My codes work but for a few pain points that perhaps you can help me understand. I want to copy files from one directory to another and rename them at the same time. for example:
c:\path\
octo.jpeg
novem.jpeg
decem.jpeg
to:
c:\newpath\
001.jpeg
002.jpeg
003.jpeg
The codes I wrote from a cursory google search are as follows but I'm not sure why I need the 'r' in the path variables. The 'files = os.listdir(srcPath)' line I'm sure I don't need. This will move the files and renames them using the 'count' variable in the for loop but I want to name each file starting at a specific number, say 65. Should I use the shutil library and copy2 method to first copy the files and then rename or is there an easier way?
import os
from os import path
srcPath = r'C:\Users\Talyn\Desktop\New folder\Keep\New folder'
destPath = r'C:\Users\Talyn\Desktop\New folder\Keep\hold'
#files = os.listdir(srcPath)
def main():
for count, filename in enumerate(os.listdir(srcPath)):
dst = '{:03d}'.format(count) + ".jpeg"
os.rename(os.path.join(srcPath, filename), os.path.join(destPath, dst))
if __name__=="__main__":
main()
From the official Python Docs:
Both string and bytes literals may optionally be prefixed with a letter 'r' or 'R'; such strings are called raw strings and treat backslashes as literal characters.
The r is telling python interpreter to treat the backslashes(\) in the path string as literal characters and not as escaping characters.
For naming the files from a specific number:
dst = '{:03d}'.format(count + your_number) + ".jpeg"
Using copyfile from shutil
copyfile(srcPath + filename, destPath + dst)
Related
I wrote a script to copy files with specific names from one folder to another.
The file name format I want to copy is 2021052444592AKC. However, the script I wrote copies all files with the ending AKC, but in the if condition I specified that it should copy only files if the filename starts with "202105" and ends with "AKC". In the folder I have other files in the same format that is"YYYYMMDD44592threeUpperCaseLetters"
Can anyone help, because I haven't found the answer to this problem, thanks in advance :)
P.S I'm using Python3 in PyCharm
import shutil
import os
os.chdir(r"C:\\")
# without a double backslash and the letter r, the compiler throws an error
dir_src = r"C:\\Users\\Adam\\Desktop\1\\"
dir_dst = r"C:\\Users\\Adam\\Desktop\\2\\"
for filename in os.listdir(dir_src):
if filename.startswith("202105") and filename.endswith("AKC"):
shutil.copy(dir_src + filename, dir_dst)
print("End")
I'm not sure exactly why your script is failing, but you might want to try a solution with a regular expression (re).
import re
pattern = re.compile(r'^202105(\d{2})44592AKC$')
os.chdir(r"C:\\")
# without a double backslash and the letter r, the compiler throws an error
dir_src = r"C:\\Users\\Adam\\Desktop\\1\\"
dir_dst = r"C:\\Users\\Adam\\Desktop\\2\\"
for filename in os.listdir(dir_src):
if pattern.match(filename):
shutil.copy(dir_src + filename, dir_dst)
print("End")
I'm relatively new to Python, and have only recently started trying to use it for data analysis. I have a list of image files in a directory that have been acquired in sequence, and they have been named as so:
IMG_E5.1.tif
IMG_E5.2.tif
IMG_E5.3.tif
...
...
IMG_E5.107.tif
I would like to replace the dot and the number following it with an underscore and a four-digit integer, while preserving the initial numbering of the file, like so:
IMG_E5_0001.tif
IMG_E5_0002.tif
IMG_E5_0003.tif
...
...
IMG_E5_0107.tif
Could you advise me on how this can be done, or if there is already an answer that I'm not aware, link me to it? Many thanks!
I managed to find a method that works for this
import os
import os.path as path
from glob import glob
# Get current working directory
file_path = os.getcwd()
file_list = []
for i in range(1, 500):
# Generate file name (with wildcards) to search for
file_name = path.abspath(file_path + "/IMG*" + "." + str(i) + ".tif")
# Search for files
file = glob(file_name)
# If found, append to list
if len(file) > 1:
file_list.append(file[0])
elif len(file) == 1:
file_list.append(file[0])
for file in file_list:
# Use the "split" function to split the string at the periods
file_name, file_num, file_ext = file.split(".")
file_new = path.abspath(file_name + "_"
+ str(file_num).zfill(4)
+ "." + file_ext)
os.rename(file, file_new)
I am still relatively inexperienced with coding, so if there is a more straightforward and efficient way to tackle this problem, do let me know. Thanks.
I'm completely new to python and I'm trying to rename a set of files using a string on a specific line within the file and use it to rename the file. Such string is found at the same line in each file.
