I'm new and studying machine learning. I stumble upon a tutorial I found online and I'd like to make the program work so I'll get a better understanding. However, I'm getting problems about loading the CSV File into the Jupyter Notebook.
I get this error:
File "<ipython-input-2-70e07fb5b537>", line 2
student_data = pd.read_csv("C:\Users\xxxx\Desktop\student-intervention-
system\student-data.csv")
^
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in
position 2-3: truncated \UXXXXXXXX escape
and here is the code:
I followed tutorials online regarding this error but none worked. Does anyone know how to fix it?
3rd attempt with r"path"
I've tried also "\" and utf-8 but none worked.
I'm using the latest version of Anaconda
Windows 7
Python 3.7
Use raw string notation for your Windows path. In python '\' have meaning in python. Try instead do string like this r"path":
student_data = pd.read_csv(r"C:\Users\xxxx\Desktop\student-intervention-
system\student-data.csv")
If it doesnt work try this way:
import os
path = os.path.join('c:' + os.sep, 'Users', 'xxxx', 'Desktop', 'student-intervention-system', 'student-data.csv')
student_data = pd.read_csv(path)
Either replace all backslashes \ with frontslashes / or place a r before your filepath string to avoid this error. It is not a matter of your folder name being too long.
As Bohun Mielecki mentioned, the \ character which is typically used to denote file structure in Windows has a different function when written within a string.
From Python3 Documentation: The backslash \ character is used to escape characters that otherwise have a special meaning, such as newline, backslash itself, or the quote character.
How this particularly affects your statement is that in the line
student_data = pd.read_csv("C:\Users\xxxx\Desktop\student-intervention-
system\student-data.csv")
\Users matches the escape sequence \Uxxxxxxxx whereby xxxxxxxx refers to a Character with 32-bit hex value xxxxxxxx. Because of this, Python tries to find a 32-bit hex value. However as the -sers from Users doesn't match the xxxxxxxx format, you get the error:
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in
position 2-3: truncated \UXXXXXXXX escape
The reason why your code works now is that you have placed a r in front of 'C:\Users\xxxx\Desktop\project\student-data.csv'. This tells python not to process the backslash character / as it usually does and read the whole string as-is.
I hope this helps you better understand your problem. If you need any more clarification, do let me know.
Source: Python 3 Documentation
I had the same problem. I tried to encode it with 'Latin-1' and it worked for me.
autos = pd.read_csv('filename',encoding = "Latin-1")
Try changing \ to /: -
import pandas as pd
student_data = pd.read_csv("C:/Users/xxxx/Desktop/student-intervention-
system/student-data.csv")
print(student data)
OR
import pandas as pd
student_data = pd.read_csv("C:/Users/xxxx/Desktop/student-intervention- system/student-data.csv"r)
print(student data)
Try this student_data = pd.read_csv("C:/Users/xxxx/Desktop/student-intervention-
system/student-data.csv").
Replacing backslashes in that code it will work for you.
You probably have a file name with a backlash... try to write the path using two backlashes instead of one.
student_data = pd.read_csv("C:\\Users\\xxxx\\Desktop\\student-intervention-system\\student-data.csv")
Try
pd.read_csv('file_name',encoding = "utf-8")
I found the problem. The problem is my folder name that is really long. I changed my folder name into "project" and the data is now finally loaded! Silly!
Please open notepad, write csv format data into the file and opt 'Save As' to save the file with format .csv.
E.g. Train.csv
Use this file, ensure you mention the same path correctly for the above saved CSV file during python coding.
Import pandas as pd
df=pd.read_csv('C:/your_path/Train.csv')
I've seen people using existing .txt/other format files to covert their format to .csv by just renaming. Which actually does nothing than changing the name of the file. It doesn't become a CSV file at all.
Hope this helps. 🙏🙏
Generally this kind of error arises if there is any space in file path...
df=pd.read_csv('/home/jovyan/binder/kidney disease.csv')
above command will create an error and will get resolved when it is
df=pd.read_csv('/home/jovyan/binder/kidney_disease.csv')
replaced space with underscore
To begin with, this has nothing to do with Jupyter. This is a pure Python problem!
In Python, as in most languages, a backslash is an special (escape) character in strings, for example "foo\"bar" is the string foo"bar without causing a syntax error. Also, "foo\nbar" is the strings foo and bar with a newline in between. There are many more escapes.
In your case, the meaning of \ in your path is literal, i.e. you actually want a backslash appearing in the string.
One option is to escape the backslash itself with another backslash: "foo\\bar" amounts to the string foo\bar. However, in your case, you have several of these, so for readability you might want to switch on "raw string mode", which disables (almost all) escapes:
r"foo\bar\baz\quux\etc"
will produce
foo\bar\baz\quuz\etc
As a matter of programming style, though, if you want your code to be portable, it is better to use os.path.join which knows the right path separator for your OS/platform:
In [1]: import os.path
In [2]: os.path.join("foo", "bar", "baz")
Out[2]: 'foo/bar/baz'
on Windows, that would produce foo\bar\baz.
import pandas as pd
data=pd.read_csv("C:\Users\ss\Desktop\file or csv file name.csv")
just place the csv file on the desktop
I am trying to check if a particular directory path exists or not.
below is my code
temp_path = '\\diwali\NSID-HYD-01\college'
meta_path = os.path.realpath(temp_path)
print(os.path.exists(meta_path))
When I am trying to execute this, it is throwing error as below
temp_path = '\\diwali\NSID-HYD-01\college'
# ^
error
SyntaxError: (unicode error) 'unicodeescape' codec can't decode bytes in position 8-9: malformed \N character escape
Help me resolve this.
Python interprets backslashes (\) inside strings as leading characters for escape codes. For example \n is a line-feed character.
If you want it to treat them as simply backslashes, add an r before the string, like so:
temp_path = r'\\diwali\NSID-HYD-01\college'
another method is using two backslashes \\ before N like this:
temp_path = '\\diwali\\NSID-HYD-01\college'
If you are get it from UI (as you mentioned in comments) you can replace \ with \\:
temp_path = '\\diwali\NSID-HYD-01\college'.replace("\\", "\\\\")
# '\\diwali\\NSID-HYD-01\\college'
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.
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
I have a path to a file containing $ signs, which I escaped with \ so open() can handle the path. But now open turns the \$ to a \\$ automatically.
For example:
open("/home/test/\$somedir\$/file.txt", "r")
result in an error message
IOError: [Errno 2] No such file or directory: '/home/test/\\$somedir\\$/file.txt'
Can I supress this. Why open() do that? I can't find anything in the docu of open, which describes this.
open() doesn't do that. It's Python, which escapes any special characters when representing a string:
>>> path = '\$'
>>> path
'\\$'
>>> print path
\$
In a regular Python string literal, a \ has special meaning, so it is escaped when echoing back the value, which can be pasted right back into a Python script or interpreter session to recreate the same value.
On Linux or Mac, you generally do not need to escape a $ value in a filename; a $ has no special meaning in a regular python string, nor in most Linux or Mac filenames:
>>> os.listdir('/tmp/$somedir$')
['test']
>>> open('/tmp/$somedir$/test')
<open file '/tmp/$somedir$/test', mode 'r' at 0x105579390>
Try using a literal representation of the string adding r before the string variable to avoid dealing with more complex scape situations, for example:
print('C:\test')
#C: est
print(r'C:\test')
#C:\test
where \t is interpreted as a tab.