YAML find and replace text Python - python

I have some files in YAML format, I need to find the text in the $title file and replace with what I specified. What the configuration file looks like approximately:
JoinGame-MOTD:
Enabled: true
Messages:
- '$title'
The YAML file may look different, so I want to make a universal code that will not get any specific string, but replace all $title with what I specified
What I was trying to do:
import sys
import yaml
with open(r'config.yml', 'w') as file:
def tr(s):
return s.replace('$title', 'Test')
yaml.dump(file, sys.stdout, transform=tr)
Please help me. It is not necessary to work with my code, I will be happy with any examples that can suit me

Might be easier to not use the yaml package at all.
with open("file.yml", "r") as fin:
with open("file_replaced.yml", "w") as fout:
for line in fin:
fout.write(line.replace('$title', 'Test'))
EDIT:
To update in place
with open("config.yml", "r+") as f:
contents = f.read()
f.seek(0)
f.write(contents.replace('$title', 'Test'))
f.truncate()

You can also read & write data in one go. os.path.join is optional, it makes sure the yaml file is read relative to path your script is stored
import re
import os
with open(os.path.join(os.path.dirname(__file__), 'temp.yaml'), 'r+') as f:
data = f.read()
f.seek(0)
new_data = data.replace('$title', 'replaced!')
f.write(new_data)
f.truncate()
In case you wish to dynamically replace other keywords besides $title, like $description or $name, you can write a function using regex like this;
def replaceString(text_to_search, keyword, replacement):
return re.sub(f"(\${keyword})[\W]", replacement, text_to_search)
replaceString('My name is $name', '$name', 'Bob')

Related

Replace a text in File with python [duplicate]

