问题描述
以下内容接受两个字符串，比较差异并将它们返回为相同和差异，并以空格分隔(保持最长字符串的长度.
the following takes in two strings, compares differences and return them both as identicals as well as their differences, separated by spaces (maintaining the length of the longest sting.
注释区域是应返回的4个字符串.
The commented area in the code, are the 4 strings that should be returned.
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)]
在将块排列为有组织的格式bla1，bla2之后，其中每个差异都以字符串形式存储，字符串的开始和结束位置为每个单独的字符串，例如['v', 33, 34].之后，我尝试插入空格以匹配必要的长度和分隔系数，这就是代码开始出现的地方.
After ranging the blocks into an organised format bla1, bla2 where each difference is stored as a string with its start and end position eg ['v', 33, 34] for each separate string. After this, I attempt to insert spaces to match the length and separation factors necessary and this is where the code starts to break.
请有人来看看！
推荐答案
我已经解决了这个问题，并且由于没有人发布回复，因此我将发布进度和解决方案.以下代码是 progress (进步) ...，它在处理偏移量较小但出现较大差异时开始中断的变体，效果很好，特别是在保持两者的间距(偏移量)方面.
I have worked through resolving this, and since no one has posted a response I will post the progress and solution. The following code is progress ... it worked well when dealing with variations that had less offset but began to break when getting into larger differences, specifically in maintaining spacing (offset) in matching up the two.
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，我忙于继续对此进行编码，并采取了子处理方法 diffutils ...这是许多艰苦的编码的绝佳替代品！
Unfortunately I have been too busy to continue coding this and have resorted to sub-processing diffutils ... this is a wonderful alternative to a lot of painstaking coding!
这篇关于python3，difflib SequenceMatcher的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！