Python

Question

从两个多行字符串中获取差异的最佳方法是什么？

a = 'testing this is working \n testing this is working 1 \n'
b = 'testing this is working \n testing this is working 1 \n testing this is working 2'

diff = difflib.ndiff(a,b)
print ''.join(diff)

这会产生：

  t  e  s  t  i  n  g     t  h  i  s     i  s     w  o  r  k  i  n  g     
     t  e  s  t  i  n  g     t  h  i  s     i  s     w  o  r  k  i  n  g     1     
+  + t+ e+ s+ t+ i+ n+ g+  + t+ h+ i+ s+  + i+ s+  + w+ o+ r+ k+ i+ n+ g+  + 2

准确获取的最佳方法是什么：

testing this is working 2?

正则表达式会是这里的解决方案吗？

Answer 1

基于@Chris_Rands 评论，您也可以使用 splitlines() 操作（如果您的字符串是多行并且您希望该行不存在于其中一行中，而存在于另一行中）：

b_s = b.splitlines()
a_s = a.splitlines()
[x for x in b_s if x not in a_s]

预期输出为：

[' testing this is working 2']

Answer 2

最简单的 Hack，学分，使用 split()。

注意：您需要确定哪个是较长的字符串，并将其用于拆分。

if len(a)>len(b): 
   res=''.join(a.split(b))             #get diff
else: 
   res=''.join(b.split(a))             #get diff

print(res.strip())                     #remove whitespace on either sides

# 驱动值

IN : a = 'testing this is working \n testing this is working 1 \n' 
IN : b = 'testing this is working \n testing this is working 1 \n testing this is working 2'

OUT : testing this is working 2

编辑： 感谢使用 replace 进行另一次破解，不需要任何 join 计算。

if len(a)>len(b): 
    res=a.replace(b,'')             #get diff
else: 
    res=b.replace(a,'')             #get diff

Answer 3

import itertools as it


"".join(y for x, y in it.zip_longest(a, b) if x != y)
# ' testing this is working 2'

或者

import collections as ct


ca = ct.Counter(a.split("\n"))
cb = ct.Counter(b.split("\n"))

diff = cb - ca
"".join(diff.keys())

Answer 4

a = 'testing this is working \n testing this is working 1 \n'
b = 'testing this is working \n testing this is working 1 \n testing this is working 2'

splitA = set(a.split("\n"))
splitB = set(b.split("\n"))

diff = splitB.difference(splitA)
diff = ", ".join(diff)  # ' testing this is working 2, more things if there were...'

基本上使每个字符串成为一组行，并取集合差异 - 即 B 中不在 A 中的所有内容。然后获取该结果并将其全部连接到一个字符串中。

编辑：这是表达@ShreyasG 所说内容的一种概括方式 - [x for x if x not in y]...

Answer 5

这基本上是@Godron629 的回答，但由于我无法发表评论，所以我将其发布在这里并稍加修改：将 difference 更改为 symmetric_difference 以便集合的顺序没关系。

a = 'testing this is working \n testing this is working 1 \n'
b = 'testing this is working \n testing this is working 1 \n testing this is working 2'

splitA = set(a.split("\n"))
splitB = set(b.split("\n"))

diff = splitB.symmetric_difference(splitA)
diff = ", ".join(diff)  # ' testing this is working 2, some more things...'

Answer 6

您可以使用以下功能：

def helper(a, b):

    for i, l_a in enumerate(a):
        if b == l_a:
            return i
    return -1

def diff(a, b):

    t_b = b
    c_i = 0
    for c in a:

        t_i = helper(t_b, c)
        if t_i != -1 and (t_i > c_i or t_i == c_i):
            c_i = t_i
            t_b = t_b[:c_i] + t_b[c_i+1:]

    t_a = a
    c_i = 0
    for c in b:

        t_i = helper(t_a, c)
        if t_i != -1 and (t_i > c_i or t_i == c_i):
            c_i = t_i
            t_a = t_a[:c_i] + t_a[c_i+1:]

    return t_b + t_a

用法示例print diff(a, b)

Python - 获取字符串之间的差异

Python - getting just the difference between strings

difflib

python-2.7