I want to loop over the contents of a text file and do a search and replace on some lines and write the result back to the file. I could first load the whole file in memory and then write it back, but that probably is not the best way to do it.
What is the best way to do this, within the following code?
f = open(file)
for line in f:
if line.contains('foo'):
newline = line.replace('foo', 'bar')
# how to write this newline back to the file
The shortest way would probably be to use the fileinput module. For example, the following adds line numbers to a file, in-place:
import fileinput
for line in fileinput.input("test.txt", inplace=True):
print('{} {}'.format(fileinput.filelineno(), line), end='') # for Python 3
# print "%d: %s" % (fileinput.filelineno(), line), # for Python 2
What happens here is:
The original file is moved to a backup file
The standard output is redirected to the original file within the loop
Thus any print statements write back into the original file
fileinput has more bells and whistles. For example, it can be used to automatically operate on all files in sys.args[1:], without your having to iterate over them explicitly. Starting with Python 3.2 it also provides a convenient context manager for use in a with statement.
While fileinput is great for throwaway scripts, I would be wary of using it in real code because admittedly it's not very readable or familiar. In real (production) code it's worthwhile to spend just a few more lines of code to make the process explicit and thus make the code readable.
There are two options:
The file is not overly large, and you can just read it wholly to memory. Then close the file, reopen it in writing mode and write the modified contents back.
The file is too large to be stored in memory; you can move it over to a temporary file and open that, reading it line by line, writing back into the original file. Note that this requires twice the storage.
I guess something like this should do it. It basically writes the content to a new file and replaces the old file with the new file:
from tempfile import mkstemp
from shutil import move, copymode
from os import fdopen, remove
def replace(file_path, pattern, subst):
#Create temp file
fh, abs_path = mkstemp()
with fdopen(fh,'w') as new_file:
with open(file_path) as old_file:
for line in old_file:
new_file.write(line.replace(pattern, subst))
#Copy the file permissions from the old file to the new file
copymode(file_path, abs_path)
#Remove original file
remove(file_path)
#Move new file
move(abs_path, file_path)
Here's another example that was tested, and will match search & replace patterns:
import fileinput
import sys
def replaceAll(file,searchExp,replaceExp):
for line in fileinput.input(file, inplace=1):
if searchExp in line:
line = line.replace(searchExp,replaceExp)
sys.stdout.write(line)
Example use:
replaceAll("/fooBar.txt","Hello\sWorld!$","Goodbye\sWorld.")
This should work: (inplace editing)
import fileinput
# Does a list of files, and
# redirects STDOUT to the file in question
for line in fileinput.input(files, inplace = 1):
print line.replace("foo", "bar"),
Based on the answer by Thomas Watnedal.
However, this does not answer the line-to-line part of the original question exactly. The function can still replace on a line-to-line basis
This implementation replaces the file contents without using temporary files, as a consequence file permissions remain unchanged.
Also re.sub instead of replace, allows regex replacement instead of plain text replacement only.
Reading the file as a single string instead of line by line allows for multiline match and replacement.
import re
def replace(file, pattern, subst):
# Read contents from file as a single string
file_handle = open(file, 'r')
file_string = file_handle.read()
file_handle.close()
