Python - 獲取字符串之間的差異

從兩個多行字符串中獲得差異的最佳方式是什麼？Python - 獲取字符串之間的差異

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)

這將產生：

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

什麼也正是得到的最好的方法：

testing this is working 2？

將正則表達式在這裏的解決方案？

來源

2017-09-27 Rekovni

'b.split（a）'。？ –

該死的@Chris_Rands。從來沒有想過這個！尼斯黑客。 –

@Chris_Rands很好的黑客，但這不是一個高效的方式去做 –

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...'

從根本上讓每串一組線，並採取了一系列的差異 - 即B的所有事情是不是A.然後把這個結果加入到一個字符串中。

編輯：這是說什麼@ShreyasG說的conveluded方式 - [X對於x如果x不以y] ...

來源

2017-09-27 17:02:20 Godron629

大廈@Chris_Rands評論，你可以使用splitlines（）操作太（如果你的字符串是多行，你想不存在之一，但其他行）：

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

預期輸出是：

[' testing this is working 2']

來源

2017-09-27 16:54:24 ShreyasG

最簡單的黑客，信貸@Chris，通過使用split()。

注意：您需要確定哪一個是較長的字符串，並將其用於分割。

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

＃司機值

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

編輯：感謝@ekhumoro使用replace另一個黑客，而無需任何所需的join計算。

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

來源

2017-09-27 16:55:53

'b.replace（a，''）'更簡單，更快速，更有意義。 – ekhumoro

哈哈，喜歡這個。另一個好的黑客。我從未想過以這種方式使用'split'或'replace'。謝謝@ekhumoro !! –

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())

來源

2017-09-27 17:00:28 pylang

這基本上是@ Godron629的答案，但因爲我不能發表評論，我在這裏稍作修改：將difference更改爲symmetric_difference，以便集合的順序無關緊要。

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...'

來源

2017-09-27 17:16:23 Keith

Python - 獲取字符串之間的差異

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