I have a python script that will create a text file and then will run a command on this newly created file.
The problem is that the command line is not recognizing the newly created file and I'm getting an error that the file is empty.
my code is something like this:
randomval is a function that will create random characters and return them as a string.
text_file = open("test.txt", "w")
text_file.write(randomval(20,10))
# do something with the `test.txt` file
but I'm getting an error that the file test.txt is empty.
Is there anyway to solve this?
This happens because unless you flush or close the file, the OS will not write any data to the disk. To make sure that the file is closed, use the with statement:
with open("test.txt", "w") as f:
f.write(randomval(20,10))
print('Whoa, at this point the file descriptor is automatically closed!')
Don't forget to close the file:
br#ymir:~$ python
Python 2.6.5 (r265:79063, Oct 1 2012, 22:04:36)
[GCC 4.4.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> file=open('foo.bar','w')
>>> file.write('42\n')
>>>
[2]+ Stopped python
br#ymir:~$ cat foo.bar
br#ymir:~$ fg
python
>>> file.close()
>>>
[2]+ Stopped python
br#ymir:~$ cat foo.bar
42
br#ymir:~$
You should, at least, flush the buffer before trying to do something with your new file (text_file.flush()). The best would be to close the file and reopen it when needed.
Do a f.close() and then open it again as text_file = open("test.txt", "r")
The stream to the file is still open!!!! Just try: text_file.close()
Related
I've just started using python and I'm creating a simple program that will ask whoever the user is a question and then from the text file I will extract a specific line and print that line and along with the line - at the end I will add their answer. here's the code.
question = input("do you want to print the line")
if "yes" in question:
print(open("tp.txt").readlines()[:10][-1],end=question)
The issue is that ,end=question) puts the users answer on a new line. I know that end= is the same as \n. So I'm just wondering is there a way or an alternative to stop 'end=' from automatically creating a new line?
print(open("tp.txt").readlines()[:10][-1],
is the way I open and read a specific line from the file
since its a 'nice' shortcut to do than rather dowith open (filename.txt,'r') as f:
The problem is that the lines returned by readlines() contain the ending newline:
$ echo 'a
> b
> c
> ' > test_file.txt
$ python3
Python 3.5.2 (default, Jul 5 2016, 12:43:10)
[GCC 5.4.0 20160609] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> with open('test_file.txt') as f:
... print(f.readlines())
...
['a\n', 'b\n', 'c\n', '\n']
See the \n? Note the difference bewteen:
>>> print('a\n')
a
>>>
And:
>>> print('a')
a
>>>
So you want to remove that:
>>> with open('test_file.txt') as f:
... for line in f:
... print(line.rstrip('\n'), end='<something>')
...
a<something>b<something>c<something><something>>>>
When I am trying to load something I dumped using cPickle, I get the error message:
ValueError: insecure string pickle
Both the dumping and loading work are done on the same computer, thus same OS: Ubuntu 8.04.
How could I solve this problem?
"are much more likely than a never-observed bug in Python itself in a functionality that's used billions of times a day all over the world": it always amazes me how cross people get in these forums.
One easy way to get this problem is by forgetting to close the stream that you're using for dumping the data structure. I just did
>>> out = open('xxx.dmp', 'w')
>>> cPickle.dump(d, out)
>>> k = cPickle.load(open('xxx.dmp', 'r'))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: insecure string pickle
Which is why I came here in the first place, because I couldn't see what I'd done wrong.
And then I actually thought about it, rather than just coming here, and realized that I should have done:
>>> out = open('xxx.dmp', 'w')
>>> cPickle.dump(d, out)
>>> out.close() # close it to make sure it's all been written
>>> k = cPickle.load(open('xxx.dmp', 'r'))
Easy to forget. Didn't need people being told that they are idiots.
