How to expire session due to inactivity in Django?

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Our Django application has the following session management requirements.

  1. Sessions expire when the user closes the browser.
  2. Sessions expire after a period of inactivity.
  3. Detect when a session expires due to inactivity and display appropriate message to the user.
  4. Warn users of a impending session expiry a few minutes before the end of the inactivity period. Along with the warning, provide users an option to extend their session.
  5. If user is working on a long business activity within the app that doesn"t involve requests being sent to the server, the session must not timeout.

After reading the documentation, Django code and some blog posts related to this, I have come up with the following implementation approach.

Requirement 1
This requirement is easily implemented by setting SESSION_EXPIRE_AT_BROWSER_CLOSE to True.

Requirement 2
I have seen a few recommendations to use SESSION_COOKIE_AGE to set the session expiry period. But this method has the following problems.

  • The session always expires at the end of the SESSION_COOKIE_AGE even if the user is actively using the application. (This can be prevented by setting the session expiry to SESSION_COOKIE_AGE on every request using a custom middleware or by saving the session on every request by setting SESSION_SAVE_EVERY_REQUEST to true. But the next problem is unavoidable due to the use of SESSION_COOKIE_AGE.)

  • Due to the way cookies work, SESSION_EXPIRE_AT_BROWSER_CLOSE and SESSION_COOKIE_AGE are mutually exclusive i.e. the cookie either expires on browser close or at the specified expiry time. If SESSION_COOKIE_AGE is used and the user closes the browser before the cookie expires, the cookie is retained and reopening the browser will allow the user (or anyone else) into the system without being re-authenticated.

  • Django relies only on the cookie being present to determine if the session is active. It doesn"t check the session expiry date stored with the session.

The following method could be used to implemented this requirement and to workaround the problems mentioned above.

  • Do not set SESSION_COOKIE_AGE.
  • Set the expiry date of the session to be "current time + inactivity period" on every request.
  • Override process_request in SessionMiddleware and check for session expiry. Discard the session if it has expired.

Requirement 3
When we detect that the session has expired (in the custom SessionMiddleware above), set an attribute on the request to indicate session expiry. This attribute can be used to display an appropriate message to the user.

Requirement 4
Use JavaScript to detect user inactivity, provide the warning and also an option to extend the session. If the user wishes to extend, send a keep alive pulse to the server to extend the session.

Requirement 5
Use JavaScript to detect user activity (during the long business operation) and send keep alive pulses to the server to prevent session from expiring.


The above implementation approach seem very elaborate and I was wondering if there might a simpler method (especially for Requirement 2).

Any insights will be highly appreciated.

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How to expire session due to inactivity in Django? around: Questions

Removing white space around a saved image in matplotlib

2 answers

I need to take an image and save it after some process. The figure looks fine when I display it, but after saving the figure, I got some white space around the saved image. I have tried the "tight" option for savefig method, did not work either. The code:

  import matplotlib.image as mpimg
  import matplotlib.pyplot as plt

  fig = plt.figure(1)
  img = mpimg.imread(path)
  plt.imshow(img)
  ax=fig.add_subplot(1,1,1)

  extent = ax.get_window_extent().transformed(fig.dpi_scale_trans.inverted())
  plt.savefig("1.png", bbox_inches=extent)

  plt.axis("off") 
  plt.show()

I am trying to draw a basic graph by using NetworkX on a figure and save it. I realized that without a graph it works, but when added a graph I get white space around the saved image;

import matplotlib.image as mpimg
import matplotlib.pyplot as plt
import networkx as nx

G = nx.Graph()
G.add_node(1)
G.add_node(2)
G.add_node(3)
G.add_edge(1,3)
G.add_edge(1,2)
pos = {1:[100,120], 2:[200,300], 3:[50,75]}

fig = plt.figure(1)
img = mpimg.imread("image.jpg")
plt.imshow(img)
ax=fig.add_subplot(1,1,1)

nx.draw(G, pos=pos)

extent = ax.get_window_extent().transformed(fig.dpi_scale_trans.inverted())
plt.savefig("1.png", bbox_inches = extent)

plt.axis("off") 
plt.show()
228

Answer #1

You can remove the white space padding by setting bbox_inches="tight" in savefig:

plt.savefig("test.png",bbox_inches="tight")

You"ll have to put the argument to bbox_inches as a string, perhaps this is why it didn"t work earlier for you.


