I modified the code based on the comments from experts in this thread. Now the script reads and writes all the individual files. The script reiterates, highlight and write the output. The current issue is, after highlighting the last instance of the search item, the script removes all the remaining contents after the last search instance in the output of each file.
Here is the modified code:
import os
import sys
import re
source = raw_input("Enter the source files path:")
listfiles = os.listdir(source)
for f in listfiles:
filepath = source+'\\'+f
infile = open(filepath, 'r+')
source_content = infile.read()
color = ('red')
regex = re.compile(r"(\b be \b)|(\b by \b)|(\b user \b)|(\bmay\b)|(\bmight\b)|(\bwill\b)|(\b's\b)|(\bdon't\b)|(\bdoesn't\b)|(\bwon't\b)|(\bsupport\b)|(\bcan't\b)|(\bkill\b)|(\betc\b)|(\b NA \b)|(\bfollow\b)|(\bhang\b)|(\bbelow\b)", re.I)
i = 0; output = ""
for m in regex.finditer(source_content):
output += "".join([source_content[i:m.start()],
"<strong><span style='color:%s'>" % color[0:],
source_content[m.start():m.end()],
"</span></strong>"])
i = m.end()
outfile = open(filepath, 'w+')
outfile.seek(0)
outfile.write(output)
print "\nProcess Completed!\n"
infile.close()
outfile.close()
raw_input()
The error message tells you what the error is:
No such file or directory: 'sample1.html'
Make sure the file exists. Or do a try statement to give it a default behavior.
The reason why you get that error is because the python script doesn't have any knowledge about where the files are located that you want to open.
You have to provide the file path to open it as I have done below. I have simply concatenated the source file path+'\\'+filename and saved the result in a variable named as filepath. Now simply use this variable to open a file in open().
import os
import sys
source = raw_input("Enter the source files path:")
listfiles = os.listdir(source)
for f in listfiles:
filepath = source+'\\'+f # This is the file path
infile = open(filepath, 'r')
Also there are couple of other problems with your code, if you want to open the file for both reading and writing then you have to use r+ mode. More over in case of Windows if you open a file using r+ mode then you may have to use file.seek() before file.write() to avoid an other issue. You can read the reason for using the file.seek() here.
Related
This question already has answers here:
How do I check whether a file exists without exceptions?
(40 answers)
Closed 1 year ago.
I am trying to write a block of code which opens a new file every time a Python3 script is run.
I am constructing the filename using an incrementing number.
For example, the following are some examples of valid filenames which should be produced:
output_0.csv
output_1.csv
output_2.csv
output_3.csv
On the next run of the script, the next filename to be used should be output_4.csv.
In C/C++ I would do this in the following way:
Enter an infinite loop
Try to open the first filename, in "read" mode
If the file is open, increment the filename number and repeat
If the file is not open, break out of the loop and re-open the file in "write" mode
This doesn't seem to work in Python 3, as opening a non-existing file in read mode causes an exception to be raised.
One possible solution might be to move the open file code block inside a try-catch block. But this doesn't seem like a particularly elegant solution.
Here is what I tried so far in code
# open a file to store output data
filename_base = "output"
filename_ext = "csv"
filename_number = 0
while True:
filename_full = f"{filename_base}_{filename_number}.{filename_ext}"
with open(filename_full, "r") as f:
if f.closed:
print(f"Writing data to {filename_full}")
break
else:
print(f"File {filename_full} exists")
filename_number += 1
with open(filename_full, "w") as f:
pass
As explained above this code crashes when trying to open a file which does not exist in "read" mode.
