Develop Reference

Open multiple files from a file

I need to open a file that have multiple absolute file directories.
EX:
Layer 1 = C:\User\Files\Menu\Menu.snt
Layer 2 = C:\User\Files\N0 - Vertical.snt
The problem is that when I try to open C:\User\Files\Menu\Menu.snt python doesn't like \U or \N
I could open using r"C:\User\Files\Menu\Menu.snt" but I can't automate this process.
file = open(config.txt, "r").read()
list = []
for line in file.split("\n"):
list.append(open(line.split("=",1)[1]).read())
It prints out:
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 33-34: malformed \N character escape

The backslash character \ is used as an escape character by the Python interpreter in order to provide special characters.
For example, \n is a "New Line" character, like you would get from pressing the Return key on your keyboard.
So if you are trying to read something like newFolder1\newFolder2, the interpreter reads it as:
newFolder1
ewFolder2
where the New Line character has been inserted between the two lines of text.
You already mentioned one workaround: using raw strings like r'my\folder\structure' and I'm a little curious why this can't be automated.
If you can automate it, you could try replacing all instances of a single backslash (\) with a double backslash (\\) in your file paths and that should work.
Alternatively, you can try looking in the os module and dynamically building your paths using os.path.join(), along with the os.sep operator.
One final point: You can save yourself some effort by replacing:
list.append(open(line.split("=",1)[1]).read())
by
list = open(line.split("=",1)[1]).readlines()

here is my solution:
file = open("config.txt", "r").readlines()
list = [open(x.split("=")[1].strip(), 'r').read() for x in file]
readlines creates a list that contains all lines in file, there is no need to split the whole string.

python csv files read and upload

I want to upload all the csv files that meet certain condition in a directory to a database. But I encounter an error at the beginning of my code.
mypath = "D:\user\01367564\Project Coordinator\Database Trying\all data csv"
csv_name_reg = r'^[0-9]{11}_HKG_[0-9]{14}_v2-0.csv$'
The error is below
File "D:\user\01367564\Project Coordinator\Database Trying\Upload_CA_Manifest.py", line 9
mypath = "D:\user\01367564\Project Coordinator\Database Trying\all data csv"
^
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 2-3: truncated \uXXXX escape
Can you help me? Thank you.

Currently your path looks like it's meant to contain a Unicode character with the \u.... Please note that on Windows you have three options for paths:
Raw strings
mypath = r"D:\user\01367564\Project Coordinator\Database Trying\all data csv"
Escaped backslashes
mypath = "D:\\user\\01367564\\Project Coordinator\\Database Trying\\all data csv"
Forward slashes
mypath = "D:/user/01367564/Project Coordinator/Database Trying/all data csv"

In Python, there are some cool backslash escapes. A "\" inside a string plus a character(s).
Some notable ones are "\n" and "\t" which are newline and tab. A non-builtin backslash escape will be turned into the actual character in the final string. "\\" will turn into one "\" during, say, a print statement.
The escape Python thinks your using is the unicode escape. "\uXXXX". To fix this all you need is to replace each backslash with a double backslash. "\\". So this string will work: "D:\\user\\01367564\\Project Coordinator\\Database Trying\\all data csv"
For a full list of Python Backslash Escapes look at the Python Docs.

Copying files raises a SyntaxError can't decode bytes

I'm trying to code a short program that makes backups of a folder whenever I run it. Currently it's like this:
import time
import shutil
import os
date = time.strftime("%d-%m-%Y")
print(date)
shutil.copy2("C:\Users\joaop\Desktop\VanillaServer\world","C:\Users\joaop\Desktop\VanillaServer\Backups")
for filename in os.listdir("C:\Users\joaop\Desktop\VanillaServer\Backups"):
if filename == world:
os.rename(filename, "Backup " + date)
However I get an error:
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 2-3: truncated \UXXXXXXXX escape
and I can't figure out why (according to documentation, I think my code is properly written)
How can I fix this/do it in a better way?

In Python, \u... denotes a Unicode sequence, so your \Users directory is interpreted as a Unicode character -- not with very much success.
>>> "\u0061"
'a'
>>> "\users"
File "<stdin>", line 1
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 0-1: truncated \uXXXX escape
To fix it, you should escape the different \ as \\, or use r"..." to make it a raw string.
>>> "C:\\Users\\joaop\\Desktop\\VanillaServer\\world"
'C:\\Users\\joaop\\Desktop\\VanillaServer\\world'
>>> r"C:\Users\joaop\Desktop\VanillaServer\world"
'C:\\Users\\joaop\\Desktop\\VanillaServer\\world'
Don't do both, though, or else they will be escaped twice:
>>> r"C:\\Users\\joaop\\Desktop\\VanillaServer\\world"
'C:\\\\Users\\\\joaop\\\\Desktop\\\\VanillaServer\\\\world'
You only have to escape them when entering the paths directly in your source; if you read those paths from a file, from user input, or from some library function, they will automatically be escaped.

