We Keep Coding

why is the error thinking Its not string?

path of file is:
"C:\Users\deana\OneDrive\Marlon's files\Programming\Python\PITT\PITT_LIbrary\Lists\test.txt"
lines of code are:
import os
os.chdir("C:/Users/deana/OneDrive/Marlon's files/Programming/Python/PITT/PITT_LIbrary/Lists")
exec(open('test.txt'))
the error is this:
Traceback (most recent call last):
File "<pyshell#14>", line 1, in <module>
exec(open('test.txt'))
TypeError: exec() arg 1 must be a string, bytes or code object
also if I try on one line as such:
exec(open(r"C:/Users/deana/OneDrive/Marlon's files/Programming/Python/PITT/PITT_LIbrary/Lists/test.txt"))
i'ts the same error. (with and without r)
super frustrationg as it reads like i'm not inputting string... but it is string!?!
also I've done this litteraly the same way before, restarted IDLE shell, no difference.
ugh! I always get stupid errors with file paths.

I should have been using os.startfile() to open this.
It was confusing by using .open(). as I was attempting to open in default app.
before, i've used exec.open() to open .py files and guess I got them confused.
exec is just used to open other scripts... need stronger coffee next time.

Try this:
import os
os.chdir("C:/Users/deana/OneDrive/Marlon's files/Programming/Python/PITT/PITT_LIbrary/Lists")
exec(open('test.txt', 'rb'))
You can convert the txt file to bytes by opening it with rb (read bytes).

Error when using Matplotlib.image in Python

Okay, so I just started this tutorial using Anaconda in PyCharm. I imported the correct libraries, but when I try to upload my image using this code:
img = mpimg.imread('FileName.png')
I get this error message:
Traceback (most recent call last):
File "<input>", line 1, in <module>
File ".../matplotlib/image.py", line 1323, in imread
with open(fname, 'rb') as fd:
IOError: [Errno 2] No such file or directory: 'FileName.png'
I tried putting in the file location like this:
img=mpimg.imread('FilePath/FileName.png')
But I got the same error. I want to follow the tutorial exactly so I'm not sure why this isn't working! Any help is appreciated!

Make sure that there is that file in the folder...
Try this command to list the files in that folder just to make sure that python atleast recognizes or reads the files.
import os
print (os.listdir('your path'))

you should give an absolute path for the image. for example
r'C:\Users\HP\Desktop\stinkbug.png' adding r to remove the Unicode error

I was providing the full path: "C:/user/username/Downloads/folder/file.ext"
When I changed it to relative, it worked for me: "Downloads/folder/file.ext"

File not found Error in reading text in python [duplicate]

This question already has answers here:
How should I write a Windows path in a Python string literal?
(5 answers)
Closed 7 months ago.
I am trying to read a text file on my hard drive via python with the following script:
fileref = open("H:\CloudandBigData\finalproj\BeautifulSoup\twitter.txt","r")
But it is giving the following error:
IOError Traceback (most recent call last)
<ipython-input-2-4f422ec273ce> in <module>()
----> 1 fileref = open("H:\CloudandBigData\finalproj\BeautifulSoup\twitter.txt","r")
IOError: [Errno 2] No such file or directory: 'H:\\CloudandBigData\x0cinalproj\\BeautifulSoup\twitter.txt'
I also tried with other way:
with open('H:\CloudandBigData\finalproj\BeautifulSoup\twitter.txt', 'r') as f:
print f.read()
Ended up with the same error. The text file is present in the directory specified.

Replace
fileref = open("H:\CloudandBigData\finalproj\BeautifulSoup\twitter.txt","r")
with
fileref = open(r"H:\CloudandBigData\finalproj\BeautifulSoup\twitter.txt","r")
Here, I have created a raw string (r""). This will cause things like "\t" to not be interpreted as a tab character.
Another way to do it without a raw string is
fileref = open("H:\\CloudandBigData\\finalproj\\BeautifulSoup\\twitter.txt","r")
This escapes the backslashes (i.e. "\\" => \).
An even better solution is to use the os module:
import os
filepath = os.path.join('H:', 'CloudandBigData', 'finalproj', 'BeautifulSoup', 'twitter.txt')
fileref = open(filepath, 'r')
This creates your path in an os-independent way so you don't have to worry about those things.
One last note... in general, I think you should use the with construct you mentioned in your question... I didn't in the answer for brevity.

