Confusing Error when Reading from a File in Python - python

I'm having a problem opening the names.txt file. I have checked that I am in the correct directory. Below is my code:
import os
print(os.getcwd())
def alpha_sort():
infile = open('names', 'r')
string = infile.read()
string = string.replace('"','')
name_list = string.split(',')
name_list.sort()
infile.close()
return 0
alpha_sort()
And the error I got:
FileNotFoundError: [Errno 2] No such file or directory: 'names'
Any ideas on what I'm doing wrong?

You mention in your question body that the file is "names.txt", however your code shows you trying to open a file called "names" (without the ".txt" extension). (Extensions are part of filenames.)
Try this instead:
infile = open('names.txt', 'r')

As a side note, make sure that when you open files you use universal mode, as windows and mac/unix have different representations of carriage returns (/r/n vs /n etc.). Universal mode gets python to handle this, so it's generally a good idea to use it whenever you need to read a file. (EDIT - should read: a text file, thanks cameron)
So the code would just look like this
infile = open( 'names.txt', 'rU' ) #capital U indicated to open the file in universal mode

This doesn't solve that issue, but you might consider using with when opening files:
with open('names', 'r') as infile:
string = infile.read()
string = string.replace('"','')
name_list = string.split(',')
name_list.sort()
return 0
This closes the file for you and handles exceptions as well.

Related

How to find first and last characters in a file using python?

I am stuck on this revision exercise which asks to copy an input file to an output file and return the first and last letters.
def copy_file(filename):
input_file = open(filename, "r")
content = input_file.read()
content[0]
content[1]
return content[0] + content[-1]
input_file.close()
Why do I get an error message which I try get the first and last letters? And how would I copy the file to the output file?
Here is the test:
input_f = "FreeAdvice.txt"
first_last_chars = copy_file(input_f)
print(first_last_chars)
print_content('cure737.txt')
Error Message:
FileNotFoundError: [Errno 2] No such file or directory: 'hjac737(my username).txt'
All the code after a return statement is never executed, a proper code editor would highlight it to you, so I recommend you use one. So the file was never closed. A good practice is to use a context manager for that : it will automatically call close for you, even in case of an exception, when you exit the scope (indentation level).
The code you provided also miss to write the file content, which may be causing the error you reported.
I explicitely used the "rt" (and "wt") mode for the files (althought they are defaults), because we want the first and last character of the file, so it supports Unicode (any character, not just ASCII).
def copy_file(filename):
with open(filename, "rt") as input_file:
content = input_file.read()
print(input_file.closed) # True
my_username = "LENORMJU"
output_file_name = my_username + ".txt"
with open(output_file_name, "wt") as output_file:
output_file.write(content)
print(output_file.closed) # True
# last: return the result
return content[0] + content[-1]
print(copy_file("so67730842.py"))
When I run this script (on itself), the file is copied and I get the output d) which is correct.

Bulk autoreplacing string in the KML file

I have a set of placemarks, which include quite a wide description included in its balloon within the property. Next each single description (former column header) is bounded in . Because of the shapefile naming restriction to 10 characters only.
https://gis.stackexchange.com/questions/15784/bypassing-10-character-limit-of-field-name-in-shapefiles
I have to retype most of these names manually.
Obviously, I use Notepad++, where I can swiftly press Ctrl+F and toggle Replace mode, as you can see below.
The green bounded strings were already replaced, the red ones still remain.
Basically, if I press "Replace All" then it works fine and quickly. Unfortunately, I have to go one by one. As you can see I have around 20 separate strings to "Replace all". Is there a possibility to do it quicker? Because all the .kml files are similar to each other, this is going to be the same everywhere. I need some tool, which will be able to do auto-replace for these headers cut by 10 characters limit. I think, that maybe Python tools might be helpful.
https://pythonhosted.org/pykml/
But in the tool above there is no information about bulk KML editing.
How can I set something like the "Replace All" tool for all my strings preferably if possible?
UPDATE:
I tried the code below:
files = []
with open("YesNF016.kml") as f:
for line in f.readlines():
if line[-1] == '\n':
files.append(line[:-1])
else:
files.append(line)
old_expression = 'ab'
new_expression = 'it worked'
for file in files:
new_file = ""
with open(file) as f:
for line in f.readlines():
new_file += line.replace(old_expression, new_expression)
with open(file, 'w') as f:
f.write(new_file)
The debugger shows:
[Errno 22] Invalid argument: ''
File "\test.py", line 13, in
with open(file) as f:
whereas line 13 is:
with open(file) as f:
The solutions here:
https://www.reddit.com/r/learnpython/comments/b9cljd/oserror_while_using_elementtree_to_parse_simple/
and
OSError: [Errno 22] Invalid argument Getting invalid argument while parsing xml in python
weren't helpful enough for me.
So you want to replace all occurence of X to Y in bunch of files ?
Pretty easy.
Just create a file_list.txt containing the list of files to edit.
python code:
files = []
with open("file_list.txt") as f:
for line in f.readlines():
if line[-1] == '\n':
files.append(line[:-1])
else:
files.append(line)
old_expression = 'ab'
new_expression = 'it worked'
for file in files:
new_file = ""
with open(file) as f:
for line in f.readlines():
new_file += line.replace(old_expression, new_expression)
with open(file, 'w') as f:
f.write(new_file)

