FileCmp | quick and easy file comparison for windows
kandi X-RAY | FileCmp Summary
kandi X-RAY | FileCmp Summary
quick and easy file comparison for windows
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QUESTION
When I run the following code on python3, the result of the filecmp() is False. Why is it so? I thought compressing twice the same file would output two files with the same exact content.
...ANSWER
Answered 2022-Feb-23 at 15:09The gzip
ed file is a structured file rather than just a file with some bytes. It has header information, compressed content, and a footer. When you create a gzip file, the header and footer is also added to the compressed file content.
In your case, you are compressing the same file twice with the same compression code and level, so the compressed content will be same. But the gzip header is going to be different. As per the documentation python allows you to configure the header values of filename
and modification_time
. If these are not specified, a default value like current time is used.
In your case, every time you compress the same file, everything remains the same but the header is different. So there is a change in the file content and filecmp
returns False
. If you want to make the output files the same, then you can use:
gzip.GzipFile(filename=None, mode=None, compresslevel=9, fileobj=None, mtime=None)
to compress the contents, with identical header information.
QUESTION
I'm trying to compile a file that makes use of Python's C API. I'm working in a conda enviroment, running on macOS Monterey. I'm compiling using GCC as following:
...ANSWER
Answered 2022-Jan-01 at 06:20This command: gcc file.o -o a.out
does not link to a python library.
You need to add (append) -lpython3
and possibly -L${CONDA_PREFIX}/lib/python3.9
to it.
QUESTION
import filecmp
import os
import tkinter as tk
import tkinter.font as tkFont
from tkinter import *
def gg():
disk = (number_entry.get())
if disk == "2" :
DIRS = [r'C:\Newtest\1', r'C:\Newtest\2', r'C:\Newtest\3',r'C:\Newtest\4',r'C:\Newtest\5',r'C:\Newtest\6',r'C:\Newtest\7',r'C:\Newtest\8',r'C:\Newtest\9',r'C:\Newtest\10']
FILES = [('copy1.txt', 'copy2.txt'), ('fcmp1.txt', 'fcmp2.txt'), ('filecp1.txt', 'filecp2.txt')]
for e, dir in enumerate(DIRS, 1):
cmp = True
for file in FILES:
try:
if not filecmp.cmp(os.path.join(dir, file[0]), os.path.join(dir, file[1])):
cmp = False
break
except Exception as ex:
print(f'Error -> {ex}', file=sys.stderr)
continue
x = (f'E_{e} compare {"pass" if cmp else "fail"}')
print(x)
result_label.configure(text=x,wraplength=300,font=fontExample3)
DIRS = [r'C:\Newtest\1', r'C:\Newtest\2', r'C:\Newtest\3',r'C:\Newtest\4',r'C:\Newtest\5',r'C:\Newtest\6',r'C:\Newtest\7',r'C:\Newtest\8',r'C:\Newtest\9',r'C:\Newtest\10']
FILES = [('copy1.txt', 'copy2.txt'), ('fcmp1.txt', 'fcmp2.txt'), ('filecp1.txt', 'filecp2.txt')]
for e, dir in enumerate(DIRS, 1):
cmp = True
window = tk.Tk()
window.title('XXX')
window.geometry('1200x700')
window.configure(background='Azure')
fontExample = tkFont.Font(family="Segoe UI", size=20)
fontExample2 = tkFont.Font(family="Segoe UI", size=10)
fontExample3 = tkFont.Font(family="Segoe UI", size=15)
header_label = tk.Label(window, text='XXXX',font=fontExample)
header_label.pack()
number_frame = tk.Frame(window)
number_frame.pack(side=tk.TOP)
number_label = tk.Label(number_frame, text='Number of disks',font=fontExample2)
number_label.pack(side=tk.LEFT)
number_entry = tk.Entry(number_frame)
number_entry.pack(side=tk.LEFT)
result_label = tk.Label(window)
result_label.pack()
calculate_btn = tk.Button(window, text='First calculate', command=gg,font=fontExample3)
calculate_btn.pack()
window.mainloop()
...ANSWER
Answered 2021-Oct-07 at 07:01If you want to add to the text that is already on the label, you can do this:
QUESTION
So, I've tried using Differ from difflib and also filecmp. Here's what I've done so far:
...ANSWER
Answered 2021-Oct-01 at 06:17Remove all spaces and newlines, convert to lower case, and compare.
QUESTION
I want to compare two directories for all ".bin" files in them. There can be some other extension type files such as ".txt", ".tar.bz2" in those directories. I want to get the common files as well as files which are not common. I tried using filecmp.dircmp(), but I am not able to use the ignore parameter with some wild card to ignore those files. Is there any solution which I can use to serve my purpose.
...ANSWER
Answered 2021-Sep-30 at 11:07Select the common subset of *.bin files in the two folders and remove the first part of the path (the folder name), then pass them to cmpfiles()
:
QUESTION
Im using python to check in hundred of websites if a web banner in GIF format exist.
I have a folder with gif files as examples and Im compairing the gif files from each web sites with my own examples files.
I use filecmp but I found that many webs compress the gif files so even the files are visually identical the filecmp wont detect as same.
Is there any python library to detect if two gif files or vídeos are similar even if the resolución has changed?
ANSWER
Answered 2021-Sep-26 at 21:11Comparing two images for similarity is a general image processing problem so the solution you develop can be as simple or complex as you want it to be. In your specific case, you'll need a method for making two images the same size and a method for comparing the images.
First, you'll probably want to convert the images to RGB or grayscale arrays for comparison.
I would suggest reducing the size of the larger image to the size of the smaller image. That is less likely to introduce artifacts than increasing the size of the smaller image. Resizing can be accomplished with the Python Pillow library.
QUESTION
I want to send a mail from within a script if the comparison of two files fails.
(The script monitors a website and stores a sample of that page as log file and compares each day to the last day version and sends a mail in case something changes)
I have the following code snippet and it works fine and sends the mail.
(Of course I have replaced my actual credentials with samples here, but the code works with my credentials entered)
The sample code (working):
...ANSWER
Answered 2021-Sep-22 at 21:32The answer described by Serge Ballesta solves the issue.
The indentation in the message string creates spaces which lead to a faulty message which cannot be processed further after received by the smtp server.
This is the right format:
QUESTION
i try to compare 2 folders if the are equal inside -
I tried it with this code
...ANSWER
Answered 2021-Jul-25 at 20:35QUESTION
I have a function with three args
- list_to_copy : list of files that I want to copy.
- list_to_avoid : list that I don't want to copy, that might exists in list_to_copy
- the destination is the path where the file is going to be.
ANSWER
Answered 2021-May-30 at 09:34from filecmp import cmp
from shutil import copy
from os import scandir
def copy_files(list_to_copy,destination):
# Copy a file from list to destination making sure file is not duplicated regarding the content.
for copied_file in list_to_copy:
for avoid_file in scandir(destination):
if cmp(copied_file,avoid_file):
break
else:
copy(copied_file, destination)
QUESTION
So I manually installed a locally downloaded python package by going into the folder directory and using the cmd command:
python setup.py install
After that it just installed itself normally. Using the python function help("modules")
in cmd also confirmed that it was installed correctly as I can see the name being given out. The two modules are called binance_d
and binance_f
ANSWER
Answered 2021-Apr-16 at 07:38I followed this document and I can get what I want. The most importance thing is that the command does not copy the generated files into the pyhton 3.9.4 folder automatically. You have to copy them manually.
1) first download the project under this link and then unpack the file.
Run these under cmd:
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