I'm trying to create a process that dynamically watches jupyter notebooks, compiles them on modification and imports them into my current file, however I can't seem to execute the updated code. It only executes the first version that was loaded.
There's a file called producer.py that calls this function repeatedly:
import fs.fs_util as fs_util
while(True):
fs_util.update_feature_list()
In fs_util.py I do the following:
from fs.feature import Feature
import inspect
from importlib import reload
import os
def is_subclass_of_feature(o):
return inspect.isclass(o) and issubclass(o, Feature) and o is not Feature
def get_instances_of_features(name):
module = __import__(COMPILED_MODULE, fromlist=[name])
module = reload(module)
feature_members = getattr(module, name)
all_features = inspect.getmembers(feature_members, predicate=is_subclass_of_feature)
return [f[1]() for f in all_features]
This function is called by:
def update_feature_list(name):
os.system("jupyter nbconvert --to script {}{} --output {}{}"
.format(PATH + "/" + s3.OUTPUT_PATH, name + JUPYTER_EXTENSION, PATH + "/" + COMPILED_PATH, name))
features = get_instances_of_features(name)
for f in features:
try:
feature = f.create_feature()
except Exception as e:
print(e)
There is other irrelevant code that checks for whether a file has been modified etc.
I can tell the file is being reloaded correctly because when I use inspect.getsource(f.create_feature) on the class it displays the updated source code, however during execution it returns older values. I've verified this by changing print statements as well as comparing the return values.
Also for some more context the file I'm trying to import:
from fs.feature import Feature
class SubFeature(Feature):
def __init__(self):
Feature.__init__(self)
def create_feature(self):
return "hello"
I was wondering what I was doing incorrectly?
So I found out what I was doing wrong.
When called reload I was reloading the module I had newly imported, which was fairly idiotic I suppose. The correct solution (in my case) was to reload the module from sys.modules, so it would be something like reload(sys.modules[COMPILED_MODULE + "." + name])
I am trying to download files in python using the wget module. I understand that it's supposed to have several progress bar modes but non actually show in the console.
I could not find any documentation for this module.
import wget
from pathlib import Path
print('Beginning download...')
url = 'https://storage.googleapis.com/meirtvmp3/archive/hebrew/mp3/sherki/daattvunot/idx_69115.mp3'
wget.download(url)
The file downloads but no progress bar shows.
The source code for the module does have a reference to it. But when I try it in my code python cant find a reference to it.
EDIT
I tested the script from the terminal and it worked as expected. I guess it's a pycharm/venv bug.
This works for me,
#create this bar_progress method which is invoked automatically from wget
def bar_progress(current, total, width=80):
progress_message = "Downloading: %d%% [%d / %d] bytes" % (current / total * 100, current, total)
# Don't use print() as it will print in new line every time.
sys.stdout.write("\r" + progress_message)
sys.stdout.flush()
#Now use this like below,
url = 'http://url_to_download'
save_path = "/home/save.file"
wget.download(url, save_path, bar=bar_progress)
It is the first time I write as I really didn't find any solution to my issue.
I want to allow my user to launch some Python program from Excel.
So i have this VBA code at some point:
lg_ErrorCode = wsh.Run(str_PythonPath & " " & str_PythonArg, 1, bl_StopVBwhilePython)
If lg_ErrorCode <> 0 Then
MsgBox "Couldn't run python script! " _
+ vbCrLf + CStr(lg_ErrorCode)
Run_Python = False
End If
str_PythonPath = "C:\Python34\python.exe C:\Users\XXXX\Documents\4_Python\Scan_FTP\test.py"
str_PythonArg = "arg1 arg2"
After multiple testing, the row in error in Python is when I try to import another module (I precise that this VBA code is working without the below row in Python):
import fct_Ftp as ftp
The architecture of the module is as follow:
4_Python
-folder: Scan_FTP
- file: test.py (The one launch from VBA)
-file: fct_Ftp.py
(For information, I change the architecture of the file, and try to copy the file at some other position just to test without success)
The import has no problem when I launch Test.py directly with:
import sys, os
sys.path.append('../')
But from VBA, this import is not working.
So I figured out this more generic solution, that dont work as well from Excel/VBA
import sys, os
def m_importingFunction():
str_absPath = os.path.abspath('')
str_absPathDad = os.path.dirname(str_absPath)
l_absPathSons = [os.path.abspath(x[0]) for x in os.walk('.')]
l_Dir = l_absPathSons + [str_absPathDad]
l_DirPy = [Dir for Dir in l_Dir if 'Python' in Dir]
for Dir in l_DirPy:
sys.path.append(Dir)
print(Dir)
m_importingFunction()
try:
import fct_Ftp as ftp
# ftp = __import__ ("fct_Ftp")
write += 'YAAAA' # write a file YAAAA from Python
except:
write += 'NOOOOOO' # write a file NOOOOO from VBA
f= open(write + ".txt","w+")
f.close()
Can you please help me as it is a very tricky questions ?