I was encountering same problem. This problem resulted due to different file path notation Python.
For example, filepath in Windows reads with backward slash like: "D:\Python\Project\file.txt"
But Python reads file path with forward slash like: "D:/Python/Project/file.txt"
I used r"filepath.txt" and "os.path.join" and "os.path.abspath" to no relief. os library also generates file path in Windows notation. Then I just resorted to IDE notation.
You don't encounter this error if "file.txt" is located in same directory, as filename is appended to working directory.
PS: I am using Python 3.6 with Spyder IDE on Windows machine.

Error opening a csv file in python from a specific directory

I am very new to python and I am not having much experience in programming.
I try to open a CSV file from a specific directory and I get error.
import csv
ifile = open('F:\Study\CEN\Mini Project\Data Sets\test.csv', "rb");
Error:
Traceback (most recent call last):
File "<pyshell#6>", line 1, in <module>
ifile = open('F:\Study\CEN\Mini Project\Data Sets\test.csv', "rb");
IOError: [Errno 22] invalid mode ('rb') or filename: 'F:\\Study\\CEN\\Mini Project\\Data Sets\test.csv'
What to do ????

Use forward slashes:
ifile = open('F:/Study/CEN/Mini Project/Data Sets/test.csv', "rb");
Or at least escape your backslashes:
ifile = open('F:\\Study\\CEN\\Mini Project\\Data Sets\\test.csv', "rb");
Another option: use os.path.join:
out = os.path.abspath(os.path.join('path', 'test.csv'))

Your problem is here:
'F:\Study\CEN\Mini Project\Data Sets\test.csv'
^^
Because you did not use a raw string, Python thinks \t is supposed to mean a tab character.
You can see that in the error message, by the way: Notice how Python translated all the backslashes into double backslashes (which is how a literal backslash needs to be represented in a normal string) in all the places except the one where "backslash plus letter" actually meant something special?
Use
ifile = open(r'F:\Study\CEN\Mini Project\Data Sets\test.csv', "rb")
(and remove the semicolons, you don't need them in Python) and it should work.

Your problem is with the "\t" AND a lack of exposure to various tools in the os.path package
The correct and easiest way to deal with this problem is to use os.path.normpath, combined with the string literal r, which ensures that backslashes are not interpreted as an escape character.
(Documentation on Lexical Analysis in python can be found here: https://docs.python.org/2/reference/lexical_analysis.html)
Open interactive python by typing "python" at the command line, and do the following to see that it's dead simple.
>>> import os
>>> path = r'F:\Study\CEN\Mini Project\Data Sets\test.csv'
>>> os.path.normpath(path)
'F:\\Study\\CEN\\Mini Project\\Data Sets\\test.csv'
normpath should be used when using hardcoded paths for scripts that may have to run on both dos and unix (eg OS X). It will ensure that the right kind of slashes are used for your particular environment
On a side note, if you are working with CSV files, you should use the petl library instead of the csv module. You'll save yourself a lot of time and hassle. Install it with pip install petl

Weird python error - I/O error No such file of directory

I am trying to open python prompt and run the below code:
>>> a=open("Andrew_Smith_(author/education_professional)_0",'w')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IOError: [Errno 2] No such file or directory:
'Andrew_Smith_(author/education_professional)_0'
I am not sure why I am getting the error. I know the file contains special characters, but I am asking it to create a new file.
Edit:
I cannot use -, as some names might contain '-'. I also don't want to use space. Is there any other alternative?

As others have said the issue is the /, it is looking for a directory named Andrew_Smith_(author to create the new file education_professional)_0 in.
bash-3.2# mkdir "Andrew_Smith_(author"
bash-3.2# python
>>> a=open("Andrew_Smith_(author/education_professional)_0", 'w')
>>>

Because you have a / character in your filename. Neither *NIX nor Windows typically allow this.

I believe there are two problems here. First is because of the '/' character. It can't distinguish between / in a file name (which isn't valid) and / as a path seperator. And second, I don't believe '(' or ')' are valid in path names either.

The problem is that you are using reserved characters -- remove the / (or replace it with, for example, -) and everything should just work.