I've get this error in Python 2.7 because of open mode 'rb':
with open(path_to_file, 'rb') as pickle_file:
obj = pickle.load(pickle_file)
So, for Python 2 'mode' should be 'r'
Also, I've wondered that Python 3 doesn't support pickle format of Python 2, and in case when you'll try to load pickle file created in Python 2 you'll get:
pickle.unpicklingerror: the string opcode argument must be quoted
Check this thread. Peter Otten says:
A corrupted pickle. The error is
raised if a string in the dump does
not both start and end with " or '.
and shows a simple way to reproduce such "corruption". Steve Holden, in the follow-up post, suggests another way to cause the problem would be to mismatch 'rb' and 'wb' (but in Python 2 and on Linux that particular mistake should pass unnoticed).
What are you doing with data between dump() and load()? It's quite common error to store pickled data in file opened in text mode (on Windows) or in database storage in the way that doesn't work properly for binary data (VARCHAR, TEXT columns in some databases, some key-value storages). Try to compare pickled data that you pass to storage and immediately retrieved from it.
If anyone has this error using youtube-dl, this issue has the fix: https://github.com/rg3/youtube-dl/issues/7172#issuecomment-242961695
richiecannizzo commented on Aug 28
brew install libav
Should fix it instantly on mac or
sudo apt-get install libav
#on linux
This error may also occur with python 2 (and early versions of python 3) if your pickle is large (Python Issue #11564):
Python 2.7.11 |Anaconda custom (64-bit)| (default, Dec 6 2015, 18:08:32)
[GCC 4.4.7 20120313 (Red Hat 4.4.7-1)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
Anaconda is brought to you by Continuum Analytics.
Please check out: http://continuum.io/thanks and https://anaconda.org
>>> import cPickle as pickle
>>> string = "X"*(2**31)
>>> pp = pickle.dumps(string)
>>> len(pp)
2147483656
>>> ss = pickle.loads(pp)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: insecure string pickle
This limitation was addressed with the introduction of pickle protocol 4 in python 3.4 (PEP 3154). Unfortunately, this feature has not been back-ported to python 2, and probably won't ever be. If this is your problem and you need to use python 2 pickle, the best you can do is reduce the size of your pickle, e.g., instead of pickling a list, pickle the elements individually into a list of pickles.
Same problem with a file that was made with python on windows, and reloaded with python on linux.
Solution : dos2unix on the file before reading in linux : works as a charm !
I got the Python ValueError: insecure string pickle message in a different way.
For me it happened after a base64 encoding a binary file and passing through urllib2 sockets.
Initially I was wrapping up a file like this
with open(path_to_binary_file) as data_file:
contents = data_file.read()
filename = os.path.split(path)[1]
url = 'http://0.0.0.0:8080/upload'
message = {"filename" : filename, "contents": contents}
pickled_message = cPickle.dumps(message)
base64_message = base64.b64encode(pickled_message)
the_hash = hashlib.md5(base64_message).hexdigest()
server_response = urllib2.urlopen(url, base64_message)
But on the server the hash kept coming out differently for some binary files
decoded_message = base64.b64decode(incoming_base64_message)
the_hash = hashlib.md5(decoded_message).hexdigest()
And unpickling gave insecure string pickle message
cPickle.loads(decoded_message)
BUT SUCCESS
What worked for me was to use urlsafe_b64encode()
base64_message = base64.urlsafe_b64encode(cPickle.dumps(message))
And decode with
base64_decoded_message = base64.urlsafe_b64decode(base64_message)
References
http://docs.python.org/2/library/base64.html
https://www.rfc-editor.org/rfc/rfc3548.html#section-3
This is what happened to me, might be a small section of population, but I want to put this out here nevertheless, for them:
Interpreter (Python3) would have given you an error saying it required the input file stream to be in bytes, and not as a string, and you may have changed the open mode argument from 'r' to 'rb', and now it is telling you the string is corrupt, and thats why you have come here.
The simplest option for such cases is to install Python2 (You can install 2.7) and then run your program with Python 2.7 environment, so it unpickles your file without issue. Basically I wasted a lot of time scanning my string seeing if it was indeed corrupt when all I had to do was change the mode of opening the file from rb to r, and then use Python2 to unpickle the file. So I'm just putting this information out there.
