我一直在研究Paul Heckel 的 Diff Algorithm,但我似乎并没有完全理解它。
我复制了 Python 代码中所示的步骤 1-5,但我无法使用算法的最后一步来显示差异。如果有人解释最后一步以及 Python 代码,我将不胜感激。
另外,我不完全理解为什么您需要参考第 4 步和第 5 步中的表格行,所以对此的解释也将是惊人的!
非常感谢
这是我当前的代码:
def find_diff(current_file_as_list, different_file_as_list):
N = current_file_as_list
O = different_file_as_list
table = {}
OA = []
NA = []
for i in O:
OA.append(i)
for i in N:
NA.append(i)
# First pass
i = 0
for line in N:
if not line in table:
table[line] = {}
table[line]["NC"] = 1
else:
if table[line]["NC"] == 1:
table[line]["NC"] = 2
else:
table[line]["NC"] = "many"
NA[i] = table[line]
i += 1
# second pass
j = 0
for line in O:
if not line in table:
table[line] = {}
table[line]["OC"] = 1
else:
if not "OC" in table[line]:
table[line]["OC"] = 1
elif table[line]["OC"] == 1:
table[line]["OC"] = 2
else:
table[line]["OC"] = "many"
table[line]["OLNO"] = j # Gets overwritten with multiple occurrences.
# Check to see if this is the intended implementation.
# Maybe only relevant for "OC" == "NC" == 1
OA[j] = table[line]
j += 1
# third pass
i = 0
for i in range(0, len(NA)):
# Check if they appear in both files
if "OC" in NA[i] and "NC" in NA[i]:
# Check if they appear exactly once
if NA[i]["OC"] == NA[i]["NC"] == 1:
olno = NA[i]["OLNO"]
NA[i], OA[olno] = olno, i
i += 1
# fourth pass
# ascending
for i in range(0, len(NA)):
for j in range(0 , len(OA)):
if NA[i] == OA[j] and i + 1 < len(NA) and j + 1 < len(OA) and NA[i + 1] == OA[j + 1]:
OA[j + 1] = table[O[i + 1]]
NA[i + 1] = table[N[j + 1]]
# fifth pass
# descending
for i in range(len(NA) - 1, 0, -1):
for j in range(len(OA) - 1, 0, -1):
if NA[i] == OA[j] and i - 1 > 0 and j - 1 > 0 and NA[i - 1] == OA[j - 1]:
OA[j - 1] = table[O[i - 1]]
NA[i - 1] = table[N[j - 1]]
# final step implementation should go here but I'm not sure how to approach it but this is my current attempt (which I am certain is wrong):
k = 0
array = []
for i in range(0, len(NA)):
if isinstance(NA[i], int):
array.append("= " + str(N[i]))
k = NA[i] + 1
elif isinstance(NA[i], dict):
array.append("+ " + N[i])
for j in range(k, len(OA)):
k = j + 1
print("j - " + str(j))
if not isinstance(OA[j], int):
array.append("- " + O[j])
else:
break
您可以将任意两个字符串或字符串列表作为输入传递给函数,例如。find_diff(“你好”,“地狱”)