Possible duplicates:

Matplotlib plots: removing axis, legends and white spaces

How to set the margins for a matplotlib figure?

Reduce left and right margins in matplotlib plot

228

Answer #2

I cannot claim I know exactly why or how my “solution” works, but this is what I had to do when I wanted to plot the outline of a couple of aerofoil sections — without white margins — to a PDF file. (Note that I used matplotlib inside an IPython notebook, with the -pylab flag.)

plt.gca().set_axis_off()
plt.subplots_adjust(top = 1, bottom = 0, right = 1, left = 0, 
            hspace = 0, wspace = 0)
plt.margins(0,0)
plt.gca().xaxis.set_major_locator(plt.NullLocator())
plt.gca().yaxis.set_major_locator(plt.NullLocator())
plt.savefig("filename.pdf", bbox_inches = "tight",
    pad_inches = 0)

I have tried to deactivate different parts of this, but this always lead to a white margin somewhere. You may even have modify this to keep fat lines near the limits of the figure from being shaved by the lack of margins.

exp

How do I merge two dictionaries in a single expression (taking union of dictionaries)?

5 answers

Carl Meyer By Carl Meyer

I have two Python dictionaries, and I want to write a single expression that returns these two dictionaries, merged (i.e. taking the union). The update() method would be what I need, if it returned its result instead of modifying a dictionary in-place.

>>> x = {"a": 1, "b": 2}
>>> y = {"b": 10, "c": 11}
>>> z = x.update(y)
>>> print(z)
None
>>> x
{"a": 1, "b": 10, "c": 11}

How can I get that final merged dictionary in z, not x?

(To be extra-clear, the last-one-wins conflict-handling of dict.update() is what I"m looking for as well.)

5839

Answer #1

How can I merge two Python dictionaries in a single expression?

For dictionaries x and y, z becomes a shallowly-merged dictionary with values from y replacing those from x.

  • In Python 3.9.0 or greater (released 17 October 2020): PEP-584, discussed here, was implemented and provides the simplest method:

    z = x | y          # NOTE: 3.9+ ONLY
    
  • In Python 3.5 or greater:

    z = {**x, **y}
    
  • In Python 2, (or 3.4 or lower) write a function:

    def merge_two_dicts(x, y):
        z = x.copy()   # start with keys and values of x
        z.update(y)    # modifies z with keys and values of y
        return z
    

    and now:

    z = merge_two_dicts(x, y)
    

Explanation

Say you have two dictionaries and you want to merge them into a new dictionary without altering the original dictionaries:

x = {"a": 1, "b": 2}
y = {"b": 3, "c": 4}

The desired result is to get a new dictionary (z) with the values merged, and the second dictionary"s values overwriting those from the first.

>>> z
{"a": 1, "b": 3, "c": 4}

A new syntax for this, proposed in PEP 448 and available as of Python 3.5, is

z = {**x, **y}

And it is indeed a single expression.

Note that we can merge in with literal notation as well:

z = {**x, "foo": 1, "bar": 2, **y}

and now:

>>> z
{"a": 1, "b": 3, "foo": 1, "bar": 2, "c": 4}

It is now showing as implemented in the release schedule for 3.5, PEP 478, and it has now made its way into the What"s New in Python 3.5 document.

However, since many organizations are still on Python 2, you may wish to do this in a backward-compatible way. The classically Pythonic way, available in Python 2 and Python 3.0-3.4, is to do this as a two-step process:

z = x.copy()
z.update(y) # which returns None since it mutates z

In both approaches, y will come second and its values will replace x"s values, thus b will point to 3 in our final result.