How to open a file of different extension with notepad in python

I have a .fhx file that I could open normally with notepad but I want to open it using Python. I have tried subprocess.popen which I got online but I keep getting errors. I also want to be able to read the contents of this file like a normal text file like how we do in f=open("blah.txt", "r") and f.read(). Could anyone guide me in the right direction ?
import subprocess
filepath = "C:\Users\Ch\Desktop\FHX\fddd.fhx"
notePath = r'C:\Windows\System32\notepad.exe'
subprocess.Popen("%s %s" % (notePath, filepath))
Solved my problem by adding encoding="utf16" to the file open command.
count = 1
filename = r'C:\Users\Ch\Desktop\FHX\27-ESDC_CM02-2.fhx'
f = open(filename, "r", encoding="utf16") #Does not work without encoding
lines = f.read().splitlines()
for line in lines:
if "WIRE SOURCE" in line:
liner = line.split()
if any('SOURCE="INPUT' in s for s in liner):
print(str(count)+") ", "SERIAL INPUT = ", liner[2].replace("DESTINATION=", ""))
count += 1
Now I'm able to get the data the way I wanted.Thanks everyone.
try with shell=True argument
subprocess.call((notePath, filepath), shell=True )
You should be passing a list of args:
import subprocess
filepath = r"C:\Users\Ch\Desktop\FHX\fddd.fhx"
notePath = r'C:\Windows\System32\notepad.exe'
subprocess.check_call([notePath, filepath])
If you want to read the contents then just open the file using open:
with open(r"C:\Users\Ch\Desktop\FHX\fddd.fhx") as f:
for line in f:
print(line)
You need to use raw string for the path also to escape the f n your file path name, if you don't you are going to get errors.
In [1]: "C:\Users\Ch\Desktop\FHX\fddd.fhx"
Out[1]: 'C:\\Users\\Ch\\Desktop\\FHX\x0cddd.fhx'
In [2]: r"C:\Users\Ch\Desktop\FHX\fddd.fhx"
Out[2]: 'C:\\Users\\Ch\\Desktop\\FHX\\fddd.fhx'

Python: Unable to open and read a file

I am totally new to python.
I was trying to read a file which I already created but getting the below error
File "C:/Python25/Test scripts/Readfile.py", line 1, in <module>
filename = open('C:\Python25\Test scripts\newfile','r')
IOError: [Errno 2] No such file or directory: 'C:\\Python25\\Test scripts\newfile
My code:
filename = open('C:\Python25\Test scripts\newfile','r')
print filename.read()
Also I tried
filename = open('C:\\Python25\\Test scripts\\newfile','r')
print filename.read()
But same errors I am getting.
Try:
fpath = r'C:\Python25\Test scripts\newfile'
if not os.path.exists(fpath):
print 'File does not exist'
return
with open(fpath, 'r') as src:
src.read()
First you validate that file, that it exists.
Then you open it. With wrapper is more usefull, it closes your file, after you finish reading. So you will not stuck with many open descriptors.
I think you're probably having this issue because you didn't include the full filename.
You should try:
filename = open('C:\Python25\Test scripts\newfile.txt','r')
print filename.read()
*Also if you're running this python file in the same location as the target file your are opening, you don't need to give the full directory, you can just call:
filename = open(newfile.txt
I had the same problem. Here's how I got it right.
your code:
filename = open('C:\\Python25\\Test scripts\\newfile','r')
print filename.read()
Try this:
with open('C:\\Python25\\Test scripts\\newfile') as myfile:
print(myfile.read())
Hope it helps.
I am using VS code. If I am not using dent it would not work for the print line. So try to have the format right then you will see the magic.
with open("mytest.txt") as myfile:
print(myfile.read())
or without format like this:
hellofile=open('mytest.txt', 'r')
print(hellofile.read())

Error when trying to read and write multiple files

I modified the code based on the comments from experts in this thread. Now the script reads and writes all the individual files. The script reiterates, highlight and write the output. The current issue is, after highlighting the last instance of the search item, the script removes all the remaining contents after the last search instance in the output of each file.
Here is the modified code:
import os
import sys
import re
source = raw_input("Enter the source files path:")
listfiles = os.listdir(source)
for f in listfiles:
filepath = source+'\\'+f
infile = open(filepath, 'r+')
source_content = infile.read()
color = ('red')
regex = re.compile(r"(\b be \b)|(\b by \b)|(\b user \b)|(\bmay\b)|(\bmight\b)|(\bwill\b)|(\b's\b)|(\bdon't\b)|(\bdoesn't\b)|(\bwon't\b)|(\bsupport\b)|(\bcan't\b)|(\bkill\b)|(\betc\b)|(\b NA \b)|(\bfollow\b)|(\bhang\b)|(\bbelow\b)", re.I)
i = 0; output = ""
for m in regex.finditer(source_content):
output += "".join([source_content[i:m.start()],
"<strong><span style='color:%s'>" % color[0:],
source_content[m.start():m.end()],
"</span></strong>"])
i = m.end()
outfile = open(filepath, 'w+')
outfile.seek(0)
outfile.write(output)
print "\nProcess Completed!\n"
infile.close()
outfile.close()
raw_input()
The error message tells you what the error is:
No such file or directory: 'sample1.html'
Make sure the file exists. Or do a try statement to give it a default behavior.
The reason why you get that error is because the python script doesn't have any knowledge about where the files are located that you want to open.
You have to provide the file path to open it as I have done below. I have simply concatenated the source file path+'\\'+filename and saved the result in a variable named as filepath. Now simply use this variable to open a file in open().
import os
import sys
source = raw_input("Enter the source files path:")
listfiles = os.listdir(source)
for f in listfiles:
filepath = source+'\\'+f # This is the file path
infile = open(filepath, 'r')
Also there are couple of other problems with your code, if you want to open the file for both reading and writing then you have to use r+ mode. More over in case of Windows if you open a file using r+ mode then you may have to use file.seek() before file.write() to avoid an other issue. You can read the reason for using the file.seek() here.

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