Many thanks to you guys.
You are able to start your program from the command line?
Why not create a batch file with excel which you then start in a shell?
just found the following . running via notebook
import
.echo("test")
Output:
/home/user/anaconda3/envs/p36/lib/python3.6/site-packages//utils.py in echo(message, file, nl, err, color)
257
258 if message:
--> 259 file.write(message)
260 file.flush()
261
UnsupportedOperation: not writable
Has someone seen this before and knows how to work around? I have to use a lib via that uses . so not is not possible.
Update:
This commit to a jupyter branch of click solves the issue:
https://github.com/elgalu/click/commit/1cb7aaba8c9dd6ec760d3e7e414d0b4e5f788543#diff-d17772ee4f65879b69a53dbc4b3d42bd
I think that Jupyter hijacks and locks the STDOUT/STDERR (at least the one click is trying to use) and if you don't provide a stream to click.echo() it will attempt writing to the STDOUT/STDERR, hence the error.
You can work around it by passing an output stream like STDOUT yourself:
import click
import sys
click.echo("test", sys.stdout)
# test
In my case using Python 3 I wanted to preserve the click styling on my message both in the Jupyter notebook and when the code was run in the terminal. I handled it this way:
from io import UnsupportedOperation
import click
item = 'Your Name'
message = click.style('"{}"'.format(item), fg='red', bold=True)
try:
click.echo(message)
except UnsupportedOperation as err:
print('Error: "{}"'.format(err))
print(message)
The color is preserved in the notebook:
I am trying to obtain the current NoteBook name when running the IPython notebook. I know I can see it at the top of the notebook. What I am after something like
currentNotebook = IPython.foo.bar.notebookname()
I need to get the name in a variable.
adding to previous answers,
to get the notebook name run the following in a cell:
%%javascript
IPython.notebook.kernel.execute('nb_name = "' + IPython.notebook.notebook_name + '"')
this gets you the file name in nb_name
then to get the full path you may use the following in a separate cell:
import os
nb_full_path = os.path.join(os.getcwd(), nb_name)
I have the following which works with IPython 2.0. I observed that the name of the notebook is stored as the value of the attribute 'data-notebook-name' in the <body> tag of the page. Thus the idea is first to ask Javascript to retrieve the attribute --javascripts can be invoked from a codecell thanks to the %%javascript magic. Then it is possible to access to the Javascript variable through a call to the Python Kernel, with a command which sets a Python variable. Since this last variable is known from the kernel, it can be accessed in other cells as well.
%%javascript
var kernel = IPython.notebook.kernel;
var body = document.body,
attribs = body.attributes;
var command = "theNotebook = " + "'"+attribs['data-notebook-name'].value+"'";
kernel.execute(command);
From a Python code cell
print(theNotebook)
Out[ ]: HowToGetTheNameOfTheNoteBook.ipynb
A defect in this solution is that when one changes the title (name) of a notebook, then this name seems to not be updated immediately (there is probably some kind of cache) and it is necessary to reload the notebook to get access to the new name.
[Edit] On reflection, a more efficient solution is to look for the input field for notebook's name instead of the <body> tag. Looking into the source, it appears that this field has id "notebook_name". It is then possible to catch this value by a document.getElementById() and then follow the same approach as above. The code becomes, still using the javascript magic
%%javascript
var kernel = IPython.notebook.kernel;
var thename = window.document.getElementById("notebook_name").innerHTML;
var command = "theNotebook = " + "'"+thename+"'";
kernel.execute(command);
Then, from a ipython cell,
In [11]: print(theNotebook)
Out [11]: HowToGetTheNameOfTheNoteBookSolBis
Contrary to the first solution, modifications of notebook's name are updated immediately and there is no need to refresh the notebook.
As already mentioned you probably aren't really supposed to be able to do this, but I did find a way. It's a flaming hack though so don't rely on this at all:
import json
import os
import urllib2
import IPython
from IPython.lib import kernel
connection_file_path = kernel.get_connection_file()
connection_file = os.path.basename(connection_file_path)
kernel_id = connection_file.split('-', 1)[1].split('.')[0]
# Updated answer with semi-solutions for both IPython 2.x and IPython < 2.x
if IPython.version_info[0] < 2:
## Not sure if it's even possible to get the port for the
## notebook app; so just using the default...
notebooks = json.load(urllib2.urlopen('http://127.0.0.1:8888/notebooks'))
for nb in notebooks:
if nb['kernel_id'] == kernel_id:
print nb['name']
break
else:
sessions = json.load(urllib2.urlopen('http://127.0.0.1:8888/api/sessions'))
for sess in sessions:
if sess['kernel']['id'] == kernel_id:
print sess['notebook']['name']
break
I updated my answer to include a solution that "works" in IPython 2.0 at least with a simple test. It probably isn't guaranteed to give the correct answer if there are multiple notebooks connected to the same kernel, etc.