Not yet on Python 3.5, but want a single expression

If you are not yet on Python 3.5 or need to write backward-compatible code, and you want this in a single expression, the most performant while the correct approach is to put it in a function:

def merge_two_dicts(x, y):
    """Given two dictionaries, merge them into a new dict as a shallow copy."""
    z = x.copy()
    z.update(y)
    return z

and then you have a single expression:

z = merge_two_dicts(x, y)

You can also make a function to merge an arbitrary number of dictionaries, from zero to a very large number:

def merge_dicts(*dict_args):
    """
    Given any number of dictionaries, shallow copy and merge into a new dict,
    precedence goes to key-value pairs in latter dictionaries.
    """
    result = {}
    for dictionary in dict_args:
        result.update(dictionary)
    return result

This function will work in Python 2 and 3 for all dictionaries. e.g. given dictionaries a to g:

z = merge_dicts(a, b, c, d, e, f, g) 

and key-value pairs in g will take precedence over dictionaries a to f, and so on.

Critiques of Other Answers

Don"t use what you see in the formerly accepted answer:

z = dict(x.items() + y.items())

In Python 2, you create two lists in memory for each dict, create a third list in memory with length equal to the length of the first two put together, and then discard all three lists to create the dict. In Python 3, this will fail because you"re adding two dict_items objects together, not two lists -

>>> c = dict(a.items() + b.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: "dict_items" and "dict_items"

and you would have to explicitly create them as lists, e.g. z = dict(list(x.items()) + list(y.items())). This is a waste of resources and computation power.

Similarly, taking the union of items() in Python 3 (viewitems() in Python 2.7) will also fail when values are unhashable objects (like lists, for example). Even if your values are hashable, since sets are semantically unordered, the behavior is undefined in regards to precedence. So don"t do this:

>>> c = dict(a.items() | b.items())

This example demonstrates what happens when values are unhashable:

>>> x = {"a": []}
>>> y = {"b": []}
>>> dict(x.items() | y.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: "list"

Here"s an example where y should have precedence, but instead the value from x is retained due to the arbitrary order of sets:

>>> x = {"a": 2}
>>> y = {"a": 1}
>>> dict(x.items() | y.items())
{"a": 2}

Another hack you should not use:

z = dict(x, **y)

This uses the dict constructor and is very fast and memory-efficient (even slightly more so than our two-step process) but unless you know precisely what is happening here (that is, the second dict is being passed as keyword arguments to the dict constructor), it"s difficult to read, it"s not the intended usage, and so it is not Pythonic.

Here"s an example of the usage being remediated in django.

Dictionaries are intended to take hashable keys (e.g. frozensets or tuples), but this method fails in Python 3 when keys are not strings.

>>> c = dict(a, **b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings

From the mailing list, Guido van Rossum, the creator of the language, wrote:

I am fine with declaring dict({}, **{1:3}) illegal, since after all it is abuse of the ** mechanism.

and

Apparently dict(x, **y) is going around as "cool hack" for "call x.update(y) and return x". Personally, I find it more despicable than cool.

It is my understanding (as well as the understanding of the creator of the language) that the intended usage for dict(**y) is for creating dictionaries for readability purposes, e.g.:

dict(a=1, b=10, c=11)

instead of

{"a": 1, "b": 10, "c": 11}

Response to comments

Despite what Guido says, dict(x, **y) is in line with the dict specification, which btw. works for both Python 2 and 3. The fact that this only works for string keys is a direct consequence of how keyword parameters work and not a short-coming of dict. Nor is using the ** operator in this place an abuse of the mechanism, in fact, ** was designed precisely to pass dictionaries as keywords.

Again, it doesn"t work for 3 when keys are not strings. The implicit calling contract is that namespaces take ordinary dictionaries, while users must only pass keyword arguments that are strings. All other callables enforced it. dict broke this consistency in Python 2:

>>> foo(**{("a", "b"): None})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() keywords must be strings
>>> dict(**{("a", "b"): None})
{("a", "b"): None}

This inconsistency was bad given other implementations of Python (PyPy, Jython, IronPython). Thus it was fixed in Python 3, as this usage could be a breaking change.

I submit to you that it is malicious incompetence to intentionally write code that only works in one version of a language or that only works given certain arbitrary constraints.