I ran into this earlier, found this thread, and assumed that I was immune to the file closing issue mentioned in a couple of these answers since I was using a with statement:
with tempfile.NamedTemporaryFile(mode='wb') as temp_file:
pickle.dump(foo, temp_file)
# Push file to another machine
_send_file(temp_file.name)
However, since I was pushing the temp file from inside the with, the file still wasn't closed, so the file I was pushing was truncated. This resulted in the same insecure string pickle error in the script that read the file on the remote machine.
Two potential fixes to this: Keep the file open and force a flush:
with tempfile.NamedTemporaryFile(mode='wb') as temp_file:
pickle.dump(foo, temp_file)
temp_file.flush()
# Push file to another machine
_send_file(temp_file.name)
Or make sure the file is closed before doing anything with it:
file_name = ''
with tempfile.NamedTemporaryFile(mode='wb', delete=False) as temp_file:
file_name = temp_file.name
pickle.dump(foo, temp_file)
# Push file to another machine
_send_file(file_name)
I'm trying to solve exercise 15's extra credit questions of Zed Shaw's Learn Python the Hard Way but I've ran into a problem. The code is as follows:
from sys import argv
script, filename = argv
txt = open(filename)
print "Here's your file %r:" % filename
print txt.read()
print "I'll also ask you to type it again:"
file_again = raw_input("> ")
txt_again = open(file_again)
print txt_again.read()
print txt_again.read()
I understand all the code that has been used, but extra credit question 7 asks:
Startup python again and use open from the prompt. Notice how you can open files and run read on them right there?
I've tried inputting everything I could think of in terminal (on a mac) after first starting up python with the 'python' command, but I can't get the code to run. What should I be doing to get this piece of code to run from the prompt?
Zed doesn't say to run this particular piece of code from within Python. Obviously, that code is getting the filename value from the parameters you used to invoke the script, and if you're just starting up the Python shell, you haven't used any parameters.
If you did:
filename = 'myfilename.txt'
txt = open(filename)
then it would work.
I just started with open(xyz.txt)
Well, yes, of course that isn't going to work, because you don't have a variable xyz, and even if you did, it wouldn't have an attribute txt. Since it's a file name, you want a string "xyz.txt", which you create by putting it in quotes: 'xyz.txt'. Notice that Python treats single and double quotes more or less the same; unlike in languages like C++ and Java, there is not a separate data type for individual characters - they're just length-1 strings.
Basically, just like in this transcript (I've added blank lines to aid readability):
pax:~$ python
Python 2.7.1+ (r271:86832, Apr 11 2011, 18:05:24)
[GCC 4.5.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> xyz = open ("minimal_main.c")
>>> print xyz.read()
int main (void) {
return 0;
}
>>> xyz.close()
>>> <CTRL-D>
pax:~$ _
All it's showing you is that you don't need a script in order to run Python commands, the command line interface can be used in much the same way.
print open('ex15_sample.txt').read()
After running python in terminal, we'll use open('filename.txt') to open the file and using the dot operator we can apply the read() function directly on it.
After running Python in terminal,
abc = open ("ex15_sample.txt")
print abc.read()
That should do.
I'm using python 2.6.4 and discovered that I can't use gzip with subprocess the way I might hope. This illustrates the problem:
May 17 18:05:36> python
Python 2.6.4 (r264:75706, Mar 10 2010, 14:41:19)
[GCC 4.1.2 20071124 (Red Hat 4.1.2-42)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import gzip
>>> import subprocess
>>> fh = gzip.open("tmp","wb")
>>> subprocess.Popen("echo HI", shell=True, stdout=fh).wait()
0
>>> fh.close()
>>>
[2]+ Stopped python
May 17 18:17:49> file tmp
tmp: data
May 17 18:17:53> less tmp
"tmp" may be a binary file. See it anyway?