# Use RE package to allow for replacement (also allowing for (multiline) REGEX)
file_string = (re.sub(pattern, subst, file_string))
# Write contents to file.
# Using mode 'w' truncates the file.
file_handle = open(file, 'w')
file_handle.write(file_string)
file_handle.close()
As lassevk suggests, write out the new file as you go, here is some example code:
fin = open("a.txt")
fout = open("b.txt", "wt")
for line in fin:
fout.write( line.replace('foo', 'bar') )
fin.close()
fout.close()
If you're wanting a generic function that replaces any text with some other text, this is likely the best way to go, particularly if you're a fan of regex's:
import re
def replace( filePath, text, subs, flags=0 ):
with open( filePath, "r+" ) as file:
fileContents = file.read()
textPattern = re.compile( re.escape( text ), flags )
fileContents = textPattern.sub( subs, fileContents )
file.seek( 0 )
file.truncate()
file.write( fileContents )
A more pythonic way would be to use context managers like the code below:
from tempfile import mkstemp
from shutil import move
from os import remove
def replace(source_file_path, pattern, substring):
fh, target_file_path = mkstemp()
with open(target_file_path, 'w') as target_file:
with open(source_file_path, 'r') as source_file:
for line in source_file:
target_file.write(line.replace(pattern, substring))
remove(source_file_path)
move(target_file_path, source_file_path)
You can find the full snippet here.
fileinput is quite straightforward as mentioned on previous answers:
import fileinput
def replace_in_file(file_path, search_text, new_text):
with fileinput.input(file_path, inplace=True) as file:
for line in file:
new_line = line.replace(search_text, new_text)
print(new_line, end='')
Explanation:
fileinput can accept multiple files, but I prefer to close each single file as soon as it is being processed. So placed single file_path in with statement.
print statement does not print anything when inplace=True, because STDOUT is being forwarded to the original file.
end='' in print statement is to eliminate intermediate blank new lines.
You can used it as follows:
file_path = '/path/to/my/file'
replace_in_file(file_path, 'old-text', 'new-text')
Create a new file, copy lines from the old to the new, and do the replacing before you write the lines to the new file.
Expanding on #Kiran's answer, which I agree is more succinct and Pythonic, this adds codecs to support the reading and writing of UTF-8:
import codecs
from tempfile import mkstemp
from shutil import move
from os import remove
def replace(source_file_path, pattern, substring):
fh, target_file_path = mkstemp()
with codecs.open(target_file_path, 'w', 'utf-8') as target_file:
with codecs.open(source_file_path, 'r', 'utf-8') as source_file:
for line in source_file:
target_file.write(line.replace(pattern, substring))
remove(source_file_path)
move(target_file_path, source_file_path)
Using hamishmcn's answer as a template I was able to search for a line in a file that match my regex and replacing it with empty string.
import re
fin = open("in.txt", 'r') # in file
fout = open("out.txt", 'w') # out file
for line in fin:
p = re.compile('[-][0-9]*[.][0-9]*[,]|[-][0-9]*[,]') # pattern
newline = p.sub('',line) # replace matching strings with empty string
print newline
fout.write(newline)
fin.close()
fout.close()
if you remove the indent at the like below, it will search and replace in multiple line.
See below for example.
def replace(file, pattern, subst):
#Create temp file
fh, abs_path = mkstemp()
print fh, abs_path
new_file = open(abs_path,'w')
old_file = open(file)
for line in old_file:
new_file.write(line.replace(pattern, subst))
#close temp file
new_file.close()
close(fh)
old_file.close()
#Remove original file
remove(file)
#Move new file
move(abs_path, file)

os.write() appends file instead of overwriting, but O_APPEND isn't used [duplicate]