It seems I cannot comment, so I have to post this as an answer.
The accepted solution by #iguananaut and the update by #mbdevpl appear not to be working with recent versions of the Notebook.
I fixed it as shown below. I checked it on Python v3.6.1 + Notebook v5.0.0 and on Python v3.6.5 and Notebook v5.5.0.
import jupyterlab
if jupyterlab.__version__.split(".")[0] == "3":
from jupyter_server import serverapp as app
key_srv_directory = 'root_dir'
else :
from notebook import notebookapp as app
key_srv_directory = 'notebook_dir'
import urllib
import json
import os
import ipykernel
def notebook_path(key_srv_directory, ):
"""Returns the absolute path of the Notebook or None if it cannot be determined
NOTE: works only when the security is token-based or there is also no password
"""
connection_file = os.path.basename(ipykernel.get_connection_file())
kernel_id = connection_file.split('-', 1)[1].split('.')[0]
for srv in app.list_running_servers():
try:
if srv['token']=='' and not srv['password']: # No token and no password, ahem...
req = urllib.request.urlopen(srv['url']+'api/sessions')
else:
req = urllib.request.urlopen(srv['url']+'api/sessions?token='+srv['token'])
sessions = json.load(req)
for sess in sessions:
if sess['kernel']['id'] == kernel_id:
return os.path.join(srv[key_srv_directory],sess['notebook']['path'])
except:
pass # There may be stale entries in the runtime directory
return None
As stated in the docstring, this works only when either there is no authentication or the authentication is token-based.
Note that, as also reported by others, the Javascript-based method does not seem to work when executing a "Run all cells" (but works when executing cells "manually"), which was a deal-breaker for me.
The ipyparams package can do this pretty easily.
import ipyparams
currentNotebook = ipyparams.notebook_name
On Jupyter 3.0 the following works. Here I'm showing the entire path on the Jupyter server, not just the notebook name:
To store the NOTEBOOK_FULL_PATH on the current notebook front end:
%%javascript
var nb = IPython.notebook;
var kernel = IPython.notebook.kernel;
var command = "NOTEBOOK_FULL_PATH = '" + nb.base_url + nb.notebook_path + "'";
kernel.execute(command);
To then display it:
print("NOTEBOOK_FULL_PATH:\n", NOTEBOOK_FULL_PATH)
Running the first Javascript cell produces no output.
Running the second Python cell produces something like:
NOTEBOOK_FULL_PATH:
/user/zeph/GetNotebookName.ipynb
Yet another hacky solution since my notebook server can change. Basically you print a random string, save it and then search for a file containing that string in the working directory. The while is needed because save_checkpoint is asynchronous.
from time import sleep
from IPython.display import display, Javascript
import subprocess
import os
import uuid
def get_notebook_path_and_save():
magic = str(uuid.uuid1()).replace('-', '')
print(magic)
# saves it (ctrl+S)
display(Javascript('IPython.notebook.save_checkpoint();'))
nb_name = None
while nb_name is None:
try:
sleep(0.1)
nb_name = subprocess.check_output(f'grep -l {magic} *.ipynb', shell=True).decode().strip()
except:
pass
return os.path.join(os.getcwd(), nb_name)
There is no real way yet to do this in Jupyterlab. But there is an official way that's now under active discussion/development as of August 2021:
https://github.com/jupyter/jupyter_client/pull/656
In the meantime, hitting the api/sessions REST endpoint of jupyter_server seems like the best bet. Here's a cleaned-up version of that approach:
from jupyter_server import serverapp
from jupyter_server.utils import url_path_join
from pathlib import Path
import re
import requests
kernelIdRegex = re.compile(r"(?<=kernel-)[\w\d\-]+(?=\.json)")
def getNotebookPath():
kernelId = kernelIdRegex.search(get_ipython().config["IPKernelApp"]["connection_file"])[0]
for jupServ in serverapp.list_running_servers():
for session in requests.get(url_path_join(jupServ["url"], "api/sessions"), params={"token": jupServ["token"]}).json():
if kernelId == session["kernel"]["id"]:
return Path(jupServ["root_dir"]) / session["notebook"]['path']
Tested working with
python==3.9
jupyter_server==1.8.0
jupyterlab==4.0.0a7
Modifying #jfb method, gives the function below which worked fine on ipykernel-5.3.4.