Using pathlib you can check with Path.is_file() which returns True when it encounters a file or a symbolic link to a file.
from pathlib import Path
filename_base = "output"
filename_ext = "csv"
filename_number = 0
filename_full = f"{filename_base}_{filename_number}.{filename_ext}"
p = Path(filename_full)
while p.is_file() or p.is_dir():
filename_number += 1
p = Path(f"{filename_base}_{filename_number}.{filename_ext}")
This loop should exit when the file isn’t there so you can open it for writing.
you can check if a file exists prior using
os.path.exists(filename)
You could use the OS module to check if the file path is a file, and then open it:
import os
file_path = './file.csv'
if(os.path.isfile(file_path)):
with open(file_path, "r") as f:
This should work:
filename_base = "output"
filename_ext = "csv"
filename_number = 0
while True:
filename_full = f"{filename_base}_{filename_number}.{filename_ext}"
try:
with open(filename_full, "r") as f:
print(f"File {filename_full} exists")
filename_number += 1
except FileNotFoundError:
print("Creating new file")
open(filename_full, 'w');
break;
You might os.path.exists to check if file already exists for example
import os
print(os.path.exists("output_0.csv"))
or harness fact that your names
output_0.csv
output_1.csv
output_2.csv
output_3.csv
are so regular, exploit glob.glob like so
import glob
existing = glob.glob("output_*.csv")
print(existing) # list of existing files
I wrote a python script that takes two files as input and then saves the difference between them as output in another file.
I bound it to a batch file .cmd (see below) and added the batch file to context menu of text files, so when I right-click on a text file and select it, a cmd window pops up and I type the address of the file to compare.
Batch file content:
#echo off
cls
python "C:\Users\User\Desktop\Difference of Two Files.py" %1
Python Code:
import sys
import os
f1 = open(sys.argv[1], 'r')
f1_name = str(os.path.basename(f1.name)).rsplit('.')[0]
f2_path = input('Enter the path of file to compare: ')
f2 = open(f2_path, 'r')
f2_name = str(os.path.basename(f2.name)).rsplit('.')[0]
f3 = open(f'{f1_name} - {f2_name} diff.txt', 'w')
file1 = set(f1.read().splitlines())
file2 = set(f2.read().splitlines())
difference = file1.difference(file2)
for i in difference:
f3.write(i + '\n')
f1.close()
f2.close()
f3.close()
Now, my question is how can I replace the typing of 2nd file path with a drag and drop solution that accepts more than one file.
I don't have any problem with python code and can extend it myself to include more files. I just don't know how to edit the batch file so instead of taking just one file by typing the path, it takes several files by drag and drop.
I would appreciate your help.
Finally, I've figured it out myself!
I Post the final code, maybe it helps somebody.
# This script prints those lines in the 1st file that are not in the other added files
# and saves the results into a 3rd file on Desktop.
import sys
import os
f1 = open(sys.argv[1], 'r')
f1_name = str(os.path.basename(f1.name)).rsplit('.')[0]
reference_set = set(f1.read().splitlines())
compare_files = input('Drag and drop files into this window to compare: ')
compare_files = compare_files.strip('"').rstrip('"')
compare_files_list = compare_files.split('\"\"')
compare_set = set()
for file in compare_files_list:
with open(os.path.abspath(file), 'r') as f2:
file_content = set(f2.read().splitlines())
compare_set.update(file_content)
f3 = open(f'C:\\Users\\User\\Desktop\\{f1_name} diff.txt', 'w')
difference = reference_set.difference(compare_set)
for i in difference:
f3.write(i + '\n')
f1.close()
f3.close()
The idea came from this fact that drag and drop to cmd, copies the file path surrounded with double-quotes into it. I used the repeated double-quotes between paths to create a list, and you can see the rest in the code.
However, there's a downside and it's that you can't drag multiple files together and you should do that one by one, but it's better than nothing. ;)
I have a .fhx file that I could open normally with notepad but I want to open it using Python. I have tried subprocess.popen which I got online but I keep getting errors. I also want to be able to read the contents of this file like a normal text file like how we do in f=open("blah.txt", "r") and f.read(). Could anyone guide me in the right direction ?
import subprocess
filepath = "C:\Users\Ch\Desktop\FHX\fddd.fhx"
notePath = r'C:\Windows\System32\notepad.exe'
subprocess.Popen("%s %s" % (notePath, filepath))
Solved my problem by adding encoding="utf16" to the file open command.
