I have a text file of messages in the form:

with open(code_path) as f:
    contents = f.readlines()
print(contents)

['22/05/2022, 21.58 - Name: message1 \n', 
'22/05/2022, 22.07 – Name2: message2\n', 
'message2 continues\n', 
'22/05/2022, 22.09 – Name: message3\n']

Currently I have the messages in strings. Some long messages are split into two. I would like to have a list of the messages in with all messages joined together (starts with the date).

This is what I want:

['22/05/2022, 21.58 - Name: message1 \n', 
'22/05/2022, 22.07 – Name2: message2   message2 continues\n', 
'22/05/2022, 22.09 – Name: message3\n']

Is there a way to do this?

I have found the strings starting with a date with:

import re

dates = [re.findall("^[0-3][0-9]/[0-3][0-9]/20[1-2][1-9]", i) for i in contents]

But I don't know how to continue.

CodePudding user response：

A basic approach would be to use a kind of cache: go through the lines,

• if the line starts with a date, append a new item to the cache
• if it doesn't, append to the most recent item.

messages = []
for line in contents:
    if re.match(r'\d{2}/\d{2}/\d{4},\s ', line):
        messages.append([line])
    else:
        messages[-1].append(line)
 
# messages
[['22/05/2022, 21.58 - Name: message1 \n'],
 ['22/05/2022, 22.07 – Name2: message2\n', 'message2 continues\n'],
 ['22/05/2022, 22.09 – Name: message3\n']]

You could then join them (e.g., [''.join(m) for m in messages]). Alternatively, it's also possible to build strings directly, but maybe you want to distinguish between primary/following lines at some point, then the list is more useful.

CodePudding user response：

You might also read all lines, and then match the line starting with a date like pattern followed by all all lines not starting with a date like pattern.

With a more specific date like pattern:

import re

with open("file") as f:
    pattern = r"^(?:0[1-9]|[12][0-9]|3[01])/(?:0[1-9]|1[012])/\d{4},.*(?:\n(?!(?:0[1-9]|[12][0-9]|3[01])/(?:0[1-9]|1[012])/\d{4},).*)*"
    print(re.findall(pattern, f.read(), re.M))

Output

[
'22/05/2022, 21.58 - Name: message1 \n', 
'22/05/2022, 22.07 – Name2: message2\n\nmessage2 continues\n', 
'22/05/2022, 22.09 – Name: message3\n'
]

With a less precise pattern, but a bit shorter:

^\d{2}/\d{2}/\d{4},.*(?:\n(?!\d{2}/\d{2}/\d{4},).*)*

Explanation

• ^ Start of string anchor
• \d{2}/\d{2}/\d{4}, Match a date like pattern followed by a comma
• .* Match the rest of the line
• (?: Non capture group to repeat as a whole part
  • \n Match a newline
  • (?!\d{2}/\d{2}/\d{4},)
  • .* Match the rest of the line
• )* Close the non capture group and optionally repeat to match all lines

Example

import re

with open("file") as f:
    pattern = r"^\d{2}/\d{2}/\d{4},.*(?:\n(?!\d{2}/\d{2}/\d{4},).*)*"
    print(re.findall(pattern, f.read(), re.M))