def getNotebookName():
display(Javascript('IPython.notebook.kernel.execute("NotebookName = " + "\'"+window.document.getElementById("notebook_name").innerHTML+"\'");'))
try:
_ = type(NotebookName)
return NotebookName
except:
return None
Note that the display javascript will take some time to reach the browser, and it will take some time to execute the JS and get back to the kernel. I know it may sound stupid, but it's better to run the function in two cells, like this:
nb_name = getNotebookName()
and in the following cell:
for i in range(10):
nb_name = getNotebookName()
if nb_name is not None:
break
However, if you don't need to define a function, the wise method is to run display(Javascript(..)) in one cell, and check the notebook name in another cell. In this way, the browser has enough time to execute the code and return the notebook name.
If you don't mind to use a library, the most robust way is:
import ipynbname
nb_name = ipynbname.name()
If you are using Visual Studio Code:
import IPython ; IPython.extract_module_locals()[1]['__vsc_ipynb_file__']
Assuming you have the Jupyter Notebook server's host, port, and authentication token, this should work for you. It's based off of this answer.
import os
import json
import posixpath
import subprocess
import urllib.request
import psutil
def get_notebook_path(host, port, token):
process_id = os.getpid();
notebooks = get_running_notebooks(host, port, token)
for notebook in notebooks:
if process_id in notebook['process_ids']:
return notebook['path']
def get_running_notebooks(host, port, token):
sessions_url = posixpath.join('http://%s:%d' % (host, port), 'api', 'sessions')
sessions_url += f'?token={token}'
response = urllib.request.urlopen(sessions_url).read()
res = json.loads(response)
notebooks = [{'kernel_id': notebook['kernel']['id'],
'path': notebook['notebook']['path'],
'process_ids': get_process_ids(notebook['kernel']['id'])} for notebook in res]
return notebooks
def get_process_ids(name):
child = subprocess.Popen(['pgrep', '-f', name], stdout=subprocess.PIPE, shell=False)
response = child.communicate()[0]
return [int(pid) for pid in response.split()]
Example usage:
get_notebook_path('127.0.0.1', 17004, '344eb91bee5742a8501cc8ee84043d0af07d42e7135bed90')
To realize why you can't get notebook name using these JS-based solutions, run this code and notice the delay it takes for the message box to appear after python has finished execution of the cell / entire notebook:
%%javascript
function sayHello() {
alert('Hello world!');
}
setTimeout(sayHello, 1000);
More info
Javascript calls are async and hence not guaranteed to complete before python starts running another cell containing the code expecting this notebook name variable to be already created... resulting in NameError when trying to access non-existing variables that should contain notebook name.
I suspect some upvotes on this page became locked before voters could discover that all %%javascript-based solutions ultimately don't work... when the producer and consumer notebook cells are executed together (or in a quick succession).
All Json based solutions fail if we execute more than one cell at a time
because the result will not be ready until after the end of the execution
(its not a matter of using sleep or waiting any time, check it yourself but remember to restart kernel and run all every test)
Based on previous solutions, this avoids using the %% magic in case you need to put it in the middle of some other code:
from IPython.display import display, Javascript
# can have comments here :)
js_cmd = 'IPython.notebook.kernel.execute(\'nb_name = "\' + IPython.notebook.notebook_name + \'"\')'
display(Javascript(js_cmd))
For python 3, the following based on the answer by #Iguananaut and updated for latest python and possibly multiple servers will work:
import os
import json
try:
from urllib2 import urlopen
except:
from urllib.request import urlopen
import ipykernel
connection_file_path = ipykernel.get_connection_file()
connection_file = os.path.basename(connection_file_path)
kernel_id = connection_file.split('-', 1)[1].split('.')[0]
running_servers = !jupyter notebook list
running_servers = [s.split('::')[0].strip() for s in running_servers[1:]]
nb_name = '???'
for serv in running_servers:
uri_parts = serv.split('?')
uri_parts[0] += 'api/sessions'
sessions = json.load(urlopen('?'.join(uri_parts)))
for sess in sessions:
if sess['kernel']['id'] == kernel_id:
nb_name = os.path.basename(sess['notebook']['path'])
break
if nb_name != '???':
break
print (f'[{nb_name}]')
just use ipynbname , which is practical
import ipynbname
nb_fname = ipynbname.name()
nb_path = ipynbname.path()
print(f"{nb_fname=}")
print(f"{nb_path=}")
I found this in https://stackoverflow.com/a/65907473/15497427