I am trying to create a to-do list application, and to store the users tasks, I'm writing them by line to a plaintext file. At multiple points I "sync" it by calling foo.readlines(), but even if I hand write test data into the file, the list returns empty and the contents of the plain text file are erased.
I tried opening the file manually and writing to it and saving it, but after running the script, it is again empty and the list returns empty.
import numpy as np
file = open('./data.txt', 'w+')
tasks = file.readlines()
print(tasks)
#writes list to a file
def writeFile(tasks):
with open('data.txt', 'w') as filehandle:
for listitem in tasks:
filehandle.write('%s\n' % listitem)
You're opening the file in write mode with "w+" at line 3. That deletes the file's contents.
You probably meant to use "r" instead of "w" in
with open('data.txt', 'w') as filehandle:
'Read' mode is the default mode if none specified
file = open('data.txt')
Opening a file in 'Read' mode
file = open('data.txt', 'r')
Opening a file in 'Write' mode (will overwrite the contents of the file if exists)
file = open('data.txt', 'w')
Opening a file in 'Append' mode (will append to the existing file without overwriting)
file = open('data.txt', 'a')
Related
I have the following code:
import re
#open the xml file for reading:
file = open('path/test.xml','r+')
#convert to string:
data = file.read()
file.write(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>",r"<xyz>ABC</xyz>\1<xyz>\2</xyz>",data))
file.close()
where I'd like to replace the old content that's in the file with the new content. However, when I execute my code, the file "test.xml" is appended, i.e. I have the old content follwed by the new "replaced" content. What can I do in order to delete the old stuff and only keep the new?
You need seek to the beginning of the file before writing and then use file.truncate() if you want to do inplace replace:
import re
myfile = "path/test.xml"
with open(myfile, "r+") as f:
data = f.read()
f.seek(0)
f.write(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>", r"<xyz>ABC</xyz>\1<xyz>\2</xyz>", data))
f.truncate()
The other way is to read the file then open it again with open(myfile, 'w'):
with open(myfile, "r") as f:
data = f.read()
with open(myfile, "w") as f:
f.write(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>", r"<xyz>ABC</xyz>\1<xyz>\2</xyz>", data))
Neither truncate nor open(..., 'w') will change the inode number of the file (I tested twice, once with Ubuntu 12.04 NFS and once with ext4).
By the way, this is not really related to Python. The interpreter calls the corresponding low level API. The method truncate() works the same in the C programming language: See http://man7.org/linux/man-pages/man2/truncate.2.html
file='path/test.xml'
with open(file, 'w') as filetowrite:
filetowrite.write('new content')
Open the file in 'w' mode, you will be able to replace its current text save the file with new contents.
Using truncate(), the solution could be
import re
#open the xml file for reading:
with open('path/test.xml','r+') as f:
#convert to string:
data = f.read()
f.seek(0)
f.write(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>",r"<xyz>ABC</xyz>\1<xyz>\2</xyz>",data))
f.truncate()
import os#must import this library
if os.path.exists('TwitterDB.csv'):
os.remove('TwitterDB.csv') #this deletes the file
else:
print("The file does not exist")#add this to prevent errors
I had a similar problem, and instead of overwriting my existing file using the different 'modes', I just deleted the file before using it again, so that it would be as if I was appending to a new file on each run of my code.
See from How to Replace String in File works in a simple way and is an answer that works with replace
fin = open("data.txt", "rt")
fout = open("out.txt", "wt")
for line in fin:
fout.write(line.replace('pyton', 'python'))
fin.close()
fout.close()
in my case the following code did the trick
with open("output.json", "w+") as outfile: #using w+ mode to create file if it not exists. and overwrite the existing content
json.dump(result_plot, outfile)
Using python3 pathlib library:
import re
from pathlib import Path
import shutil
shutil.copy2("/tmp/test.xml", "/tmp/test.xml.bak") # create backup
filepath = Path("/tmp/test.xml")
content = filepath.read_text()
filepath.write_text(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>",r"<xyz>ABC</xyz>\1<xyz>\2</xyz>", content))
Similar method using different approach to backups:
from pathlib import Path
filepath = Path("/tmp/test.xml")
filepath.rename(filepath.with_suffix('.bak')) # different approach to backups
content = filepath.read_text()
filepath.write_text(re.sub(r"<string>ABC</string>(\s+)<string>(.*)</string>",r"<xyz>ABC</xyz>\1<xyz>\2</xyz>", content))
Let's say that I have a txt file that I have to get all in lowercase. I tried this
def lowercase_txt(file):
file = file.casefold()
with open(file, encoding = "utf8") as f:
f.read()
Here I get "'str' object has no attribute 'read'"
then I tried
def lowercase_txt(file):
with open(poem_filename, encoding="utf8") as f:
f = f.casefold()
f.read()
and here '_io.TextIOWrapper' object has no attribute 'casefold'
What can I do?
EDIT: I re-runned this exact code and now there are no errors (dunno why) but the file doesn't change at all, all the letters stay the way they are.