I have the following code:
import re
#open the xml file for reading:
file = open('path/test.xml','r+')
#convert to string:
data = file.read()
file.write(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>",r"<xyz>ABC</xyz>\1<xyz>\2</xyz>",data))
file.close()
where I'd like to replace the old content that's in the file with the new content. However, when I execute my code, the file "test.xml" is appended, i.e. I have the old content follwed by the new "replaced" content. What can I do in order to delete the old stuff and only keep the new?
You need seek to the beginning of the file before writing and then use file.truncate() if you want to do inplace replace:
import re
myfile = "path/test.xml"
with open(myfile, "r+") as f:
data = f.read()
f.seek(0)
f.write(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>", r"<xyz>ABC</xyz>\1<xyz>\2</xyz>", data))
f.truncate()
The other way is to read the file then open it again with open(myfile, 'w'):
with open(myfile, "r") as f:
data = f.read()
with open(myfile, "w") as f:
f.write(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>", r"<xyz>ABC</xyz>\1<xyz>\2</xyz>", data))
Neither truncate nor open(..., 'w') will change the inode number of the file (I tested twice, once with Ubuntu 12.04 NFS and once with ext4).
By the way, this is not really related to Python. The interpreter calls the corresponding low level API. The method truncate() works the same in the C programming language: See http://man7.org/linux/man-pages/man2/truncate.2.html
file='path/test.xml'
with open(file, 'w') as filetowrite:
filetowrite.write('new content')
Open the file in 'w' mode, you will be able to replace its current text save the file with new contents.
Using truncate(), the solution could be
import re
#open the xml file for reading:
with open('path/test.xml','r+') as f:
#convert to string:
data = f.read()
f.seek(0)
f.write(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>",r"<xyz>ABC</xyz>\1<xyz>\2</xyz>",data))
f.truncate()
import os#must import this library
if os.path.exists('TwitterDB.csv'):
os.remove('TwitterDB.csv') #this deletes the file
else:
print("The file does not exist")#add this to prevent errors
I had a similar problem, and instead of overwriting my existing file using the different 'modes', I just deleted the file before using it again, so that it would be as if I was appending to a new file on each run of my code.
See from How to Replace String in File works in a simple way and is an answer that works with replace
fin = open("data.txt", "rt")
fout = open("out.txt", "wt")
for line in fin:
fout.write(line.replace('pyton', 'python'))
fin.close()
fout.close()
in my case the following code did the trick
with open("output.json", "w+") as outfile: #using w+ mode to create file if it not exists. and overwrite the existing content
json.dump(result_plot, outfile)
Using python3 pathlib library:
import re
from pathlib import Path
import shutil
shutil.copy2("/tmp/test.xml", "/tmp/test.xml.bak") # create backup
filepath = Path("/tmp/test.xml")
content = filepath.read_text()
filepath.write_text(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>",r"<xyz>ABC</xyz>\1<xyz>\2</xyz>", content))
Similar method using different approach to backups:
from pathlib import Path
filepath = Path("/tmp/test.xml")
filepath.rename(filepath.with_suffix('.bak')) # different approach to backups
content = filepath.read_text()
filepath.write_text(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>",r"<xyz>ABC</xyz>\1<xyz>\2</xyz>", content))

