How to delete the first line of a text file?

Question

I have been searching online, but have not found any good solution.

Here is my text file:

[54, 95, 45, -97, -51, 84, 0, 32, -55, 14, 50, 54, 68, -3,          


        

Answer 1

Assuming you have enough memory to hold everything in memory:

with open('file.txt', 'r') as fin:
    data = fin.read().splitlines(True)
with open('file.txt', 'w') as fout:
    fout.writelines(data[1:])

We could get fancier, opening the file, reading and then seeking back to the beginning eliminating the second open, but really, this is probably good enough.

Answer 2

This solution will work for big files that don't fit into memory by reading and writing one line at a time:

import os
from shutil import move
from tempfile import NamedTemporaryFile

# Take off the first line which has the system call and params
file_path = 'xxxx'
temp_path = None
with open(file_path, 'r') as f_in:
    with NamedTemporaryFile(mode='w', delete=False) as f_out:
        temp_path = f_out.name
        next(f_in)  # skip first line
        for line in f_in:
            f_out.write(line)

os.remove(file_path)
move(temp_path, file_path)

Answer 3

You can do it much easier but simply stating what is the first line to be read:

    with open(filename, "r") as f:
        rows = f.readlines()[1:]

Answer 4

Bash will be faster for that purpose. You can use these in you python script:

subprocess.Popen.communicate()

I wrote a function for running a subprocess cmd for shell:

def popen_method(call):
    subprocess_call = Popen([call], shell=True, stdout=PIPE, stderr=PIPE)
    out, err = subprocess_call.communicate()
    if err:
        raise yourError(
            '\n============= WARNING/ERROR ===============\n{}\n===========================================\n'.format(
                err.rstrip()))
    return out

You call it like this:

testing = "sed -i /var/output/ip_list.csv -e '1 s/^.*$/host_id,ip,last_updated/g'"
popen_method(testing)

or use:

from sh import sed

then run the sed command:

sed -i /var/output/ip_list.csv -e '1 s/^.*$/host_id,ip,last_updated/g'

This will replace whatever you had on the first line with host_id,ip,last_updated.

Answer 5

Here is a memory-efficient (?) solution which makes use of shutil:

import shutil

source_file = open('file.txt', 'r')
source_file.readline()
# this will truncate the file, so need to use a different file name:
target_file = open('file.txt.new', 'w')

shutil.copyfileobj(source_file, target_file)

Answer 6

Safer to use one open for read & write, if you want to use the file from another thread/process:

def pop(self, file):
    with open(file, 'r+') as f: # open file in read / write mode
        firstLine = f.readline() # read the first line and throw it out
        data = f.read() # read the rest
        f.seek(0) # set the cursor to the top of the file
        f.write(data) # write the data back
        f.truncate() # set the file size to the current size
        return firstLine

fifo = pop('filename.txt')