首先检查字符串中每行的几位数，如果相等，则将这些行的一部分一起打印

如何解决首先检查字符串中每行的几位数，如果相等，则将这些行的一部分一起打印

例如，我有一个字符串

textstring= """
0000 Onn ch=1 n=60 v=50   
0000 Onn ch=1 n=67 v=50
9600 Off ch=1 n=67 v=00
9600 Off ch=1 n=60 v=00
9600 Onn ch=1 n=62 v=50
9600 Onn ch=1 n=69 v=50
1920 Off ch=1 n=69 v=00
1920 Off ch=1 n=62 v=00
"""   

当前四行数字相等时，例如9600，如何一起打印67/60/62/69？ （来自n=之后的四行）

我尝试了以下类似方法，但是我认为它没有按预期运行

for i,char in enumerate(textstring):
    if char=="O" and (textstring[i+1]=="f" or textstring[i+1]=="n"):
        if textstring[i-5]==textstring[i+18] and textstring[i-4]==textstring[i+19] and textstring[i-3]==textstring[i+20] and textstring[i-2]==textstring[i+21]:
            if char=="n" and textstring[i+1]=="=":  #we look for "n=" in the text
                note=int (textstring[i+2]+textstring[i+3])  #we restore the int from string after "n=",the use it as a note
                note2=int (textstring[i+25]+textstring[i+26])
                print(note)
                print(note2)

解决方法

最好在词典中将具有相同代码的行分组，例如：

from collections import defaultdict
from pprint import pprint

d = defaultdict(list)
for line in textstring.splitlines():
  cells = line.split()
  if len(cells) > 1:
    line_id,*values = cells
    d[line_id].append(values)
pprint(d)

这将输出：

defaultdict(<class 'list'>,{'0000': [['Onn','ch=1','n=60','v=50'],['Onn','n=67','v=50']],'1920': [['Off','n=69','v=00'],['Off','n=62','v=00']],'9600': [['Off','v=50']]})

然后您可以轻松设置此字典的格式，例如：

for k,v in d.items():
  print(k,"/".join(e[2][2:] for e in v))

输出为：

0000 60/67
9600 67/60/62/69
1920 69/62

,

通常，如果您有对象集合，则将对象放入类将始终帮助您操纵它们以执行所需的操作。

我在下面放置了示例代码，向您展示了一种非常基本的方法来将对象解析为一个类，然后使用该类来执行您想要的操作。

textstring = """
0000 Onn ch=1 n=60 v=50   
0000 Onn ch=1 n=67 v=50
9600 Off ch=1 n=67 v=00
9600 Off ch=1 n=60 v=00
9600 Onn ch=1 n=62 v=50
9600 Onn ch=1 n=69 v=50
1920 Off ch=1 n=69 v=00
1920 Off ch=1 n=62 v=00
"""   

# You appear to have a list of objects,so lets make a class that represents the object.
# The code inside here is very basic and assumes perfect input.
class MyThing():
    def __init__(self,description: str):
        fields = description.split()
        self.Id: int = int(fields[0])
        self.State: str = fields[1]
        self.Channel: int = int(fields[2][3:])
        self.N: int = int(fields[3][2:])
        self.V: int = int(fields[4][2:])

    def __str__(self):
        # I use .format here but you can use f'' format strings in python >3.6
        return '{:0>4d} {} ch={} n={} v={}'.format(self.Id,self.State,self.Channel,self.N,self.V)

if __name__ == '__main__':
    objects = (MyThing(desc) for desc in textstring.splitlines() if desc != '')

    # create a grouping so you can manipulate objects by matching id's
    grouping = defaultdict(list)
    for thing in objects:
        grouping[thing.Id].append(thing)

    for key in grouping.keys():
        # print out the items that match the group
        print('{:0>4d} {}'.format(key,'/'.join((str(thing.N) for thing in grouping[key]))))

输出：

0000 60/67
9600 67/60/62/69
1920 69/62

,

这是我使用itertools的建议：

import itertools

textstring = """
0000 Onn ch=1 n=60 v=50   
0000 Onn ch=1 n=67 v=50
9600 Off ch=1 n=67 v=00
9600 Off ch=1 n=60 v=00
9600 Onn ch=1 n=62 v=50
9600 Onn ch=1 n=69 v=50
1920 Off ch=1 n=69 v=00
1920 Off ch=1 n=62 v=00
"""

def header(line):
    """Return the first four digits. You can change how this is done."""
    return line.split()[0] if line else None

def concatenate(group):
    """Return the 'n=' part. Be smarter about that,this is ugly."""
    return [entry[16:18] for entry in group]

for head,group in itertools.groupby(lines,header):
    group_list = concatenate(group)
    print("/".join(group_list))

这将产生输出：

60/67
67/60/62/69
69/62

为满足您的需求，我会在末尾添加一个filter：

filter = "9600"

for head,header):
    if head != filter:
        continue
    group_list = concatenate(group)
    print("/".join(group_list))

输出： 67/60/62/69