束位置(例如，每个单独字符串的['v', 33, 34] )。在此之后，我尝试插入空格以匹配所需的长度和分离因子，这就是代码开始中断的地方。

如果有人能看一看就好了！

Answer 1

我一直致力于解决这个问题，而且由于没有人发布响应，我将发布progress和解决方案。以下代码是progress ..。它很好地处理变化，有较少的偏移，但开始打破进入更大的差异，特别是在保持间距(偏移)，以匹配两者。

from difflib import SequenceMatcher
import pdb


t1 = 'betty:  backstreetvboysareback"give.jpg"LAlarrygarryhannyhref="ang"_self'

t2 = 'betty:  backstreetvboysareback"lol.jpg"LAlarrygarryhannyhref="ang"_self'

#t2 = 'bettyv:  backstreetvboysareback"lifeislike"LAlarrygarryhannyhref="in.php"_selff'

#t2 = 'LA'
#t2 = 'c give.'
#t2 = 'give.'




#t1 = 'betty :  backstreetvboysareback" i e      "LAlarrygarryhannyhref=" n    "_self'
#t2 = 'betty :  backstreetvboysareback" i e      "LAlarrygarryhannyhref=" n    "_self'

#o1 = '                                g v .jpg                          g           '
#o2 = '     v                          l f islike                        i .php      '



matcher = SequenceMatcher(None, t1, t2)
blocks = matcher.get_matching_blocks()

#print(len(blocks))

bla1 = []
bla2 = []

#bla = (string), (first pos), (second pos), (pos1 + pos2), (pos + pos2 total positions added togeather)
dnt = False
for i in range(len(blocks)):

    if i == 0:
      if blocks[i].a != 0 and dnt == False:
        bla1.append([t1[blocks[i].a:blocks[i].b], 0, blocks[i].a, 0, 0])
        bla2.append([t2[blocks[i].a:blocks[i].b], 0, blocks[i].b, 0, 0])
        dnt = True

      if blocks[i].b != 0 and dnt == False:
        bla2.append([t2[blocks[i].a:blocks[i].b], 0, blocks[i].b, 0, 0])
        bla1.append([t1[blocks[i].a:blocks[i].b], 0, blocks[i].a, 0, 0])
        dnt = True

    if i != len(blocks)-1:
        print(blocks[i])

        bla1.append([t1[blocks[i].a + blocks[i].size:blocks[i+1].a], blocks[i].a + blocks[i].size, blocks[i+1].a, 0, 0])
        bla2.append([t2[blocks[i].b + blocks[i].size:blocks[i+1].b], blocks[i].b + blocks[i].size, blocks[i+1].b, 0, 0])

#pdb.set_trace()

ttl = 0
for i in range(len(bla1)):
  cnt = bla1[i][2] - bla1[i][1]
  if cnt != 0:
    bla1[i][3] = cnt
  ttl = ttl + cnt
  bla1[i][4] = ttl

ttl = 0
for i in range(len(bla2)):
  cnt = bla2[i][2] - bla2[i][1]
  if cnt != 0:
    bla2[i][3] = cnt
  ttl = ttl + cnt
  bla2[i][4] = ttl

print(bla1)
print(bla2)

tt1 = ''
dif = 0
i = 0
while True:

  if i == 0:
    if bla1[i][3] >= bla2[i][3]: dif = bla1[i][3]
    if bla1[i][3] < bla2[i][3]: dif = bla2[i][3]  
    tt1 += t1[:bla1[i][1]] + '_'*dif

  if i <= len(bla1) -1:

    if bla1[i][3] >= bla2[i][3]: dif = bla1[i][3]
    if bla1[i][3] < bla2[i][3]: dif = bla2[i][3]

    if len(bla1) != 1:
      if i == 0: tt1 += t1[bla1[i][1] + bla1[i][3]:bla1[i+1][1]]
      if i != 0 and i != len(bla1)-1: tt1 += '_'*dif + t1[bla1[i][1] + bla1[i][3]:bla1[i+1][1]]
      if i == len(bla1)-1: tt1 += '_'*dif + t1[bla1[i][1] + bla1[i][3]:len(t1)]

    i = i+1
    print('t1 = ' + tt1)

  else:
    break

tt2 = ''
i = 0
dif = 0
while True:

  if i == 0:

    if bla1[i][3] >= bla2[i][3]: dif = bla1[i][3]
    if bla1[i][3] < bla2[i][3]: dif = bla2[i][3]   
    tt2 += t2[:bla2[i][1]] + '_'*dif

  if i <= len(bla2) -1:

    if bla1[i][3] >= bla2[i][3]: dif = bla1[i][3]
    if bla1[i][3] < bla2[i][3]: dif = bla2[i][3]    

    if len(bla2) != 1:
      if i == 0: tt2 += t2[bla2[i][1] + bla2[i][3]:bla2[i+1][1]]
      if i != 0 and i != len(bla1)-1: tt2 += '_'*dif + t2[bla2[i][1] + bla2[i][3]:bla2[i+1][1]]
      if i == len(bla2)-1: tt2 += '_'*dif + t2[bla2[i][1] + bla2[i][3]:len(t2)]

    i = i+1
    print('t2 = ' + tt2)

  else:
    break

  print()

解决方案：

不幸的是，我太忙了，无法继续编写这些代码，并求助于子处理巴氏 .这是一个非常好的选择，许多艰苦的编码！