def Delete_con():
contact_to_delete= input("choose name to delete from contact")
to_Delete=list(contact_to_delete)
with open("phonebook1.txt", "r+") as file:
content = file.read()
for line in content:
if not any(line in line for line in to_Delete):
content.write(line)
I get zero error. but the line is not deleted. This function ask the user what name he or she wants to delete from the text file.
This should help.
def Delete_con():
contact_to_delete= input("choose name to delete from contact")
contact_to_delete = contact_to_delete.lower() #Convert input to lower case
with open("phonebook1.txt", "r") as file:
content = file.readlines() #Read lines from text
content = [line for line in content if contact_to_delete not in line.lower()] #Check if user input is in line
with open("phonebook1.txt", "w") as file: #Write back content to text
file.writelines(content)
Assuming that:
you want the user to supply just the name, and not the full 'name:number' pair
your phonebook stores one name:number pair per line
I'd do something like this:
import os
from tempfile import NamedTemporaryFile
def delete_contact():
contact_name = input('Choose name to delete: ')
# You probably want to pass path in as an argument
path = 'phonebook1.txt'
base_dir = os.path.dirname(path)
with open(path) as phonebook, \
NamedTemporaryFile(mode='w+', dir=base_dir, delete=False) as tmp:
for line in phonebook:
# rsplit instead of split supports names containing ':'
# if numbers can also contain ':' you need something smarter
name, number = line.rsplit(':', 1)
if name != contact_name:
tmp.write(line)
os.replace(tmp.name, path)
Using a tempfile like this means that if something goes wrong while processing the file you aren't left with a half-written phonebook, you'll still have the original file unchanged. You're also not reading the entire file into memory with this approach.
os.replace() is Python 3.3+ only, if you're using something older you can use os.rename() as long as you're not using Windows.
Here's the tempfile documentation. In this case, you can think of NamedTemporaryFile(mode='w+', dir=base_dir, delete=False) as something like open('tmpfile.txt', mode='w+'). NamedTemporaryFile saves you from having to find a unique name for your tempfile (so that you don't overwrite an existing file). The dir argument creates the tempfile in the same directory as phonebook1.txt which is a good idea because os.replace() can fail when operating across two different filesystems.
Related
I am writing a program where it asks you what text file the user wants to read then it begins to read whatever file name the user inputs. Here is what I have so far:
import sys
import os
import re
#CHOOSE FILE
print "Welcome to the Parsing Database"
raw_input=raw_input("enter file name to parse: ")
#ASSIGN HEADERS AND SEQUENCES
f=open("raw_input", "r")
header=[]
sequence=[]
string=""
for line in f:
if ">" in line and string=="":
header.append(line[:-2])
elif ">" in line and string!="":
sequence.append(string)
header.append(line[:-2])
string=""
else:
string=string+line[:-2]
sequence.append(string)
The first two lines work but then it says it cannot find the file that I inputted to read. Please help! Thanks.
Off the top of my head, I think that f = open("raw_input", "r") needs to be f=open(raw_input, "r"), because you are trying to reference the string contained in the variable raw_input, as opposed to trying to open a file named raw_input. Also you should probably change the name of the variable to something more readable, because raw_input() is a function used in your code as well as a variable, which makes it hard to read. Are there any other specific problems you are having with your code?
f=open("raw_input", "r")
"raw_input" is a plain string. You have to referente to it as raw_input.
Also, there's no lines if you don't use .read() with open() method so you can't parse them. Read lines from a file given from raw_input can be done doing that:
import sys
import os
import re
#CHOOSE FILE
print "Welcome to the Parsing Database"
raw_input_file=raw_input("enter file name to parse: ")
#ASSIGN HEADERS AND SEQUENCES
testfile = open(raw_input_file, "r")
secuence = []
for line in testfile.read().splitlines():
secuence.append(line)
for i in secuence:
print i
testfile.close()
I have a file called usernames.py that may contain a list or does exist at all:
usernames.py
['user1', 'user2', 'user3']
In Python I now want to read this file if it exists and append to the list a new user or create a list with that user i.e. ['user3']
This is what I have tried:
with open(path + 'usernames.py', 'w+') as file:
file_string = host_file.read()
file_string.append(instance)
file.write(file_string)
This gives me an error unresolved 'append'. How can I achieve this? Python does not know it is a list and if the file does not exist even worst as I have nothing to convert to a list.