ConfigParser - Write to existing section

I am a bit stuck at the ConfigParser.
I want to add a specific setting to a existing section.
I do:
import ConfigParser
Config = ConfigParser.ConfigParser()
Config
Config.read("/etc/yum.repos.d/epel.repo")
Config.sections()
Config.set('epel','priority',10)
with open('/etc/yum.repos.d/epel.repo', 'w') as fout:
Then it shows:
...
File "<stdin>", line 2
^
IndentationError: expected an indented block
>>>
Edit #1
Now i tried it with the iniparse module. I did:
from iniparse import INIConfig
cfg = INIConfig(open('/etc/yum.repos.d/epel.repo'))
cfg.epel.priority=10
f = open('/etc/yum.repos.d/epel.repo', 'w')
print >>f, cfg
f.close()
Unfortunately it deletes the old content. How can i solve this?
Edit #2
It looks like that it works now.
f = open('/etc/yum.repos.d/epel.repo', 'wb')
did the trick.
Simply,
with open('epel.cfg', 'wb') as configfile:
config.write(configfile)
See here for examples and documentation.
The method you're looking for is Config.write.
See, for example, the first example in the docs
It should accept a file-like object to write the config data to. e.g.:
with open('new_config.cfg', 'w') as fout:
Config.write(fout)

Add file name as last column of CSV file

I have a Python script which modifies a CSV file to add the filename as the last column:
import sys
import glob
for filename in glob.glob(sys.argv[1]):
file = open(filename)
data = [line.rstrip() + "," + filename for line in file]
file.close()
file = open(filename, "w")
file.write("\n".join(data))
file.close()
Unfortunately, it also adds the filename to the header (first) row of the file. I would like the string "ID" added to the header instead. Can anybody suggest how I could do this?
Have a look at the official csv module.
Here are a few minor notes on your current code:
It's a bad idea to use file as a variable name, since that shadows the built-in type.
You can close the file objects automatically by using the with syntax.
Don't you want to add an extra column in the header line, called something like Filename, rather than just omitting a column in the first row?
If your filenames have commas (or, less probably, newlines) in them, you'll need to make sure that the filename is quoted - just appending it won't do.
That last consideration would incline me to use the csv module instead, which will deal with the quoting and unquoting for you. For example, you could try something like the following code:
import glob
import csv
import sys
for filename in glob.glob(sys.argv[1]):
data = []
with open(filename) as finput:
for i, row in enumerate(csv.reader(finput)):
to_append = "Filename" if i == 0 else filename
data.append(row+[to_append])
with open(filename,'wb') as foutput:
writer = csv.writer(foutput)
for row in data:
writer.writerow(row)
That may quote the data slightly differently from your input file, so you might want to play with the quoting options for csv.reader and csv.writer described in the documentation for the csv module.
As a further point, you might have good reasons for taking a glob as a parameter rather than just the files on the command line, but it's a bit surprising - you'll have to call your script as ./whatever.py '*.csv' rather than just ./whatever.py *.csv. Instead, you could just do:
for filename in sys.argv[1:]:
... and let the shell expand your glob before the script knows anything about it.
One last thing - the current approach you're taking is slightly dangerous, in that if anything fails when writing back to the same filename, you'll lose data. The standard way of avoiding this is to instead write to a temporary file, and, if that was successful, rename the temporary file over the original. So, you might rewrite the whole thing as:
import csv
import sys
import tempfile
import shutil
for filename in sys.argv[1:]:
tmp = tempfile.NamedTemporaryFile(delete=False)
with open(filename) as finput:
with open(tmp.name,'wb') as ftmp:
writer = csv.writer(ftmp)
for i, row in enumerate(csv.reader(finput)):
to_append = "Filename" if i == 0 else filename
writer.writerow(row+[to_append])
shutil.move(tmp.name,filename)
You can try:
data = [file.readline().rstrip() + ",id"]
data += [line.rstrip() + "," + filename for line in file]
You can try changing your code, but using the csv module is recommended. This should give you the result you want:
import sys
import glob
import csv
filename = glob.glob(sys.argv[1])[0]
yourfile = csv.reader(open(filename, 'rw'))
csv_output=[]
for row in yourfile:
if len(csv_output) != 0: # skip the header
row.append(filename)
csv_output.append(row)
yourfile = csv.writer(open(filename,'w'),delimiter=',')
yourfile.writerows(csv_output)
Use the CSV module that comes with Python.
import csv
import sys
def process_file(filename):
# Read the contents of the file into a list of lines.
f = open(filename, 'r')
contents = f.readlines()
f.close()
# Use a CSV reader to parse the contents.
reader = csv.reader(contents)
# Open the output and create a CSV writer for it.
f = open(filename, 'wb')
writer = csv.writer(f)
# Process the header.
header = reader.next()
header.append('ID')
writer.writerow(header)
# Process each row of the body.
for row in reader:
row.append(filename)
writer.writerow(row)
# Close the file and we're done.
f.close()
# Run the function on all command-line arguments. Note that this does no
# checking for things such as file existence or permissions.
map(process_file, sys.argv[1:])
You can run this as follows:
blair#blair-eeepc:~$ python csv_add_filename.py file1.csv file2.csv
you can use fileinput to do in place editing
import sys
import glob
import fileinput
for filename in glob.glob(sys.argv[1]):
for line in fileinput.FileInput(filename,inplace=1) :
if fileinput.lineno()==1:
print line.rstrip() + " ID"
else
print line.rstrip() + "," + filename

