The following worked for my needs:

get_ipython().__class__.__name__

It returns 'TerminalInteractiveShell' on a terminal IPython, 'ZMQInteractiveShell' on Jupyter (notebook AND qtconsole) and fails (NameError) on a regular Python interpreter. The function get_ipython() is available on the global namespace by default when IPython is started.

Wrapping it in a simple function:

def is_notebook() -> bool:
    try:
        shell = get_ipython().__class__.__name__
        if shell == 'ZMQInteractiveShell':
            return True   # Jupyter notebook or qtconsole
        elif shell == 'TerminalInteractiveShell':
            return False  # Terminal running IPython
        else:
            return False  # Other type (?)
    except NameError:
        return False      # Probably standard Python interpreter

The above was tested with Python 3.5.2, IPython 5.1.0 and Jupyter 4.2.1 on macOS 10.12 and Ubuntu 14.04.4 LTS

EDIT: This still works fine in 2022 on newer Python/IPython/Jupyter/OS versions

You can check whether python is in interactive mode with the following snippet [1]:

def is_interactive():
    import __main__ as main
    return not hasattr(main, '__file__')

I have found this method very useful because I do a lot of prototyping in the notebook. For testing purposes, I use default parameters. Otherwise, I read the parameters from sys.argv.

from sys import argv

if is_interactive():
    params = [<list of default parameters>]
else:
    params = argv[1:]

Following the implementation of autonotebook, you can tell whether you are in a notebook using the following code.

def in_notebook():
    try:
        from IPython import get_ipython
        if 'IPKernelApp' not in get_ipython().config:  # pragma: no cover
            return False
    except ImportError:
        return False
    except AttributeError:
        return False
    return True

To check if you're in a notebook, which can be important e.g. when determining what sort of progressbar to use, this worked for me:

def in_ipynb():
    try:
        cfg = get_ipython().config 
        if cfg['IPKernelApp']['parent_appname'] == 'ipython-notebook':
            return True
        else:
            return False
    except NameError:
        return False

Recently I encountered a bug in Jupyter notebook which needs a workaround, and I wanted to do this without loosing functionality in other shells. I realized that keflavich's solution does not work in this case, because get_ipython() is available only directly from the notebook, and not from imported modules. So I found a way to detect from my module whether it is imported and used from a Jupyter notebook or not:

import sys

def in_notebook():
    """
    Returns ``True`` if the module is running in IPython kernel,
    ``False`` if in IPython shell or other Python shell.
    """
    return 'ipykernel' in sys.modules

# later I found out this:

def ipython_info():
    ip = False
    if 'ipykernel' in sys.modules:
        ip = 'notebook'
    elif 'IPython' in sys.modules:
        ip = 'terminal'
    return ip

Comments are appreciated if this is robust enough.

Similar way it is possible to get some info about the client, and IPython version as well:

import sys

if 'ipykernel' in sys.modules:
    ip = sys.modules['ipykernel']
    ip_version = ip.version_info
    ip_client = ip.write_connection_file.__module__.split('.')[0]

# and this might be useful too:

ip_version = IPython.utils.sysinfo.get_sys_info()['ipython_version']

Tested for python 3.7.3

CPython implementations have the name __builtins__ available as part of their globals which btw. can be retrieved by the function globals().
If a script is running in an Ipython environment then __IPYTHON__ should be an attribute of __builtins__.
The code below therefore returns True if run under Ipython or else it gives False

hasattr(__builtins__,'__IPYTHON__')

All you have to do is to place these two cells at the beginning of your notebook:

Cell 1: (marked as "code"):

is_notebook = True

Cell 2: (marked as "Raw NBConvert"):

is_notebook = False

The first cell will always be executed, but the second cell will only be executed when you export the notebook as a Python script.

Later, you can check:

if is_notebook:
    notebook_code()
else:
    script_code()

Hope this helps.