Try this:
import os
filename = 'data'
if os.path.isfile(filename):
with open(filename, 'r') as f:
l = eval(f.readline())
else:
l = []
l.append(instance)
with open(filename, 'w') as f:
f.write(str(l))
BUT this is quite unsafe if you don't know where the file is from as it could include any code to do anything!
It would be better not to use a python file for persistence -- what happens if someone slips you a usernames.py that has exploit code in it? Consider a csv file or a pickle, or just a text file with one user per line.
That said, if you don't open it as a python file, something like this should work:
from os.path import join
with open( join(path, 'usernames.py'), 'r+') as file:
file_string = file.read()
file_string = file_string.strip().strip('[').strip(']')
file_data = [ name.strip().strip('"').strip("'") for name in file_string.split(',' )]
file_data.append( instance )
file.fseek(0)
file.write(str(file_data))
If usernames contain commas or end in quotes, you have to be more careful.
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.
I want to read the contents of a zip file into memory rather than extracting them to disc, find a particular file in the archive, open the file and extract a line from it.
Can a StringIO instance be opened and parsed? Suggestions? Thanks in advance.
zfile = ZipFile('name.zip', 'r')
for name in zfile.namelist():
if fnmatch.fnmatch(name, '*_readme.xml'):
name = StringIO.StringIO()
print name # prints StringIO instances
open(name, 'r') # IO Error: No such file or directory...
I found a few similar posts, but none that seem to address this issue: Extracting a zipfile to memory?
IMO just using read is enough:
zfile = ZipFile('name.zip', 'r')
files = []
for name in zfile.namelist():
if fnmatch.fnmatch(name, '*_readme.xml'):
files.append(zfile.read(name))
This will make a list with contents of files that match the pattern.
Test:
You can then parse contents afterwards by iterating through the list:
for file in files:
print(file[0:min(35,len(file))].decode()) # "parsing"
Or better use a functor:
import zipfile as zip
import os
import fnmatch
zip_name = os.sys.argv[1]
zfile = zip.ZipFile(zip_name, 'r')
def parse(contents, member_name = ""):
if len(member_name) > 0:
print( "Parsed `{}`:".format(member_name) )
print(contents[0:min(35, len(contents))].decode()) # "parsing"
for name in zfile.namelist():
if fnmatch.fnmatch(name, '*.cpp'):
parse(zfile.read(name), name)
This way there is no data kept in memory for no reason and memory foot print is smaller. It might be important if the files are big.
Don't overthink it. It Just Works:
import zipfile
# 1) I want to read the contents of a zip file ...
with zipfile.ZipFile('A-Zip-File.zip') as zipper:
# 2) ... find a particular file in the archive, open the file ...
with zipper.open('A-Particular-File.txt') as fp:
# 3) ... and extract a line from it.
first_line = fp.readline()
print first_line
The question you link shows you that you need to read the file. Depending on your use case that may already be enough. In your code you replace the loop variable holding a filename with an empty string buffer. Try something like this:
zfile = ZipFile('name.zip', 'r')
for name in zfile.namelist():
if fnmatch.fnmatch(name, '*_readme.xml'):
ex_file = zfile.open(name) # this is a file like object
content = ex_file.read() # now file-contents are a single string
If you really want a buffer that you can manipulate, then simply instantiate it with the contents:
buf = StringIO(zfile.open(name).read())
You may also want to look at BytesIO and note that there are differences between Python 2 and 3.