As an example:
10 files in the same path
the string is found in line number 14 and it starts at character number 40 and it is 50 characters in length
Then use the extracted string to rename the respective file
I'm trying to use this code, but I fail to figure out how to get it working:
for filename in os.listdir(path):
if filename.startswith("out"):
with open(filename) as openfile:
fourteenline = linecache.getline(path, 14)
os.rename(filename, fourteenline.strip())
Take care providing the full path to the file, in case your not allready working in this folder (use os.path.join()). Furthermore, when using linecache, you dont need to open the file.
import os, linecache
for filename in os.listdir(path):
if not filename.startswith("out"): continue # less deep
file_path = os.path.join(path, filename) # folderpath + filename
fourteenline = linecache.getline(file_path, 14) # maybe 13 for 0-based index?
new_file_name = fourteenline[40:40+50].rstrip() # staring at 40 with length of 50
os.rename(file_path, os.path.join(path, new_file_name))
useful ressources:
Reading specific lines only (Python)
Understanding Python's slice notation
How to rename a file using Python
python docs - string.strip()
I am trying to open a file that has a random time date stamp as part of its name. Most of the filename is known eg filename = 2017_01_23_624.txt is my test name. The date and numbers is the part I am trying to replace with something unknown as it will change.
My question relates to opening an existing file and not creating a new filename. The filenames are created by a separate program and I must be able to open them. I also want to change them once they are open.
I have been trying to construct the filename string as follow but get invalid syntax.
filename = "testfile" + %s + ".txt"
print(filename)
fo = open(filename)
You can't open a file with a partial filename containing a wildcard for it to match. What you would have to do it look at all the files in your directory and pick the one that matches best.
Simple example:
import os
filename = "filename2017_" # known section
direc = r"directorypath"
matches = [ fname for fname in os.listdir(direc) if fname.startswith(filename) ]
print(matches)
You can however use the glob module (Thanks to bzimor) to pattern match your files. See glob docs for more info.
import glob, os
# ? - represents any single character in the filename
# * - represents any number of characters in the filename
direc = r"directorypath"
pattern = 'filename2017-??-??-*.txt'
matches = glob.glob(os.path.join(direc, pattern))
print(matches)
Similar to first solution is that you still get back a list of filenames to choose from to then open. But with the glob module you can more accurately match your files if you so need. It all depends on how tight you want it to be.
My application offers the ability to the user to export its results. My application exports text files with name Exp_Text_1, Exp_Text_2 etc. I want it so that if a file with the same file name pre-exists in Desktop then to start counting from this number upwards. For example if a file with name Exp_Text_3 is already in Desktop, then I want the file to be created to have the name Exp_Text_4.
This is my code:
if len(str(self.Output_Box.get("1.0", "end"))) == 1:
self.User_Line_Text.set("Nothing to export!")
else:
import os.path
self.txt_file_num = self.txt_file_num + 1
file_name = os.path.join(os.path.expanduser("~"), "Desktop", "Exp_Txt" + "_" + str(self.txt_file_num) + ".txt")
file = open(file_name, "a")
file.write(self.Output_Box.get("1.0", "end"))
file.close()
self.User_Line_Text.set("A text file has been exported to Desktop!")
you likely want os.path.exists:
>>> import os
>>> help(os.path.exists)
Help on function exists in module genericpath:
exists(path)
Test whether a path exists. Returns False for broken symbolic links
a very basic example would be create a file name with a formatting mark to insert the number for multiple checks:
import os
name_to_format = os.path.join(os.path.expanduser("~"), "Desktop", "Exp_Txt_{}.txt")
#the "{}" is a formatting mark so we can do file_name.format(num)
num = 1
while os.path.exists(name_to_format.format(num)):
num+=1
new_file_name = name_to_format.format(num)
this would check each filename starting with Exp_Txt_1.txt then Exp_Txt_2.txt etc. until it finds one that does not exist.
However the format mark may cause a problem if curly brackets {} are part of the rest of the path, so it may be preferable to do something like this:
import os
def get_file_name(num):
return os.path.join(os.path.expanduser("~"), "Desktop", "Exp_Txt_" + str(num) + ".txt")
num = 1
while os.path.exists(get_file_name(num)):
num+=1
new_file_name = get_file_name(num)
EDIT: answer to why don't we need get_file_name function in first example?
First off if you are unfamiliar with str.format you may want to look at Python doc - common string operations and/or this simple example:
text = "Hello {}, my name is {}."
x = text.format("Kotropoulos","Tadhg")
print(x)
print(text)
The path string is figured out with this line:
name_to_format = os.path.join(os.path.expanduser("~"), "Desktop", "Exp_Txt_{}.txt")
But it has {} in the place of the desired number. (since we don't know what the number should be at this point) so if the path was for example:
name_to_format = "/Users/Tadhg/Desktop/Exp_Txt_{}.txt"
then we can insert a number with:
print(name_to_format.format(1))
print(name_to_format.format(2))
and this does not change name_to_format since str objects are Immutable so the .format returns a new string without modifying name_to_format. However we would run into a problem if out path was something like these:
name_to_format = "/Users/Bob{Cat}/Desktop/Exp_Txt_{}.txt"
#or
name_to_format = "/Users/Bobcat{}/Desktop/Exp_Txt_{}.txt"
#or
name_to_format = "/Users/Smiley{:/Desktop/Exp_Txt_{}.txt"
Since the formatting mark we want to use is no longer the only curly brackets and we can get a variety of errors:
KeyError: 'Cat'
IndexError: tuple index out of range
ValueError: unmatched '{' in format spec
So you only want to rely on str.format when you know it is safe to use. Hope this helps, have fun coding!