Backslashes are used for escape characters so when the interpreter sees the \ in your file path string it attempts to use them as an escape character (which are things like \n for new line and \t for tabs).
There are 2 ways around this, using raw strings or double slashing your file path so the interpeter ignores the escape sequence. Use a r to specify a raw string or \\. Now the choice in which you use is up to you but personally I prefer raw strings.
#with raw strings
shutil.copy2(r"C:\Users\joaop\Desktop\VanillaServer\world",r"C:\Users\joaop\Desktop\VanillaServer\Backups")
for filename in os.listdir(r"C:\Users\joaop\Desktop\VanillaServer\Backups"):
if filename == world:
os.rename(filename, "Backup " + date)
#with double slashes
shutil.copy2("C:\\Users\\joaop\\Desktop\\VanillaServer\\world","C:\\Users\\joaop\\Desktop\\VanillaServer\\Backups")
for filename in os.listdir("C:\\Users\\joaop\\Desktop\\VanillaServer\\Backups"):
if filename == world:
os.rename(filename, "Backup " + date)

How do I load fiducial node data from Slicer in Python? [duplicate]

This question already has answers here:
How should I write a Windows path in a Python string literal?
(5 answers)
Closed 2 years ago.
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I am using Python 3.1 on a Windows 7 machine. Russian is the default system language, and utf-8 is the default encoding.
Looking at the answer to a previous question, I have attempting using the "codecs" module to give me a little luck. Here's a few examples:
>>> g = codecs.open("C:\Users\Eric\Desktop\beeline.txt", "r", encoding="utf-8")
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 2-4: truncated \UXXXXXXXX escape (<pyshell#39>, line 1)
>>> g = codecs.open("C:\Users\Eric\Desktop\Site.txt", "r", encoding="utf-8")
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 2-4: truncated \UXXXXXXXX escape (<pyshell#40>, line 1)
>>> g = codecs.open("C:\Python31\Notes.txt", "r", encoding="utf-8")
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 11-12: malformed \N character escape (<pyshell#41>, line 1)
>>> g = codecs.open("C:\Users\Eric\Desktop\Site.txt", "r", encoding="utf-8")
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 2-4: truncated \UXXXXXXXX escape (<pyshell#44>, line 1)
My last idea was, I thought it might have been the fact that Windows "translates" a few folders, such as the "users" folder, into Russian (though typing "users" is still the correct path), so I tried it in the Python31 folder. Still, no luck. Any ideas?

The problem is with the string
"C:\Users\Eric\Desktop\beeline.txt"
Here, \U in "C:\Users... starts an eight-character Unicode escape, such as \U00014321. In your code, the escape is followed by the character 's', which is invalid.
You either need to duplicate all backslashes:
"C:\\Users\\Eric\\Desktop\\beeline.txt"
Or prefix the string with r (to produce a raw string):
r"C:\Users\Eric\Desktop\beeline.txt"

Typical error on Windows because the default user directory is C:\user\<your_user>, so when you want to pass this path as a string argument into a Python function, you get a Unicode error, just because the \u is a Unicode escape. If the next 8 characters after the \u are not numeric this produces an error.
To solve it, just double the backslashes: C:\\user\\<\your_user>...
This will ensure that Python treats the single backslashes as single backslashes.

Prefixing with 'r' works very well, but it needs to be in the correct syntax. For example:
passwordFile = open(r'''C:\Users\Bob\SecretPasswordFile.txt''')
No need for \\ here - maintains readability and works well.

With Python 3 I had this problem:
self.path = 'T:\PythonScripts\Projects\Utilities'
produced this error:
self.path = 'T:\PythonScripts\Projects\Utilities'
^
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in
position 25-26: truncated \UXXXXXXXX escape
the fix that worked is:
self.path = r'T:\PythonScripts\Projects\Utilities'
It seems the '\U' was producing an error and the 'r' preceding the string turns off the eight-character Unicode escape (for a raw string) which was failing. (This is a bit of an over-simplification, but it works if you don't care about unicode)
Hope this helps someone

Or you could replace '\' with '/' in the path.

path = pd.read_csv(**'C:\Users\mravi\Desktop\filename'**)
The error is because of the path that is mentioned
Add 'r' before the path
path = pd.read_csv(**r'C:\Users\mravi\Desktop\filename'**)
This would work fine.

I had this same error in python 3.2.
I have script for email sending and:
csv.reader(open('work_dir\uslugi1.csv', newline='', encoding='utf-8'))
when I remove first char in file uslugi1.csv works fine.

Refer to openpyxl document, you can do changes as followings.
from openpyxl import Workbook
from openpyxl.drawing.image import Image
wb = Workbook()
ws = wb.active
ws['A1'] = 'Insert a xxx.PNG'
# Reload an image
img = Image(**r**'x:\xxx\xxx\xxx.png')
# Insert to worksheet and anchor next to cells
ws.add_image(img, 'A2')
wb.save(**r**'x:\xxx\xxx.xlsx')

I had same error, just uninstalled and installed again the numpy package, that worked!

I had this error.
I have a main python script which calls in functions from another, 2nd, python script.
At the end of the first script I had a comment block designated with ''' '''.
I was getting this error because of this commenting code block.
I repeated the error multiple times once I found it to ensure this was the error, & it was.
I am still unsure why.

Python - Must add r when opening a file

I have several .py files and I can open my file everywhere, except in my test.py file (I test scripts and functions there) instead of this:
file = open("C:\Users\User\Desktop\key_values.txt", "r")
I need to use this (with r) to avoid error:
file = open(r"C:\Users\User\Desktop\key_values.txt", "r")
I get this error: (when I try to open a file without r in my test.py script)
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 2-3: truncated \UXXXXXXXX escape
Any idea why is this happening ?

Backslash is an escape character, so you can include characters like "\n" (new line) and "\t" (tab). The r before the string means means "my backslashes are not escape characters".
Interestingly, it looks like your string "C:\Users\User\Desktop\key_values.txt" works ok in python 2 because none of the backslashes are part of anything looking like a known escape sequence. But in python 3, "\Uxxxx" indicates a unicode character. So maybe that is why some of your python files can cope and some can't.

The other answers are OK.. but this a time saving trick:
Try using slashes instead of backslashes:
file = open("C:/Users/User/Desktop/key_values.txt", "r")
It works in Windows. Tried with Python 2.7
Hope this helps