Question

从两个多线字符串中获得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？

正则表达式会成为解决方案吗？

Answer 1

最简单的黑客，使用split()来归功@Chris。

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

#driver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@ekhumoro使用replace进行另一次黑客攻击，无需任何join计算。

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

Answer 2

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，并取出设定的差异 - 即B中所有不在A中的东西。然后取出该结果并将其全部加入一个字符串中。

编辑：这是一种流行的方式来说明@ShreyasG所说的话 - [x代表x如果x不在y中] ......

Answer 3

这基本上是@ Godron629的答案，但由于我无法发表评论，我将在此处稍作修改：将difference更改为symmetric_difference，以便集合的顺序不会无所谓。

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 4

在@Chris_Rands注释的基础上，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一个而另一个中）：

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 5

import itertools as it


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

可选地

import collections as ct


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

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

Answer 6

您可以使用以下功能：

def __slave(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 = __slave(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 = __slave(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

使用示例打印差异（a，b）

Python - 获取字符串之间的差异

6 个答案: