python diff SequenceMatcher - 修补列表

Question

我修补一个列表，使其看起来像另一个列表：

a = [x for x in "qabxcd"]
b = [x for x in "abycdf"]
c = a[:]
s = SequenceMatcher(None, a, b)
for tag, i1, i2, j1, j2 in s.get_opcodes():
    print ("%7s a[%d:%d] (%s) b[%d:%d] (%s)" % 
    (tag, i1, i2, a[i1:i2], j1, j2, b[j1:j2]))
    if tag == "delete":
        del c[i1:i2]
    elif tag == "replace":
        c[i1:i2] = b[j1-1:j2-1]
    elif tag == "insert":
        c[i1:i2] = b[j1:j2]
print c
print b
print c == b
a == b

但是列表不相等：

 delete a[0:1] (['q']) b[0:0] ([])
  equal a[1:3] (['a', 'b']) b[0:2] (['a', 'b'])
replace a[3:4] (['x']) b[2:3] (['y'])
  equal a[4:6] (['c', 'd']) b[3:5] (['c', 'd'])
 insert a[6:6] ([]) b[5:6] (['f'])
['a', 'b', 'x', 'b', 'd', 'f']
['a', 'b', 'y', 'c', 'd', 'f']
False

有什么问题吗？

Answer 1

所有的动作都会改变索引。当我愿意这样做时，我必须计算变化：

a = [x for x in "abyffgh fg99"]
b = [x for x in "999aby99ff9h9"]
c = a[:]

s = SequenceMatcher(None, a, b)

i = 0
for tag, i1, i2, j1, j2 in s.get_opcodes():
    print ("%7s a[%d:%d] (%s) b[%d:%d] (%s) c[%d:%d] (%s)" % 
    (tag, i1, i2, a[i1:i2], j1, j2, b[j1:j2], i1, i2, c[i1 + i:i2 + i]))
    if tag == "delete":
        del c[i1 + i:i2 + i]
        i -= i2 - i1
    elif tag == "replace":
        c[i1 + i:i2 + i] = b[j1:j2]
        i -= i2 - i1 - j2 + j1
    elif tag == "insert":
        c[i1 + i:i2 + i] = b[j1:j2]
        i += j2 - j1
    print c
    print i
print c
print b
print c == b
a == b

输出：

['9', '9', '9', 'a', 'b', 'y', '9', '9', 'f', 'f', '9', 'h', ' ', 'f', 'g', '9', '9']
5
 delete a[7:10] ([' ', 'f', 'g']) b[12:12] ([]) c[7:10] ([' ', 'f', 'g'])
['9', '9', '9', 'a', 'b', 'y', '9', '9', 'f', 'f', '9', 'h', '9', '9']
1
  equal a[10:11] (['9']) b[12:13] (['9']) c[10:11] (['h'])
['9', '9', '9', 'a', 'b', 'y', '9', '9', 'f', 'f', '9', 'h', '9', '9']
1
 delete a[11:12] (['9']) b[13:13] ([]) c[11:12] (['9'])
['9', '9', '9', 'a', 'b', 'y', '9', '9', 'f', 'f', '9', 'h', '9']
-1
['9', '9', '9', 'a', 'b', 'y', '9', '9', 'f', 'f', '9', 'h', '9']
['9', '9', '9', 'a', 'b', 'y', '9', '9', 'f', 'f', '9', 'h', '9']
True

Answer 2

我想我可以明白为什么：

s.get_opcodes()

返回的5元组在容器的初始状态下是有效的，也就是说，如果你的对象发生变化，它们必须进行调整：这就是删除操作的情况，特别是，它发生了变化索引（这就是为什么

'x'

没有变成

'y'

）。

据我所知，删除操作是唯一更改索引的操作，因此我将用标记替换已删除的项目（我使用“#”）并在最后将其删除：

>>> c = a[:]
>>> for tag, i1, i2, j1, j2 in s.get_opcodes():
    print ("%7s a[%d:%d] (%s) b[%d:%d] (%s)" % 
    (tag, i1, i2, a[i1:i2], j1, j2, b[j1:j2]))
    if tag == "delete":
        c[i1:i2] = ['#' for i in range(i1, i2)]
    elif tag == "replace":
        c[i1:i2] = b[j1:j2]
    elif tag == "insert":
        c[i1:i1] = b[j1:j2]
    print c


 delete a[0:1] (['q']) b[0:0] ([])
['#', 'a', 'b', 'x', 'c', 'd']
  equal a[1:3] (['a', 'b']) b[0:2] (['a', 'b'])
['#', 'a', 'b', 'x', 'c', 'd']
replace a[3:4] (['x']) b[2:3] (['y'])
['#', 'a', 'b', 'y', 'c', 'd']
  equal a[4:6] (['c', 'd']) b[3:5] (['c', 'd'])
['#', 'a', 'b', 'y', 'c', 'd']
 insert a[6:6] ([]) b[5:6] (['f'])
['#', 'a', 'b', 'y', 'c', 'd', 'f']
>>> c = [i for i in c if i != '#']
>>> c
['a', 'b', 'y', 'c', 'd', 'f']
>>> 

Answer 3

OP 的示例可以通过将

c

初始化为空字符串来稍微简化。那么就不需要处理

"delete"

的情况了：

a = "qabxcd"
b = "abycdf"
c = ""  # <-- initialize to empty string

s = SequenceMatcher(None, a, b)
for tag, i1, i2, j1, j2 in s.get_opcodes():
    ...
    if tag == "equal":
        c += a[i1:i2]
    elif tag in ["replace", "insert"]:
        c += b[j1:j2]

print(c)
print(b)
assert b == c