Thank you to everyone that contributed solutions. This is what ended up working for me:
zfile = ZipFile('name.zip', 'r')
for name in zfile.namelist():
if fnmatch.fnmatch(name, '*_readme.xml'):
zopen = zfile.open(name)
for line in zopen:
if re.match('(.*)<foo>(.*)</foo>(.*)', line):
print line
I have a program which writes a user's highscore to a text file. The file is named by the user when they choose a playername.
If the file with that specific username already exists, then the program should append to the file (so that you can see more than one highscore). And if a file with that username doesn't exist (for example, if the user is new), it should create a new file and write to it.
Here's the relevant, so far not working, code:
try:
with open(player): #player is the varible storing the username input
with open(player, 'a') as highscore:
highscore.write("Username:", player)
except IOError:
with open(player + ".txt", 'w') as highscore:
highscore.write("Username:", player)
The above code creates a new file if it doesn't exist, and writes to it. If it exists, nothing has been appended when I check the file, and I get no errors.
Have you tried mode 'a+'?
with open(filename, 'a+') as f:
f.write(...)
Note however that f.tell() will return 0 in Python 2.x. See https://bugs.python.org/issue22651 for details.
It's not clear to me exactly where the high-score that you're interested in is stored, but the code below should be what you need to check if the file exists and append to it if desired. I prefer this method to the "try/except".
import os
player = 'bob'
filename = player+'.txt'
if os.path.exists(filename):
append_write = 'a' # append if already exists
else:
append_write = 'w' # make a new file if not
highscore = open(filename,append_write)
highscore.write("Username: " + player + '\n')
highscore.close()
Just open it in 'a' mode:
a Open for writing. The file is created if it does not exist. The stream is positioned at the end of the file.
with open(filename, 'a') as f:
f.write(...)
To see whether you're writing to a new file, check the stream position. If it's zero, either the file was empty or it is a new file.
with open('somefile.txt', 'a') as f:
if f.tell() == 0:
print('a new file or the file was empty')
f.write('The header\n')
else:
print('file existed, appending')
f.write('Some data\n')
If you're still using Python 2, to work around the bug, either add f.seek(0, os.SEEK_END) right after open or use io.open instead.
Notice that if the file's parent folder doesn't exist you'll get the same error:
IOError: [Errno 2] No such file or directory:
Below is another solution which handles this case:
(*) I used sys.stdout and print instead of f.write just to show another use case
# Make sure the file's folder exist - Create folder if doesn't exist
folder_path = 'path/to/'+folder_name+'/'
if not os.path.exists(folder_path):
os.makedirs(folder_path)
print_to_log_file(folder_path, "Some File" ,"Some Content")
Where the internal print_to_log_file just take care of the file level:
# If you're not familiar with sys.stdout - just ignore it below (just a use case example)
def print_to_log_file(folder_path ,file_name ,content_to_write):
#1) Save a reference to the original standard output
original_stdout = sys.stdout
#2) Choose the mode
write_append_mode = 'a' #Append mode
file_path = folder_path + file_name
if (if not os.path.exists(file_path) ):
write_append_mode = 'w' # Write mode
#3) Perform action on file
with open(file_path, write_append_mode) as f:
sys.stdout = f # Change the standard output to the file we created.
print(file_path, content_to_write)
sys.stdout = original_stdout # Reset the standard output to its original value
Consider the following states:
'w' --> Write to existing file
'w+' --> Write to file, Create it if doesn't exist
'a' --> Append to file
'a+' --> Append to file, Create it if doesn't exist
In your case I would use a different approach and just use 'a' and 'a+'.
Using the pathlib module (python's object-oriented filesystem paths)
Just for kicks, this is perhaps the latest pythonic version of the solution.
from pathlib import Path
path = Path(f'{player}.txt')
path.touch() # default exists_ok=True
with path.open('a') as highscore:
highscore.write(f'Username:{player}')