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Python-仅获取字符串之间的差异

difflib python-2.7 python

Rekovni · 技术社区 · 8 年前

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)

这将产生:

  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

testing this is working 2 ?

正则表达式是这里的解决方案吗?

6 回复 | 直到 8 年前

Kaushik NP 8 年前

最简单的黑客,信用卡 @Chris ,通过使用 split() .

注: 您需要确定哪个字符串较长,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

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: 幸亏 @ekhumoro replace ,无需任何 join 需要计算。

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

Godron629 8 年前

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...'

本质上,把每个字符串都变成一组行,取集合差,即B中不在a中的所有东西,然后取结果,将其全部合并到一个字符串中。

Keith 8 年前

这基本上是@Godron629的答案,但由于我无法发表评论,我在这里发布了一个小小的修改:改变 difference 对于 symmetric_difference 所以集合的顺序无关紧要。

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...'

pylang 8 年前

import itertools as it


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

import collections as ct


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

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

ShreyasG 8 年前

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']

Michael Scott Asato Cuthbert 4 年前

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

使用示例 打印差异(a,b)