Search and replace a line in a file in Python

I want to loop over the contents of a text file and do a search and replace on some lines and write the result back to the file. I could first load the whole file in memory and then write it back, but that probably is not the best way to do it.
What is the best way to do this, within the following code?
f = open(file)
for line in f:
if line.contains('foo'):
newline = line.replace('foo', 'bar')
# how to write this newline back to the file
The shortest way would probably be to use the fileinput module. For example, the following adds line numbers to a file, in-place:
import fileinput
for line in fileinput.input("test.txt", inplace=True):
print('{} {}'.format(fileinput.filelineno(), line), end='') # for Python 3
# print "%d: %s" % (fileinput.filelineno(), line), # for Python 2
What happens here is:
The original file is moved to a backup file
The standard output is redirected to the original file within the loop
Thus any print statements write back into the original file
fileinput has more bells and whistles. For example, it can be used to automatically operate on all files in sys.args[1:], without your having to iterate over them explicitly. Starting with Python 3.2 it also provides a convenient context manager for use in a with statement.
While fileinput is great for throwaway scripts, I would be wary of using it in real code because admittedly it's not very readable or familiar. In real (production) code it's worthwhile to spend just a few more lines of code to make the process explicit and thus make the code readable.
There are two options:
The file is not overly large, and you can just read it wholly to memory. Then close the file, reopen it in writing mode and write the modified contents back.
The file is too large to be stored in memory; you can move it over to a temporary file and open that, reading it line by line, writing back into the original file. Note that this requires twice the storage.
I guess something like this should do it. It basically writes the content to a new file and replaces the old file with the new file:
from tempfile import mkstemp
from shutil import move, copymode
from os import fdopen, remove
def replace(file_path, pattern, subst):
#Create temp file
fh, abs_path = mkstemp()
with fdopen(fh,'w') as new_file:
with open(file_path) as old_file:
for line in old_file:
new_file.write(line.replace(pattern, subst))
#Copy the file permissions from the old file to the new file
copymode(file_path, abs_path)
#Remove original file
remove(file_path)
#Move new file
move(abs_path, file_path)
Here's another example that was tested, and will match search & replace patterns:
import fileinput
import sys
def replaceAll(file,searchExp,replaceExp):
for line in fileinput.input(file, inplace=1):
if searchExp in line:
line = line.replace(searchExp,replaceExp)
sys.stdout.write(line)
Example use:
replaceAll("/fooBar.txt","Hello\sWorld!$","Goodbye\sWorld.")
This should work: (inplace editing)
import fileinput
# Does a list of files, and
# redirects STDOUT to the file in question
for line in fileinput.input(files, inplace = 1):
print line.replace("foo", "bar"),
Based on the answer by Thomas Watnedal.
However, this does not answer the line-to-line part of the original question exactly. The function can still replace on a line-to-line basis
This implementation replaces the file contents without using temporary files, as a consequence file permissions remain unchanged.
Also re.sub instead of replace, allows regex replacement instead of plain text replacement only.
Reading the file as a single string instead of line by line allows for multiline match and replacement.
import re
def replace(file, pattern, subst):
# Read contents from file as a single string
file_handle = open(file, 'r')
file_string = file_handle.read()
file_handle.close()
# Use RE package to allow for replacement (also allowing for (multiline) REGEX)
file_string = (re.sub(pattern, subst, file_string))
# Write contents to file.
# Using mode 'w' truncates the file.
file_handle = open(file, 'w')
file_handle.write(file_string)
file_handle.close()
As lassevk suggests, write out the new file as you go, here is some example code:
fin = open("a.txt")
fout = open("b.txt", "wt")
for line in fin:
fout.write( line.replace('foo', 'bar') )
fin.close()
fout.close()
If you're wanting a generic function that replaces any text with some other text, this is likely the best way to go, particularly if you're a fan of regex's:
import re
def replace( filePath, text, subs, flags=0 ):
with open( filePath, "r+" ) as file:
fileContents = file.read()
textPattern = re.compile( re.escape( text ), flags )
fileContents = textPattern.sub( subs, fileContents )
file.seek( 0 )
file.truncate()
file.write( fileContents )
A more pythonic way would be to use context managers like the code below:
from tempfile import mkstemp
from shutil import move
from os import remove
def replace(source_file_path, pattern, substring):
fh, target_file_path = mkstemp()
with open(target_file_path, 'w') as target_file:
with open(source_file_path, 'r') as source_file:
for line in source_file:
target_file.write(line.replace(pattern, substring))
remove(source_file_path)
move(target_file_path, source_file_path)
You can find the full snippet here.
fileinput is quite straightforward as mentioned on previous answers:
import fileinput
def replace_in_file(file_path, search_text, new_text):
with fileinput.input(file_path, inplace=True) as file:
for line in file:
new_line = line.replace(search_text, new_text)
print(new_line, end='')
Explanation:
fileinput can accept multiple files, but I prefer to close each single file as soon as it is being processed. So placed single file_path in with statement.
print statement does not print anything when inplace=True, because STDOUT is being forwarded to the original file.
end='' in print statement is to eliminate intermediate blank new lines.
You can used it as follows:
file_path = '/path/to/my/file'
replace_in_file(file_path, 'old-text', 'new-text')
Create a new file, copy lines from the old to the new, and do the replacing before you write the lines to the new file.
Expanding on #Kiran's answer, which I agree is more succinct and Pythonic, this adds codecs to support the reading and writing of UTF-8:
import codecs
from tempfile import mkstemp
from shutil import move
from os import remove
def replace(source_file_path, pattern, substring):
fh, target_file_path = mkstemp()
with codecs.open(target_file_path, 'w', 'utf-8') as target_file:
with codecs.open(source_file_path, 'r', 'utf-8') as source_file:
for line in source_file:
target_file.write(line.replace(pattern, substring))
remove(source_file_path)
move(target_file_path, source_file_path)
Using hamishmcn's answer as a template I was able to search for a line in a file that match my regex and replacing it with empty string.
import re
fin = open("in.txt", 'r') # in file
fout = open("out.txt", 'w') # out file
for line in fin:
p = re.compile('[-][0-9]*[.][0-9]*[,]|[-][0-9]*[,]') # pattern
newline = p.sub('',line) # replace matching strings with empty string
print newline
fout.write(newline)
fin.close()
fout.close()
if you remove the indent at the like below, it will search and replace in multiple line.
See below for example.
def replace(file, pattern, subst):
#Create temp file
fh, abs_path = mkstemp()
print fh, abs_path
new_file = open(abs_path,'w')
old_file = open(file)
for line in old_file:
new_file.write(line.replace(pattern, subst))
#close temp file
new_file.close()
close(fh)
old_file.close()
#Remove original file
remove(file)
#Move new file
move(abs_path, file)

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