python字典和双端队列根据某些条件打印所需的输出

问题描述

我有CSV File，其中包含一些来自采矿的数据
我想按照要求的格式打印

required Format
A -> B -> C -> D -> E -> F
A -> B -> C -> I
X -> Y -> Z
X -> Y -> P -> Q
A -> B -> K -> L

a.csv文件

## code
from collections import deque
import pandas as pd

data = pd.read_csv("a.csv")

data['Start'] = data['Start'].str.replace(' ','_')
data['End'] = data['End'].str.replace(' ','_')

fronts = dict()
backs = dict()

sequences = []
position_counter = 0

selector = data.apply(lambda row: row.str.extractall("([\w+\d]+)"),axis=1)

for relation in selector:
    front,back = relation[0]
    llist = deque((front,back))
    
    finb =  front in backs.keys()
    
    if finb:
        position = backs[front]
        llist2 = sequences[position]
        back_llist2 = llist2.pop()
        llist = llist2 + llist
        sequences[position] = llist
        backs[llist[-1]] = position
        if front in fronts.keys():
            del fronts[front]
        if back_llist2 in backs.keys():
            del backs[back_llist2]

    if not finb: 
        sequences.append(llist)
        fronts[front] = position_counter
        backs[back] = position_counter
        position_counter += 1

data = []
for s in sequences:
    data.append(' -> '.join(str(el) for el in s)) 
data

我得到的是：
'A-> B-> C-> D-> E-> F'
'C-> I'
'A-> N-> A'
'X-> Y-> Z'
'Y-> P-> Q'
'B-> K-> L'
'X1-> Y1'

解决方法

您需要在现有路径中搜索新行的开始元素。如果找到，请追加到现有路径或复制该路径并追加新的end元素。

尝试以下代码：

ss = '''
A B
B C
C D
D E
E F
C I
A N
N A
X Y
Y Z
Y P
P Q
B K
K L
X1 Y1
'''.strip()

lst = []
for r in ss.split('\n'):
   lst.append(r.split())

################

paths = []
for e in lst:  # each row in source data
   pnew = []  # new path
   for p in paths:
       if e[0] in p:  # if start in existing path
           if p.index(e[0]) == len(p)-1:  # if end of path
               p.append(e[1])  # add to path
           else:
               pnew.append(p[:p.index(e[0])+1]+[e[1]])  # copy path then add
           break
   else:  # loop completed,not found
      paths.append(list(e)) # create new path
   if len(pnew):  # copied path
      paths.extend(pnew) # add copied path

print('\n'.join([' => '.join(e) for e in paths]))

输出

A => B => C => D => E => F
A => B => C => I
A => N => A
X => Y => Z
X => Y => P => Q
A => B => K => L
X1 => Y1

根据源数据，A->N->A和X1->Y1是正确的。我不知道为什么要将它们排除在所需的输出中。