在列表中搜索模式：python 正则表达式

Question

在数据分析并获得所需结果后，我将该结果附加到列表
现在我需要检索或分离结果（搜索模式并获取它）

代码：

data = []
data.append('\n'.join([' -> '.join(e) for e in paths]))

列表包含此数据：

CH_Trans -> St_1 -> WDL
TRANSFER_Trn -> St_1
Access_Ltd -> MPL_Limited
IPIPI -> TLC_Pvt_Ltd
234er -> Three_Star_Services -> Asian_Pharmas -> PPP_Channel
Sonata_Ltd -> Three_Star_Services
Arc_Estates -> Russian_Hosp
A -> B -> C -> D -> E -> F
G -> H
ZN_INTBNKOUT_SET -> -2008_1 -> X
ZZ_1_ -> AA_2 -> AA_3 -> ZZ_1_
XYZ- -> ABC -> XYZ-
SSS -> BBB -> SSS
Rock_8CC -> Russ -> By_sus -> Rock_8CC

Note : Display or Retrieve Pattern Which has more than two symbol of type[->]
( Txt -> txt -> txt )

我正在尝试通过正则表达式完成它

for i in data:
    regex = ("\w+\s->\s\w+\s->\s\w+")             
    match = re.findall(regex, i,re.MULTILINE)
    print(match)

Regex Expression I Tried But Unable to get Requried Result
#\w+\s->\s\w+\s->\s\w+
#\w+\s[-][>]\s\w+\s[-][>]\s\w+
#\w+\s[-][>]\s\w+\s[-][>]\s\w+\s[-][>]\s\w+

Result I Got
['CH_Trans-> St_1-> WDL', '234er -> Three_Star_Services -> Asian_Pharmas',
 'A -> B -> C', 'D -> E -> F', 'ZZ_1_ -> AA_2 -> AA_3', 
'SSS -> BBB -> SSS', 'Rock_8CC -> Russ -> By_sus']

需要的结果我要获取的是

----Pattern I------
CH_Trans -> St_1 -> WDL
234er -> Three_Star_Services -> Asian_Pharmas -> PPP_Channel
A -> B -> C -> D -> E -> F
ZN_INTBNKOUT_SET -> -2008_1 -> X


# Pattern II Consists of Patterns which are same i.e[ Fist_ele & Last_Ele Is Same]
----Pattern II------
ZZ_1_ -> AA_2 -> AA_3 -> ZZ_1_
XYZ- -> ABC -> XYZ-
SSS -> BBB -> SSS
Rock_8CC -> Russ -> By_sus -> Rock_8CC

Answer 1

请您尝试以下作为起点：

regex = r'^\S+(?:\s->\s\S+){2,}$'
for i in data:
    m = re.match(regex, i)
    if (m):
        print(m.group())

结果（模式 I + 模式 II）：

CH_Trans -> St_1 -> WDL
234er -> Three_Star_Services -> Asian_Pharmas -> PPP_Channel
A -> B -> C -> D -> E -> F
ZN_INTBNKOUT_SET -> -2008_1 -> X
ZZ_1_ -> AA_2 -> AA_3 -> ZZ_1_
XYZ- -> ABC -> XYZ-
SSS -> BBB -> SSS
Rock_8CC -> Russ -> By_sus -> Rock_8CC

正则表达式^\S+(?:\s->\s\S+){2,}$的解释：

^\S+       start with non-blank string
(?: ... )  grouping
\s->\s\S+  a blank followed by "->" followed by a blank and non-blank string
{2,}       repeats the previous pattern (or group) two or more times
$          end of the string

关于模式二，请说：

regex = r'^(\S+)(?:\s->\s\S+){1,}\s->\s\1$'
for i in data:
    m = re.match(regex, i)
    if (m):
        print(m.group())

结果：

ZZ_1_ -> AA_2 -> AA_3 -> ZZ_1_
XYZ- -> ABC -> XYZ-
SSS -> BBB -> SSS
Rock_8CC -> Russ -> By_sus -> Rock_8CC

正则表达式r'^(\S+)(?:\s->\s\S+){1,}\s->\s\1$'的解释：

- ^(\S+)     captures the 1st element and assigns \1 to it
- (?: ... )  grouping
- \s->\s\S+  a blank followed by "->" followed by a blank and non-blank string
- {1,}       repeats the previous pattern (or group) one or more times
- \s->\s\1   a blank followed by "->" followed by a blank and the 1st element \1
- $          end of the string

为了得到模式 I 的结果，我们可能需要从第一个结果中减去模式 II 的列表。
如果我们可以说：

regex = r'^(\S+)(?:\s->\s\S+){2,}(?<!\1)$'

它将排除最后一个元素与第一个元素不同的字符串，然后我们可以直接获得模式 I 的结果，但正则表达式导致错误说 "group references in lookbehind assertions" 到目前为止。