This will rewrite the file. Warning: if there is some type of error in the middle of processing (power failure, you spill coffee on your computer, etc.) you could lose your file. So, you might want to first make a backup of your file:
def lowercase_txt(file_name):
"""
file_name is the full path to the file to be opened
"""
with open(file_name, 'r', encoding = "utf8") as f:
contents = f.read() # read contents of file
contents = contents.lower() # convert to lower case
with open(file_name, 'w', encoding = "utf8") as f: # open for output
f.write(contents)
For example:
lowercase_txt('/mydirectory/test_file.txt')
Update
The following version opens the file for reading and writing. After the file is read, the file position is reset to the start of the file before the contents is rewritten. This might be a safer option.
def lowercase_txt(file_name):
"""
file_name is the full path to the file to be opened
"""
with open(file_name, 'r+', encoding = "utf8") as f:
contents = f.read() # read contents of file
contents = contents.lower() # convert to lower case
f.seek(0, 0) # position back to start of file
f.write(contents)
f.truncate() # in case new encoded content is shorter than older
This question already has answers here:
Why doesn't calling a string method (such as .replace or .strip) modify (mutate) the string?
(3 answers)
Closed 3 years ago.
I am trying to display my python file in html and therefore I would like to replace every time the file jumps to a newline with < br> but the program I've written is not working.
I've looked on here and tried changing the code around a bit I have gotten different results but not the ones I need.
with open(path, "r+") as file:
contents = file.read()
contents.replace("\n", "<br>")
print(contents)
file.close()
I want to have the file display < br> every time I have a new line but instead the code dosen't change anything to the file.
Here is an example program that works:
path = "example"
contents = ""
with open(path, "r") as file:
contents = file.read()
new_contents = contents.replace("\n", "<br>")
with open(path, "w") as file:
file.write(new_contents)
Your program doesn't work because the replace method does not modify the original string; it returns a new string.
Also, you need to write the new string to the file; python won't do it automatically.
Hope this helps :)
P.S. a with statement automatically closes the file stream.
Your code reads from the file, saves the contents to a variable and replaces the newlines. But the result is not saved anywhere. And to write the result into a file you must open the file for writing.
with open(path, "r+") as file:
contents = file.read()
contents = contents.replace("\n", "<br>")
with open(path, "w+") as file:
contents = file.write(contents)
there are some issues in this code snippet.
contents.replace("\n", "<br>") will return a new object which replaced \n with <br>, so you can use html_contents = contents.replace("\n", "<br>") and print(html_contents)
when you use with the file descriptor will close after leave the indented block.
Try this:
import re
with open(path, "r") as f:
contents = f.read()
contents = re.sub("\n", "<br>", contents)
print(contents)
Borrowed from this post:
import tempfile
def modify_file(filename):
#Create temporary file read/write
t = tempfile.NamedTemporaryFile(mode="r+")
#Open input file read-only
i = open(filename, 'r')
#Copy input file to temporary file, modifying as we go
for line in i:
t.write(line.rstrip()+"\n")
i.close() #Close input file
t.seek(0) #Rewind temporary file to beginning
o = open(filename, "w") #Reopen input file writable
#Overwriting original file with temporary file contents
for line in t:
o.write(line)
t.close() #Close temporary file, will cause it to be deleted
I'm trying to print a list or dict of file names into a text file. it's currently only returning the first item on the list. the items are fetched from s3 Aws.I'm using Python 2.6
for obj in bucket.objects.filter(Prefix=prefix):
s = obj.key
with open('test.txt', 'w') as f:
f.write(s)
The problem here is that for every item, you create a new file (in case the file already exists, you remove the content so to speak), and then write s to it.
So you should swap the order of things here:
with open('test.txt', 'w') as f: # first open the file
for obj in bucket.objects.filter(Prefix=prefix): # then iterate
f.write(obj.key)
So we keep the file handle open, and each item will be written. A potential problem is that you will not write a new line after you written the key of an object. We can do this by writing a new line as well:
with open('test.txt', 'w') as f:
for obj in bucket.objects.filter(Prefix=prefix):
f.write(obj.key)
f.write('\n')
whenever you open a file for writing, the previous content is erased and new text is written. So in this case you are erasing whatever you wrote to the file in the next iteration. you can do it in this way or open the file in "append" mode and continue with what you have written.
f= open("test.txt", "w")
for obj in bucket.objects.filter(Prefix=prefix):
s = obj.key
f.write(s)
f.write('\n)
f.close()
def readfile(file):
edges = [] # to contain tuples of all edges
with open(file) as f:
for line in f:
I'm trying to pass in a text file called file, then read it, but it doesn't work. Even if I cast file to a string it doesn't work.
Python says
with open(file) as f: IOError: [Errno 22] invalid mode ('r') or
filename: "<type 'file'>"
How do I open a file, passed to open as a variable?
Rename file by filename as file is a built-in name for python :
def readfile(filename):
edges = [] # to contain tuples of all edges
with open(filename) as f:
for line in f:
open() use 'r' as default mode.
First of all , as file is a type in python you shouldn't use it as a variable name or file name , second you need to put the file name in quote inside the open function and note that open function use read mod as default ! :
with open('file_name.txt','r') as f:
for line in f: