python3，difflib序列匹配器

Question

以下内容接受两个字符串，比较差异并将它们作为相同以及它们的差异返回，用空格分隔（保持最长字符串的长度。

代码中的注释区，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)]

在将块排列成有组织的格式bla1、bla2 之后，其中每个差异都存储为带有其开始和结束位置的字符串，例如['v', 33, 34] 用于每个单独的字符串。在此之后，我尝试插入空格以匹配必要的长度和分隔因子，这就是代码开始中断的地方。

如果有人可以看一下，请！

Answer 1

我已经解决了这个问题，由于没有人发布回复，我将发布进度和解决方案。以下代码是 progress ...它在处理偏移量较小但在出现较大差异时开始中断的变化时效果很好，特别是在保持间距（偏移量）以匹配两者时。

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，无法继续编写此代码，并求助于子处理diffutils ...这是许多艰苦编码的绝佳替代方案！