May 17 18:17:58> zcat tmp
zcat: tmp: not in gzip format
Here's what it looks like inside less
HI
^_<8B>^H^Hh<C0><F1>K^B<FF>tmp^#^C^#^#^#^#^#^#^#^#^#
which looks like it put in the stdout as text and then put in an empty gzip file. Indeed, if I remove the "Hi\n", then I get this:
May 17 18:22:34> file tmp
tmp: gzip compressed data, was "tmp", last modified: Mon May 17 18:17:12 2010, max compression
What is going on here?
UPDATE:
This earlier question is asking the same thing: Can I use an opened gzip file with Popen in Python?
You can't use file-likes with subprocess, only real files. The fileno() method of GzipFile returns the FD of the underlying file, so that's what the echo redirects to. The GzipFile then closes, writing an empty gzip file.
just pipe that sucker
from subprocess import Popen,PIPE
GZ = Popen("gzip > outfile.gz",stdin=PIPE,shell=True)
P = Popen("echo HI",stdout=GZ.stdin,shell=True)
# these next three must be in order
P.wait()
GZ.stdin.close()
GZ.wait()
I'm not totally sure why this isn't working (perhaps the output redirection is not calling python's write, which is what gzip works with?) but this works:
>>> fh.write(subprocess.Popen("echo Hi", shell=True, stdout=subprocess.PIPE).stdout.read())
You don't need to use subprocess to write to the gzip.GzipFile. Instead, write to it like any other file-like object. The result is automagically gzipped!
import gzip
with gzip.open("tmp.gz", "wb") as fh:
fh.write('echo HI')
When I am trying to load something I dumped using cPickle, I get the error message:
ValueError: insecure string pickle
Both the dumping and loading work are done on the same computer, thus same OS: Ubuntu 8.04.
How could I solve this problem?
"are much more likely than a never-observed bug in Python itself in a functionality that's used billions of times a day all over the world": it always amazes me how cross people get in these forums.
One easy way to get this problem is by forgetting to close the stream that you're using for dumping the data structure. I just did
>>> out = open('xxx.dmp', 'w')
>>> cPickle.dump(d, out)
>>> k = cPickle.load(open('xxx.dmp', 'r'))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: insecure string pickle
Which is why I came here in the first place, because I couldn't see what I'd done wrong.
And then I actually thought about it, rather than just coming here, and realized that I should have done:
>>> out = open('xxx.dmp', 'w')
>>> cPickle.dump(d, out)
>>> out.close() # close it to make sure it's all been written
>>> k = cPickle.load(open('xxx.dmp', 'r'))
Easy to forget. Didn't need people being told that they are idiots.
I've get this error in Python 2.7 because of open mode 'rb':
with open(path_to_file, 'rb') as pickle_file:
obj = pickle.load(pickle_file)
So, for Python 2 'mode' should be 'r'
Also, I've wondered that Python 3 doesn't support pickle format of Python 2, and in case when you'll try to load pickle file created in Python 2 you'll get:
pickle.unpicklingerror: the string opcode argument must be quoted
Check this thread. Peter Otten says:
A corrupted pickle. The error is
raised if a string in the dump does
not both start and end with " or '.
and shows a simple way to reproduce such "corruption". Steve Holden, in the follow-up post, suggests another way to cause the problem would be to mismatch 'rb' and 'wb' (but in Python 2 and on Linux that particular mistake should pass unnoticed).
What are you doing with data between dump() and load()? It's quite common error to store pickled data in file opened in text mode (on Windows) or in database storage in the way that doesn't work properly for binary data (VARCHAR, TEXT columns in some databases, some key-value storages). Try to compare pickled data that you pass to storage and immediately retrieved from it.
