如何读取某些内容有换行符的文本文件？

Question

I have a text file of this form:

06/01/2016, 10:40 pm - abcde
07/01/2016, 12:04 pm - abcde
07/01/2016, 12:05 pm - abcde
07/01/2016, 12:05 pm - abcde
07/01/2016, 6:14 pm - abcde

fghe
07/01/2016, 6:20 pm - abcde
07/01/2016, 7:58 pm - abcde

fghe

ijkl
07/01/2016, 7:58 pm - abcde

You can see that every line is separated by a line break, but some row contents have line breaks in them. So, simply separating by line doesn't parse every line properly.

As an example, for the 5th entry, I want my output to be 07/01/2016, 6:14 pm - abcde fghe

Here is my current code:

with open('file.txt', 'r') as text_file:
data = []
for line in text_file:
    row = line.strip()
    data.append(row)

Answer 1

Considering that ',' can only appear as a separator, we may check if the line has a comma and concatenate it to the last row if it doesn't:

data = []

with open('file.txt', 'r') as text_file:
    for line in text_file:
        row = line.strip()
        if ',' not in row:
            data[-1] += '\n' + row
        else:
            data.append(row)

Answer 2

Given your example input, you can use a regex with a forward lookahead:

pat=re.compile(r'^(\d\d\/\d\d\/\d\d\d\d.*?)(?=^^\d\d\/\d\d\/\d\d\d\d|\Z)', re.S | re.M)

with open (fn) as f:
    pprint([m.group(1) for m in pat.finditer(f.read())])    

Prints:

['06/01/2016, 10:40 pm - abcde\n',
 '07/01/2016, 12:04 pm - abcde\n',
 '07/01/2016, 12:05 pm - abcde\n',
 '07/01/2016, 12:05 pm - abcde\n',
 '07/01/2016, 6:14 pm - abcde\n\nfghe\n',
 '07/01/2016, 6:20 pm - abcde\n',
 '07/01/2016, 7:58 pm - abcde\n\nfghe\n\nijkl\n',
 '07/01/2016, 7:58 pm - abcde\n']

---

With the Dropbox example, prints:

['11/11/2015, 3:16 pm - IK: 12\n',
 '13/11/2015, 12:10 pm - IK: Hi.\n\nBut this is not about me.\n\nA donation, however small, will go a long way.\n\nThank you.\n',
 '13/11/2015, 12:11 pm - IK: Boo\n',
 '15/11/2015, 8:36 pm - IR: Root\n',
 '15/11/2015, 8:36 pm - IR: LaTeX?\n',
 '15/11/2015, 8:43 pm - IK: Ws\n']

If you want to delete the \\n in what is captured, just add m.group(1).strip().replace('\\n', '') to the list comprehension above.

---

Explanation of regex:

^(\d\d\/\d\d\/\d\d\d\d.*?)(?=^^\d\d\/\d\d\/\d\d\d\d|\Z)

^                                                       start of line   
    ^  ^  ^  ^   ^                                      pattern for a date  
                       ^                                capture the rest...  
                           ^                            until (look ahead)
                                      ^ ^ ^             another date
                                                  ^     or
                                                     ^  end of string

Answer 3

You could use regular expressions (using the re module) to check for dates like this:

import re
with open('file.txt', 'r') as text_file:
  data = []
  for line in text_file:
    row = line.strip()
    if re.match(r'\d{2}/\d{2}/\d{4}.*'):  
      data.append(row)  # date: new record
    else:
      data[-1] += '\n' + row  # no date: append to last record

# '\d{2}': two digits
# '.*': any character, zero or more times

Answer 4

Simple testing for length:

#!python3
#coding=utf-8

data = """06/01/2016, 10:40 pm - abcde
07/01/2016, 12:04 pm - abcde
07/01/2016, 12:05 pm - abcde
07/01/2016, 12:05 pm - abcde
07/01/2016, 6:14 pm - abcde

fghe
07/01/2016, 6:20 pm - abcde
07/01/2016, 7:58 pm - abcde

fghe

ijkl
07/01/2016, 7:58 pm - abcde"""

lines = data.split("\n")
out = []
for l in lines:
    c = l.strip()
    if c:
        if len(c) < 10:
            out[-1] += c
        else:
            out.append(c)
    #skip empty

for o in out:
    print(o)

results in:

06/01/2016, 10:40 pm - abcde
07/01/2016, 12:04 pm - abcde
07/01/2016, 12:05 pm - abcde
07/01/2016, 12:05 pm - abcde
07/01/2016, 6:14 pm - abcdefghe
07/01/2016, 6:20 pm - abcde
07/01/2016, 7:58 pm - abcdefgheijkl
07/01/2016, 7:58 pm - abcde

Does not contain the line breaks in the data!

---

But this one liner regular expression should do it (split on linebreak followed by digit), at least for the sample data (breaks when data contains linebreak followed by digit):

#!python3
#coding=utf-8

text_file = """06/01/2016, 10:40 pm - abcde
07/01/2016, 12:04 pm - abcde
07/01/2016, 12:05 pm - abcde
07/01/2016, 12:05 pm - abcde
07/01/2016, 6:14 pm - abcde

fghe
07/01/2016, 6:20 pm - abcde
07/01/2016, 7:58 pm - abcde

fghe

ijkl
07/01/2016, 7:58 pm - abcde"""

import re
data = re.split("\n(?=\d)", text_file)

print(data)

for d in data:
    print(d)

Output:

   ['06/01/2016, 10:40 pm - abcde', '07/01/2016, 12:04 pm - abcde', '07/01/2016, 12:05 pm - abcde', '07/01/2016, 12:05 pm - abcde', '07/01/2016, 6:14 pm - abcde\n\
nfghe', '07/01/2016, 6:20 pm - abcde', '07/01/2016, 7:58 pm - abcde\n\nfghe\n\nijkl', '07/01/2016, 7:58 pm - abcde']
06/01/2016, 10:40 pm - abcde
07/01/2016, 12:04 pm - abcde
07/01/2016, 12:05 pm - abcde
07/01/2016, 12:05 pm - abcde
07/01/2016, 6:14 pm - abcde

fghe
07/01/2016, 6:20 pm - abcde
07/01/2016, 7:58 pm - abcde

fghe

ijkl
07/01/2016, 7:58 pm - abcde

---

(fixed with lookahead)