The following captures the cases of https://stackoverflow.com/a/50234148/1491619 without needing to parse the output of ps

def pythonshell():
    """Determine python shell

    pythonshell() returns

    'shell' (started python on command line using "python")
    'ipython' (started ipython on command line using "ipython")
    'ipython-notebook' (e.g., running in Spyder or started with "ipython qtconsole")
    'jupyter-notebook' (running in a Jupyter notebook)

    See also https://stackoverflow.com/a/37661854
    """

    import os
    env = os.environ
    shell = 'shell'
    program = os.path.basename(env['_'])

    if 'jupyter-notebook' in program:
        shell = 'jupyter-notebook'
    elif 'JPY_PARENT_PID' in env or 'ipython' in program:
        shell = 'ipython'
        if 'JPY_PARENT_PID' in env:
            shell = 'ipython-notebook'

    return shell

I would recommend avoiding to detect specific frontend because there are too many of them. Instead you can just test if you are running from within iPython environment:

def is_running_from_ipython():
    from IPython import get_ipython
    return get_ipython() is not None

Above will return False if you are invoking running_from_ipython from usual python command line. When you invoke it from Jupyter Notebook, JupyterHub, iPython shell, Google Colab etc then it will return True.

A very simple and efficient solution is to check if the top of the call stack refers to IPython environment, as follows:

import traceback

def is_in_notebook():
    rstk = traceback.extract_stack(limit=1)[0]
    return rstk[0].startswith("<ipython")

This code works for both Python 2 and 3, on IPython or Jupyter, with no need to check, set or change the environment.

As far as I know, Here has 3 kinds of ipython that used ipykernel

1. ipython qtconsole ("qtipython" for short)
2. IPython in spyder ("spyder" for short)
3. IPython in jupyter notebook ("jn" for short)

use 'spyder' in sys.modules can distinguish spyder

but for qtipython and jn are hard to distinguish because

they have same sys.modules and same IPython config:get_ipython().config

I find a different between qtipython and jn:

first run os.getpid() in IPython shell to get the pid number

then run ps -ef|grep [pid number]

my qtipython pid is 8699

yanglei   8699  8693  4 20:31 ?        00:00:01 /home/yanglei/miniconda2/envs/py3/bin/python -m ipykernel_launcher -f /run/user/1000/jupyter/kernel-8693.json

my jn pid is 8832

yanglei   8832  9788 13 20:32 ?        00:00:01 /home/yanglei/miniconda2/bin/python -m ipykernel_launcher -f /run/user/1000/jupyter/kernel-ccb962ec-3cd3-4008-a4b7-805a79576b1b.json

the different of qtipython and jn is the ipython's json name, jn's json name are longer than qtipython's

so, we can auto-detect all Python Environment using the following code:

import sys,os
def jupyterNotebookOrQtConsole():
    env = 'Unknow'
    cmd = 'ps -ef'
    try:
        with os.popen(cmd) as stream:
            if not py2:
                stream = stream._stream
            s = stream.read()
        pid = os.getpid()
        ls = list(filter(lambda l:'jupyter' in l and str(pid) in l.split(' '), s.split('\n')))
        if len(ls) == 1:
            l = ls[0]
            import re
            pa = re.compile(r'kernel-([-a-z0-9]*)\.json')
            rs = pa.findall(l)
            if len(rs):
                r = rs[0]
                if len(r)<12:
                    env = 'qtipython'
                else :
                    env = 'jn'
        return env
    except:
        return env
    
pyv = sys.version_info.major
py3 = (pyv == 3)
py2 = (pyv == 2)
class pyi():
    '''
    python info
    
    plt : Bool
        mean plt avaliable
    env :
        belong [cmd, cmdipython, qtipython, spyder, jn]
    '''
    pid = os.getpid()
    gui = 'ipykernel' in sys.modules
    cmdipython = 'IPython' in sys.modules and not gui
    ipython = cmdipython or gui
    spyder = 'spyder' in sys.modules
    if gui:
        env = 'spyder' if spyder else jupyterNotebookOrQtConsole()
    else:
        env = 'cmdipython' if ipython else 'cmd'
    
    cmd = not ipython
    qtipython = env == 'qtipython'
    jn = env == 'jn'
    
    plt = gui or 'DISPLAY' in os.environ 

print('Python Envronment is %s'%pyi.env)

The source code is here: Detection Python Environment, Especially distinguish Spyder, Jupyter notebook, Qtconsole.py

I am using Django Shell Plus to launch IPython, and I wanted to make 'running in notebook' available as a Django settings value. get_ipython() is not available when loading settings, so I use this (which is not bulletproof, but good enough for the local development environments it's used in):

import sys

if '--notebook' in sys.argv:
    ENVIRONMENT = "notebook"
else:
    ENVIRONMENT = "dev"