count = 1
filename = r'C:\Users\Ch\Desktop\FHX\27-ESDC_CM02-2.fhx'
f = open(filename, "r", encoding="utf16") #Does not work without encoding
lines = f.read().splitlines()
for line in lines:
if "WIRE SOURCE" in line:
liner = line.split()
if any('SOURCE="INPUT' in s for s in liner):
print(str(count)+") ", "SERIAL INPUT = ", liner[2].replace("DESTINATION=", ""))
count += 1
Now I'm able to get the data the way I wanted.Thanks everyone.
try with shell=True argument
subprocess.call((notePath, filepath), shell=True )
You should be passing a list of args:
import subprocess
filepath = r"C:\Users\Ch\Desktop\FHX\fddd.fhx"
notePath = r'C:\Windows\System32\notepad.exe'
subprocess.check_call([notePath, filepath])
If you want to read the contents then just open the file using open:
with open(r"C:\Users\Ch\Desktop\FHX\fddd.fhx") as f:
for line in f:
print(line)
You need to use raw string for the path also to escape the f n your file path name, if you don't you are going to get errors.
In [1]: "C:\Users\Ch\Desktop\FHX\fddd.fhx"
Out[1]: 'C:\\Users\\Ch\\Desktop\\FHX\x0cddd.fhx'
In [2]: r"C:\Users\Ch\Desktop\FHX\fddd.fhx"
Out[2]: 'C:\\Users\\Ch\\Desktop\\FHX\\fddd.fhx'
I'm having a problem opening the names.txt file. I have checked that I am in the correct directory. Below is my code:
import os
print(os.getcwd())
def alpha_sort():
infile = open('names', 'r')
string = infile.read()
string = string.replace('"','')
name_list = string.split(',')
name_list.sort()
infile.close()
return 0
alpha_sort()
And the error I got:
FileNotFoundError: [Errno 2] No such file or directory: 'names'
Any ideas on what I'm doing wrong?
You mention in your question body that the file is "names.txt", however your code shows you trying to open a file called "names" (without the ".txt" extension). (Extensions are part of filenames.)
Try this instead:
infile = open('names.txt', 'r')
As a side note, make sure that when you open files you use universal mode, as windows and mac/unix have different representations of carriage returns (/r/n vs /n etc.). Universal mode gets python to handle this, so it's generally a good idea to use it whenever you need to read a file. (EDIT - should read: a text file, thanks cameron)
So the code would just look like this
infile = open( 'names.txt', 'rU' ) #capital U indicated to open the file in universal mode
This doesn't solve that issue, but you might consider using with when opening files:
with open('names', 'r') as infile:
string = infile.read()
string = string.replace('"','')
name_list = string.split(',')
name_list.sort()
return 0
This closes the file for you and handles exceptions as well.
I am trying to create bulk text files based on list. A text file has number of lines/titles and aim is to create text files. Following is how my titles.txt looks like along with non-working code and expected output.
titles = open("C:\\Dropbox\\Python\\titles.txt",'r')
for lines in titles.readlines():
d_path = 'C:\\titles'
output = open((d_path.lines.strip())+'.txt','a')
output.close()
titles.close()
titles.txt
Title-A
Title-B
Title-C
new blank files to be created under directory c:\\titles\\
Title-A.txt
Title-B.txt
Title-C.txt
It's a little difficult to tell what you're attempting here, but hopefully this will be helpful:
import os.path
with open('titles.txt') as f:
for line in f:
newfile = os.path.join('C:\\titles',line.strip()) + '.txt'
ff = open( newfile, 'a')
ff.close()
If you want to replace existing files with blank files, you can open your files with mode 'w' instead of 'a'.
The following should work.
import os
titles='C:/Dropbox/Python/titles.txt'
d_path='c:/titles'
with open(titles,'r') as f:
for l in f:
with open(os.path.join(d_path,l.strip()),'w') as _:
pass