Jason
Jason

Reputation: 3307

How to insert a new line before the first line in a file using python?

More details below:

1st line

2nd line

3rd line

4th line

...

Now want to insert a new line named zero line before 1st line. File looks like below:

zero line

1st line

2nd line

3rd line

4th line

...

I know sed command can do this work, but how to do it using python? Thanks

Upvotes: 4

Views: 14124

Answers (5)

Zorro
Zorro

Reputation: 1519

with open(filename, 'r+') as f:
    lines = f.readlines()     
    lines.insert(0, 'zero line\n')   
    f.seek(0)                 
    f.writelines(lines)  

Upvotes: 1

dting
dting

Reputation: 39307

this might be of interest

http://net4geeks.com/index.php?option=com_content&task=view&id=53&Itemid=11

adapted to your question:

# read the current contents of the file
f = open('filename')
text = f.read()
f.close()
# open the file again for writing
f = open('filename', 'w')
f.write("zero line\n\n")
# write the original contents
f.write(text)
f.close()
  • Open the file and read the contents into 'text'.

  • Close the file

  • Reopen the file with argument 'w' to write

  • Write text to prepend to the file

  • Write the original contents of the file to the file

  • Close file

Read the warnings in the link.

edit:

But note that this isn't entirely safe, if your Python session crashes after opening the file the second time and before closing it again, you will lose data.

Upvotes: 5

jfs
jfs

Reputation: 414915

Here's an implementation that fixes some deficiencies in other approaches presented sofar:

It mimics fileinput's error handling:

import os

def prepend(filename, data, bufsize=1<<15):
    # backup the file
    backupname = filename + os.extsep+'bak'
    try: os.unlink(backupname) # remove previous backup if it exists
    except OSError: pass
    os.rename(filename, backupname)

    # open input/output files,  note: outputfile's permissions lost
    with open(backupname) as inputfile, open(filename, 'w') as outputfile:
        # prepend
        outputfile.write(data)
        # copy the rest
        buf = inputfile.read(bufsize)
        while buf:
            outputfile.write(buf)
            buf = inputfile.read(bufsize)

    # remove backup on success
    try: os.unlink(backupname)
    except OSError: pass

prepend('file', '0 line\n')

You could use cat utility if it is available to copy the files. It might be more efficient:

import os
from subprocess import PIPE, Popen

def prepend_cat(filename, data, bufsize=1<<15):
    # backup the file
    backupname = filename + os.extsep+'bak'
    try: os.unlink(backupname)
    except OSError: pass
    os.rename(filename, backupname)

    # $ echo $data | cat - $backupname > $filename
    with open(filename, 'w') as outputfile: #note: outputfile's permissions lost
        p = Popen(['cat', '-', backupname], stdin=PIPE, stdout=outputfile)
        p.communicate(data)

    # remove backup on success
    if p.poll() == 0:
        try: os.unlink(backupname)
        except OSError: pass

prepend_cat('file', '0 line\n')

Upvotes: 4

kurumi
kurumi

Reputation: 25609

you can use fileinput

>>> import fileinput
>>> for linenum,line in enumerate( fileinput.FileInput("file",inplace=1) ):
...   if linenum==0 :
...     print "new line"
...     print line.rstrip()
...   else:
...     print line.rstrip()
...

Upvotes: 7

Jim
Jim

Reputation: 3284

code

L = list()
f = open('text2.txt', 'r')
for line in f.readlines():
        L.append(line)
L.insert(0,"Zero\n")
f.close()

fi = open('text2.txt', 'w')
for line in xrange(len(L)):
        fi.write(L[line])

fi.close()

text2.txt

Hello
The second line
3
4
5
6

output

Zero
Hello
The second line
3
4
5
6

This can be memory heavy and time consuming for large files however.

If you are worried about something like 31st line, I would just do a mod%10 on the num, to get a more accurate version.

Let me know if this helps, or if you want a better version. Also, if you want a better formatting, look into ljust and rjust for left and right justify.

Upvotes: 0

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