Assuming you have control of the Jupyter Notebook you could:

1. set an environment value in a cell that uses this as a flag in your code. Place a unique comment in that cell (or all cells you want to exclude)

   # exclude_from_export
   %set_env is_jupyter=1

2. Export the notebook as a python script to be used in a different context. The export would exclude the commented cell(s) and subsequently the code that sets the environment value. Note: replace your_notebook.ipynb with the name of your actual notebook file.

   jupyter nbconvert --to script --RegexRemovePreprocessor.patterns="['^# exclude_from_export']" your_notebook.ipynb

This will generate a file that will not have the jupyter environment flag set allowing for code that uses it to deterministically execute.

The following would let one know where the python code is being run... be it in Jupyter standalone, Python standalone, Spyder, VSCode or Jupyter notbeook within VSCode.

import os
import IPython as ipy

# add string sources only
sources = str(os.environ.keys()) + \
          ipy.get_ipython().__class__.__name__

# make pattern of unique keys
checks = {'SPYDER': 'Spyder', 'QTIPYTHON': 'qt IPython', 'VSCODE': 
          'VS Code', 'ZMQINTERACTIVEshell': 'Jupyter', }

results = []
msg = []

for k, v in checks.items():
    u = str(k.upper())
    if u in sources.upper():
        results.append(checks[k])

if not results:
    msg.append("Unknown IDE")
else:
    msg.append("Program working ")
    while results:
        msg.append(f"in {results.pop()}")
        if results:
            msg.append(' with')

print(''.join(msg))

This works:

def is_interactive():
    try:
        __IPYTHON__
        return True
    except NameError:
        return False

How about something like this:

import sys

inJupyter = sys.argv[-1].endswith('json')

print(inJupyter);

I wanted to use a solution that would work if ipython wasn't installed, and I came up with this 1(?) liner which I found to be pretty useful.
No rescues are required for this approach.

Solution

if locals().get('get_ipython', lambda: None).__class__.__name__ == 'TerminalInteractiveShell':
    print('ipython interactive shell detected!')

Explanation

• If ipython is installed and loaded, then the function will be in locals().

  locals().get('get_ipython')
  # -> <bound method InteractiveShell.get_ipython of 
  # <IPython.terminal.interactiveshell.TerminalInteractiveShell object at 0x107164a70>>
  

• We can then call this function get_ipython(), and get the class name of the "interactive shell instance".

  locals().get('get_ipython')()
  # -> <IPython.terminal.interactiveshell.TerminalInteractiveShell at 0x107164a70>
  

• If get_ipython is not available as a function, then we can return an empty lambda function for the class name comparison

  in ipython session:

  locals().get('get_ipython', lambda: None).__class__.__name__
  # -> 'TerminalInteractiveShell'
  

  not in ipython session:

  ☯ ~ python3 -c "print(locals().get('get_ipython', lambda: None)().__class__.__name__)"
  # -> NoneType
  

You can try to eval('__IPYTHON__'), borrowed from pydantic:

def in_ipython() -> bool:
    """
    Check whether we're in an ipython environment, including jupyter notebooks.
    """
    try:
        eval('__IPYTHON__')
    except NameError:
        return False
    else:  # pragma: no cover
        return True

Check the parent process

So far the only solution that worked for me is to check the parent processes using psutil:

import psutil
def runninginJupyterNotebook():
    for i in psutil.Process().parent().cmdline():
        if "jupyter-notebook" in i:
            return True
    else:
        return False

Or you can set a variable in one line:

RUNNING_IN_JUPYTER = any(["jupyter-notebook" in i for i in psutil.Process().parent().cmdline()])

RUNNING_IN_JUPYTER is True if you are running a Jupyter Notebook.

Note that it will be true also if you are running a Colab notebook.

in MacOS

import psutil
def runninginJupyterNotebook():
    for i in psutil.Process().parent().cmdline():
        if "jupyter-notebook" in i:
            return True
    else:
        return False

Or you can set a variable in one line:

RUNNING_IN_JUPYTER = any([i.endswith("bin/jupyter-notebook") for i in psutil.Process().parent().cmdline()])

Comparison to the other solutions:

get_ipython().__class__.__name__

All the solutions using get_ipython() work only if you don't care if it is running on a QtConsole.

Credits: https://stackoverflow.com/a/65498256/2132157