Writing string to a file on a new line every time

Question:

I want to append a newline to my string every time I call file.write(). What’s the easiest way to do this in Python?

Asked By: kaushik

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Answers:

You can do this in two ways:

f.write("text to writen")

or, depending on your Python version (2 or 3):

print >>f, "text to write"         # Python 2.x
print("text to write", file=f)     # Python 3.x
Answered By: Greg Hewgill

Use “n”:

file.write("My Stringn")

See the Python manual for reference.

Answered By: halfdan

You can use:

file.write(your_string + 'n')
Answered By: Krishna K

If you use it extensively (a lot of written lines), you can subclass ‘file’:

class cfile(file):
    #subclass file to have a more convienient use of writeline
    def __init__(self, name, mode = 'r'):
        self = file.__init__(self, name, mode)

    def wl(self, string):
        self.writelines(string + 'n')

Now it offers an additional function wl that does what you want:

with cfile('filename.txt', 'w') as fid:
    fid.wl('appends newline charachter')
    fid.wl('is written on a new line')

Maybe I am missing something like different newline characters (n, r, …) or that the last line is also terminated with a newline, but it works for me.

Answered By: mathause

Just a note, file isn’t supported in Python 3 and was removed. You can do the same with the open built-in function.

f = open('test.txt', 'w')
f.write('testn')
Answered By: user1767754
file_path = "/path/to/yourfile.txt"
with open(file_path, 'a') as file:
    file.write("This will be added to the next linen")

or

log_file = open('log.txt', 'a')
log_file.write("This will be added to the next linen")

This is the solution that I came up with trying to solve this problem for myself in order to systematically produce n’s as separators. It writes using a list of strings where each string is one line of the file, however it seems that it may work for you as well. (Python 3.+)

#Takes a list of strings and prints it to a file.
def writeFile(file, strList):
    line = 0
    lines = []
    while line < len(strList):
        lines.append(cheekyNew(line) + strList[line])
        line += 1
    file = open(file, "w")
    file.writelines(lines)
    file.close()

#Returns "n" if the int entered isn't zero, otherwise "".
def cheekyNew(line):
    if line != 0:
        return "n"
    return ""
Answered By: democidist

you could do:

file.write(your_string + 'n')

as suggested by another answer, but why using string concatenation (slow, error-prone) when you can call file.write twice:

file.write(your_string)
file.write("n")

note that writes are buffered so it amounts to the same thing.

Unless write to binary files, use print. Below example good for formatting csv files:

def write_row(file_, *columns):
    print(*columns, sep='t', end='n', file=file_)

Usage:

PHI = 45
with open('file.csv', 'a+') as f:
    write_row(f, 'header', 'phi:', PHI, 'serie no. 2')
    write_row(f)  # additional empty line
    write_row(f, data[0], data[1])

You can also use partial as a more pythonic way of creating this kind of wrappers. In the example below, row is print with predefined kwargs.

from functools import partial


with open('file.csv', 'a+') as f:
    row = partial(print, sep='t', end='n', file=f)

    row('header', 'phi:', PHI, 'serie no. 2', end='nn')
    row(data[0], data[1])

Notes:

Answered By: Mesco

Another solution that writes from a list using fstring

lines = ['hello','world']
with open('filename.txt', "w") as fhandle:
  for line in lines:
    fhandle.write(f'{line}n')

And as a function

def write_list(fname, lines):
    with open(fname, "w") as fhandle:
      for line in lines:
        fhandle.write(f'{line}n')

write_list('filename.txt', ['hello','world'])
Answered By: citynorman

I really didn’t want to type n every single time and @matthause’s answer didn’t seem to work for me, so I created my own class

class File():

    def __init__(self, name, mode='w'):
        self.f = open(name, mode, buffering=1)
        
    def write(self, string, newline=True):
        if newline:
            self.f.write(string + 'n')
        else:
            self.f.write(string)

And here it is implemented

f = File('console.log')

f.write('This is on the first line')
f.write('This is on the second line', newline=False)
f.write('This is still on the second line')
f.write('This is on the third line')

This should show in the log file as

This is on the first line
This is on the second lineThis is still on the second line
This is on the third line
Answered By: joshua

Ok, here is a safe way of doing it.

with open('example.txt', 'w') as f:
 for i in range(10):
  f.write(str(i+1))
  f.write('n')


This writes 1 to 10 each number on a new line.

Answered By: KIDUS Sisay

You can decorate method write in specific place where you need this behavior:

#Changed behavior is localized to single place.
with open('test1.txt', 'w') as file:    
    def decorate_with_new_line(method):
        def decorated(text):
            method(f'{text}n')
        return decorated
    file.write = decorate_with_new_line(file.write)
    
    file.write('This will be on line 1')
    file.write('This will be on line 2')
    file.write('This will be on line 3')

#Standard behavior is not affected. No class was modified.
with open('test2.txt', 'w') as file:
        
    file.write('This will be on line 1')
    file.write('This will be on line 1')
    file.write('This will be on line 1')  
Answered By: Pijukatel

Using append (a) with open() on a print() statement looks easier for me:

save_url  = ".test.txt"

your_text = "This will be on line 1"
print(your_text, file=open(save_url, "a+"))

another_text = "This will be on line 2"
print(another_text, file=open(save_url, "a+"))

another_text = "This will be on line 3"
print(another_text, file=open(save_url, "a+"))
Answered By: Nwoye CID

Usually you would use n but for whatever reason in Visual Studio Code 2019 Individual it won’t work. But you can use this:

# Workaround to n not working
print("lorem ipsum", file=f)  # Python 3.0 onwards only
print >>f, "Text"             # Python 2.0 and under
Answered By: vey

If write is a callback, you may need a custom writeln.

  def writeln(self, string):
        self.f.write(string + 'n')

Itself inside a custom opener. See answers and feedback for this question : subclassing file objects (to extend open and close operations) in python 3

(Context Manager)

I faced this when using ftplib to "retrieve lines" from a file that was "record based" (FB80):

with open('somefile.rpt', 'w') as fp:
     ftp.retrlines('RETR USER.REPORT', fp.write)

and ended up with one long record with no newlines, this is likely a problem with ftplib, but obscure.

So this became:

with OpenX('somefile.rpt') as fp:
     ftp.retrlines('RETR USER.REPORT', fp.writeln) 

It does the job. This is a use case a few people will be looking for.

Complete declaration (only the last two lines are mine):

class OpenX:
    def __init__(self, filename):
        self.f = open(filename, 'w')

    def __enter__(self):
        return self.f

    def __exit__(self, exc_type, exc_value, traceback):
        self.f.close()

    def writeln(self, string):
        self.f.write(string + 'n')
Answered By: mckenzm

in order to suspport multiple operating systems use:
file.write(f'some strings and/or {variable}. {os.linesep}')

Answered By: Assaf-ge

You could use C-style string formatting:

file.write("%sn" % "myString")

More about String Formatting.

Answered By: Vincent B

Actually, when you use the multiline syntax, like so:

f.write("""
line1
line2
line2""")

You don’t need to add n!

Answered By: Eryk
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