Prepend a line to an existing file in Python

Question

I need to add a single line to the first line of a text file and it looks like the only options available to me are more lines of code than I would expect from python. Something like this:

f = open('filename','r')
temp = f.read()
f.close()

f = open('filename', 'w')
f.write("#testfirstline")

f.write(temp)
f.close()

Is there no easier way? Additionally, I see this two-handle example more often than opening a single handle for reading and writing ('r+') - why is that?

Answer 1

Use pathlib.Path read_text() and write_text() to prepend to a file concisely:

from pathlib import Path

file = Path("filename")
original = file.read_text()
file.write_text(f"new first line\n{original}")

Answer 2

I would stick with separate reads and writes, but we certainly can express each more concisely:

with open('filename', 'r') as original:
    data = original.read()
with open('filename', 'w') as modified:
    modified.write("new first line\n" + data)

Answer 3

What about using OS library.

In this particular case with OS Linux and using the sed function

import os

header = 'Your first line'
cmd_header = (r"sed -i '1s/^/" + f"{header}" + r"\n/'" + f" {file_name}")
os.cmd(cmd_header)

This will create a new line into the specified file_name location and insert the header string. This process does not rewrite the whole file structure and can process big files.

Answer 4

this worked for me

def prepend(str, file):
    with open(file, "r") as fr:
        read = fr.read()
        with open(file, "w") as fw:
            fw.write(str + read)
            fw.close()

Answer 5

If you rewrite it like this:

with open('filename') as f:
    read_data = f.read()
with open('filename', 'w') as f:
    f.write("#testfirstline\n" + read_data)

It's rather short and simple. For 'r+' the file needs to exist already.

Answer 6

As I suggested in this answer, you can do it using the following:

def prepend_text(filename: Union[str, Path], text: str):
    with fileinput.input(filename, inplace=True) as file:
        for line in file:
            if file.isfirstline():
                print(text)
            print(line, end="")

Answer 7

Am I not seeing something or couldn't we just use a buffer large-enough to read-in the input file in parts (instead of the whole content) and with this buffer traverse the file while it is open and keep exchanging file<->buffer contents?

This seems much more efficient (for big files especially) than reading the whole content in memory, modifying it in memory and writing it back to the same file or (even worse) a different one. Sorry that now I don't have time to implement a sample snippet, I'll get back to this later, but maybe you get the idea.

Answer 8

If you wish to prepend in the file after a specific text then you can use the function below.

def prepend_text(file, text, after=None):
    ''' Prepend file with given raw text '''
    f_read = open(file, 'r')
    buff = f_read.read()
    f_read.close()
    f_write = open(file, 'w')
    inject_pos = 0
    if after:
        pattern = after
        inject_pos = buff.find(pattern)+len(pattern)
    f_write.write(buff[:inject_pos] + text + buff[inject_pos:])
    f_write.close()

So first you open the file, read it and save it all into one string. Then we try to find the character number in the string where the injection will happen. Then with a single write and some smart indexing of the string we can rewrite the whole file including the injected text now.

Answer 9

Here's a 3 liner that I think is clear and flexible. It uses the list.insert function, so if you truly want to prepend to the file use l.insert(0, 'insert_str'). When I actually did this for a Python Module I am developing, I used l.insert(1, 'insert_str') because I wanted to skip the '# -- coding: utf-8 --' string at line 0. Here is the code.

f = open(file_path, 'r'); s = f.read(); f.close()
l = s.splitlines(); l.insert(0, 'insert_str'); s = '\n'.join(l)
f = open(file_path, 'w'); f.write(s); f.close()

Answer 10

This does the job without reading the whole file into memory, though it may not work on Windows

def prepend_line(path, line):
    with open(path, 'r') as old:
        os.unlink(path)
        with open(path, 'w') as new:
            new.write(str(line) + "\n")
            shutil.copyfileobj(old, new)

Answer 11

Python makes a lot of things easy and contains libraries and wrappers for a lot of common operations, but the goal is not to hide fundamental truths.

The fundamental truth you are encountering here is that you generally can't prepend data to an existing flat structure without rewriting the entire structure. This is true regardless of language.

There are ways to save a filehandle or make your code less readable, many of which are provided in other answers, but none change the fundamental operation: You must read in the existing file, then write out the data you want to prepend, followed by the existing data you read in.

By all means save yourself the filehandle, but don't go looking to pack this operation into as few lines of code as possible. In fact, never go looking for the fewest lines of code -- that's obfuscation, not programming.

Answer 12

with open("file", "r+") as f: s = f.read(); f.seek(0); f.write("prepend\n" + s)

Answer 13

Other approach:

with open("infile") as f1:
    with open("outfile", "w") as f2:
        f2.write("#test firstline")
        for line in f1:
            f2.write(line)

or a one liner:

open("outfile", "w").write("#test firstline\n" + open("infile").read())

Thanks for the opportunity to think about this problem :)

Cheers

Answer 14

One possibility is the following:

import os
open('tempfile', 'w').write('#testfirstline\n' + open('filename', 'r').read())
os.rename('tempfile', 'filename')

Answer 15

You can save one write call with this:

f.write('#testfirstline\n' + temp)

When using 'r+', you would have to rewind the file after reading and before writing.