将文件夹名称用作文本文件中的列

Question

The lazy me is thinking about adding a column to some textfiles.

The textfiles are in directories and I would like to add the directory name to the text file.

Like the text file text.txt in the folder the_peasant :

has a wart    
was dressed up like a witch     
has a false nose

would become:

the_peasant has a wart    
the_peasant was dressed up like a witch    
the_peasant has a false nose

Then I have similar text files in other folders called "the_king" etc.

I would think this is a combination of the find command, bash scripting and sed but I cant see it through. Any ideas?

Answer 1

This might work for you:

find . -name text.txt | sed 's|.*/\(.*\)/.*|sed -i "s@^@\1 @" & |' | sh

or if you have GNU sed:

find . -name text.txt | sed 's|.*/\(.*\)/.*|sed -i "s@^@\1 @" & |e' 

Answer 2

Simple python script for this (should work from any folder, as long as you pass the fullpath to the target file, obviously):

#!/usr/bin/python
if __name__ == '__main__':
    import sys
    import os

    # Get full filepath and directory name
    filename = os.path.abspath(sys.argv[1])
    dirname = os.path.split(os.path.dirname(filename))[1]

    # Read current file contents
    my_file = open(filename, 'r')
    lines = my_file.readlines()
    my_file.close()

    # Rewrite lines, adding folder name to the start
    output_lines = [dirname + ' ' + line for line in lines]
    my_file = open(filename, 'w')
    my_file.write('\n'.join(output_lines))
    my_file.close()

Answer 3

Here is what I came up with:

find /path/to/dir -type f | sed -r 'p;s:.*/(.*)/.*:\1:' | xargs -n 2 sh -c 'sed -i "s/^/$1 /" $0'

Here is an example of how the commands would be constructed, assuming the following files exist:

/home/the_peasant/a.txt
/home/the_peasant/b.txt
/home/the_peasant/farmer/c.txt

First find /home/the_peasant -type f would output those files exactly as above.

Next, the sed command would output a file name, followed by the directory name, like this:

/home/the_peasant/a.txt
the_peasant
/home/the_peasant/b.txt
the_peasant
/home/the_peasant/farmer/c.txt
farmer

The xargs would group every two lines and pass them to the sh command, so you would end up with the following three commands:

$ sh -c 'sed -i "s/^/$1 /" $0' /home/the_peasant/a.txt the_peasant
$ sh -c 'sed -i "s/^/$1 /" $0' /home/the_peasant/b.txt the_peasant
$ sh -c 'sed -i "s/^/$1 /" $0' /home/the_peasant/farmer/c.txt farmer

And finally this will result in the following sed commands which will add the folder name to the beginning of each line:

$ sed -i "s/^/the_peasant /" /home/the_peasant/a.txt
$ sed -i "s/^/the_peasant /" /home/the_peasant/b.txt
$ sed -i "s/^/farmer /" /home/the_peasant/farmer/c.txt

Answer 4

Obligatory single liner using find and perl

find . -maxdepth 1 -mindepth 1 -type d | perl -MFile::Basename -ne 'chomp; my $dir = basename($_); for my $file (glob "$dir/*") { print qq{sed -i "s/^/$dir /" $file\n} }' | tee rename_commands.sh

sh rename_commands.sh

Assumes perl and sed are in your $PATH. Generates a file of sed commands to do the actual change so you can review what is to be done.

In my test, that command file looks like so:

sed -i "s/^/foo /" foo/text1
sed -i "s/^/foo /" foo/text2
sed -i "s/^/bar /" bar/belvedere
sed -i "s/^/bar /" bar/robin

Answer 5

The directory tree:

% tree .
.
├── the_king
│   └── text.txt
├── the_knight
│   └── text.txt
├── the_peasant
│   └── text.txt
└── wart.py
3 directories, 4 files

Directories and contents before:

% find . -name 'text.txt' -print -exec cat {} \;       
./the_king/text.txt
has a wart    
was dressed up like a witch     
has a false nose
./the_knight/text.txt
has a wart    
was dressed up like a witch     
has a false nose
./the_peasant/text.txt
has a wart    
was dressed up like a witch     
has a false nose

Code (wart.py):

#!/usr/bin/env python

import os

text_file = 'text.txt'
cwd = os.path.curdir # '.'

# Walk thru each directory starting at '.' and if the directory contains
# 'text.txt', print each line of the file prefixed by the name containing
# directory.
for root, dirs, files in os.walk(cwd):
    if text_file in files: # We only care IF the file is in this directory.
        print 'Found %s!' % root
        filepath = os.path.join(root, text_file) # './the_peasant/text.txt'
        root_base = os.path.basename(root)       # './the_peasant' => 'the_peasant'
        output = ''
        with open(filepath, 'r') as reader:      # Open file for read/write
            for line in reader:                  # Iterate the lines of the file
                new_line = "%s %s" % (root_base, line)
                print new_line,
                output += new_line               # Append to the output

        with open(filepath, 'w') as writer:
            writer.write(output)                 # Write to the file

        print

Which outputs:

Found ./the_king!
the_king has a wart    
the_king was dressed up like a witch     
the_king has a false nose

Found ./the_knight!
the_knight has a wart    
the_knight was dressed up like a witch     
the_knight has a false nose

Found ./the_peasant!
the_peasant has a wart    
the_peasant was dressed up like a witch     
the_peasant has a false nose

Directories and contents after:

% find . -name 'text.txt' -print -exec cat {} \;
./the_king/text.txt
the_king has a wart    
the_king was dressed up like a witch     
the_king has a false nose
./the_knight/text.txt
the_knight has a wart    
the_knight was dressed up like a witch     
the_knight has a false nose
./the_peasant/text.txt
the_peasant has a wart    
the_peasant was dressed up like a witch     
the_peasant has a false nose

This was fun! Thanks for the challenge!

Answer 6

I would.

• get the file path eg fpath = "example.txt"
• find the directory of that file using the below
• read in the file and write to a new file appending dir_name to the row just read before writing.

Accessing the directory can be done by using

import os
fpath = "example.txt"
dir_name = os.path.dirname(fpath)

Answer 7

Are you running the script in the appropriate folder? Then you can use the os module to find the current folder. Say you wanted to take just the end of the directory tree, you could use os.path, like:

import os, os.path

curDirectory = os.getcwd()
baseDir = os.path.basename()

inFile = open("filename.txt").xreadlines()
outFile = open("filename.out", "w")

for line in inFile:
    outFile.write("%s %s" % (baseDir, line))
outFile.close()

Answer 8

Edit: noticed something wasn't correct. I removed the dir loop - its recursively walking now. Sorry for the mix up.

Using os.walk

import os.path
directory = os.path.curdir
pattern = ".py";
for (path,dirs,files) in os.walk(directory):
    for file in files:
        if not file.endswith(pattern):
            continue
        filename = os.path.join(path,file)
        #print "file: ",filename
        #continue
        with open(filename,"r") as f:
            for line in f.readlines():
                print "{0} {1}".format(filename,line)
            f.close()

Output:

list1.py   # LAB(replace solution)
list1.py   # return
list1.py   # LAB(end solution)

Answer 9

Here's a one-ish-liner in bash and awk:

find . -type f -print0 |
while read -r -d "" path; do
  mv "$path" "$path.bak"
  awk -v dir="$(basename "$(dirname "$path")")" '{print dir, $0}' "$path.bak" > "$path"
done