If anyone has this error using youtube-dl, this issue has the fix: https://github.com/rg3/youtube-dl/issues/7172#issuecomment-242961695
richiecannizzo commented on Aug 28
brew install libav
Should fix it instantly on mac or
sudo apt-get install libav
#on linux
This error may also occur with python 2 (and early versions of python 3) if your pickle is large (Python Issue #11564):
Python 2.7.11 |Anaconda custom (64-bit)| (default, Dec 6 2015, 18:08:32)
[GCC 4.4.7 20120313 (Red Hat 4.4.7-1)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
Anaconda is brought to you by Continuum Analytics.
Please check out: http://continuum.io/thanks and https://anaconda.org
>>> import cPickle as pickle
>>> string = "X"*(2**31)
>>> pp = pickle.dumps(string)
>>> len(pp)
2147483656
>>> ss = pickle.loads(pp)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: insecure string pickle
This limitation was addressed with the introduction of pickle protocol 4 in python 3.4 (PEP 3154). Unfortunately, this feature has not been back-ported to python 2, and probably won't ever be. If this is your problem and you need to use python 2 pickle, the best you can do is reduce the size of your pickle, e.g., instead of pickling a list, pickle the elements individually into a list of pickles.
Same problem with a file that was made with python on windows, and reloaded with python on linux.
Solution : dos2unix on the file before reading in linux : works as a charm !
I got the Python ValueError: insecure string pickle message in a different way.
For me it happened after a base64 encoding a binary file and passing through urllib2 sockets.
Initially I was wrapping up a file like this
with open(path_to_binary_file) as data_file:
contents = data_file.read()
filename = os.path.split(path)[1]
url = 'http://0.0.0.0:8080/upload'
message = {"filename" : filename, "contents": contents}
pickled_message = cPickle.dumps(message)
base64_message = base64.b64encode(pickled_message)
the_hash = hashlib.md5(base64_message).hexdigest()
server_response = urllib2.urlopen(url, base64_message)
But on the server the hash kept coming out differently for some binary files
decoded_message = base64.b64decode(incoming_base64_message)
the_hash = hashlib.md5(decoded_message).hexdigest()
And unpickling gave insecure string pickle message
cPickle.loads(decoded_message)
BUT SUCCESS
What worked for me was to use urlsafe_b64encode()
base64_message = base64.urlsafe_b64encode(cPickle.dumps(message))
And decode with
base64_decoded_message = base64.urlsafe_b64decode(base64_message)
References
http://docs.python.org/2/library/base64.html
https://www.rfc-editor.org/rfc/rfc3548.html#section-3
This is what happened to me, might be a small section of population, but I want to put this out here nevertheless, for them:
Interpreter (Python3) would have given you an error saying it required the input file stream to be in bytes, and not as a string, and you may have changed the open mode argument from 'r' to 'rb', and now it is telling you the string is corrupt, and thats why you have come here.
The simplest option for such cases is to install Python2 (You can install 2.7) and then run your program with Python 2.7 environment, so it unpickles your file without issue. Basically I wasted a lot of time scanning my string seeing if it was indeed corrupt when all I had to do was change the mode of opening the file from rb to r, and then use Python2 to unpickle the file. So I'm just putting this information out there.
I ran into this earlier, found this thread, and assumed that I was immune to the file closing issue mentioned in a couple of these answers since I was using a with statement:
with tempfile.NamedTemporaryFile(mode='wb') as temp_file:
pickle.dump(foo, temp_file)
# Push file to another machine
_send_file(temp_file.name)
However, since I was pushing the temp file from inside the with, the file still wasn't closed, so the file I was pushing was truncated. This resulted in the same insecure string pickle error in the script that read the file on the remote machine.
Two potential fixes to this: Keep the file open and force a flush:
with tempfile.NamedTemporaryFile(mode='wb') as temp_file:
pickle.dump(foo, temp_file)
temp_file.flush()
# Push file to another machine
_send_file(temp_file.name)
Or make sure the file is closed before doing anything with it:
file_name = ''
with tempfile.NamedTemporaryFile(mode='wb', delete=False) as temp_file:
file_name = temp_file.name
pickle.dump(foo, temp_file)
# Push file to another machine
_send_file(file_name)