More comments:

dict(x.items() + y.items()) is still the most readable solution for Python 2. Readability counts.

My response: merge_two_dicts(x, y) actually seems much clearer to me, if we"re actually concerned about readability. And it is not forward compatible, as Python 2 is increasingly deprecated.

{**x, **y} does not seem to handle nested dictionaries. the contents of nested keys are simply overwritten, not merged [...] I ended up being burnt by these answers that do not merge recursively and I was surprised no one mentioned it. In my interpretation of the word "merging" these answers describe "updating one dict with another", and not merging.

Yes. I must refer you back to the question, which is asking for a shallow merge of two dictionaries, with the first"s values being overwritten by the second"s - in a single expression.

Assuming two dictionaries of dictionaries, one might recursively merge them in a single function, but you should be careful not to modify the dictionaries from either source, and the surest way to avoid that is to make a copy when assigning values. As keys must be hashable and are usually therefore immutable, it is pointless to copy them:

from copy import deepcopy

def dict_of_dicts_merge(x, y):
    z = {}
    overlapping_keys = x.keys() & y.keys()
    for key in overlapping_keys:
        z[key] = dict_of_dicts_merge(x[key], y[key])
    for key in x.keys() - overlapping_keys:
        z[key] = deepcopy(x[key])
    for key in y.keys() - overlapping_keys:
        z[key] = deepcopy(y[key])
    return z

Usage:

>>> x = {"a":{1:{}}, "b": {2:{}}}
>>> y = {"b":{10:{}}, "c": {11:{}}}
>>> dict_of_dicts_merge(x, y)
{"b": {2: {}, 10: {}}, "a": {1: {}}, "c": {11: {}}}

Coming up with contingencies for other value types is far beyond the scope of this question, so I will point you at my answer to the canonical question on a "Dictionaries of dictionaries merge".

Less Performant But Correct Ad-hocs

These approaches are less performant, but they will provide correct behavior. They will be much less performant than copy and update or the new unpacking because they iterate through each key-value pair at a higher level of abstraction, but they do respect the order of precedence (latter dictionaries have precedence)

You can also chain the dictionaries manually inside a dict comprehension:

{k: v for d in dicts for k, v in d.items()} # iteritems in Python 2.7

or in Python 2.6 (and perhaps as early as 2.4 when generator expressions were introduced):

dict((k, v) for d in dicts for k, v in d.items()) # iteritems in Python 2

itertools.chain will chain the iterators over the key-value pairs in the correct order:

from itertools import chain
z = dict(chain(x.items(), y.items())) # iteritems in Python 2

Performance Analysis

I"m only going to do the performance analysis of the usages known to behave correctly. (Self-contained so you can copy and paste yourself.)

from timeit import repeat
from itertools import chain

x = dict.fromkeys("abcdefg")
y = dict.fromkeys("efghijk")

def merge_two_dicts(x, y):
    z = x.copy()
    z.update(y)
    return z

min(repeat(lambda: {**x, **y}))
min(repeat(lambda: merge_two_dicts(x, y)))
min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
min(repeat(lambda: dict(chain(x.items(), y.items()))))
min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))

In Python 3.8.1, NixOS:

>>> min(repeat(lambda: {**x, **y}))
1.0804965235292912
>>> min(repeat(lambda: merge_two_dicts(x, y)))
1.636518670246005
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
3.1779992282390594
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
2.740647904574871
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
4.266070580109954
$ uname -a
Linux nixos 4.19.113 #1-NixOS SMP Wed Mar 25 07:06:15 UTC 2020 x86_64 GNU/Linux

Resources on Dictionaries

5839

Answer #2

In your case, what you can do is:

z = dict(list(x.items()) + list(y.items()))

This will, as you want it, put the final dict in z, and make the value for key b be properly overridden by the second (y) dict"s value:

>>> x = {"a":1, "b": 2}
>>> y = {"b":10, "c": 11}
>>> z = dict(list(x.items()) + list(y.items()))
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python 2, you can even remove the list() calls. To create z:

>>> z = dict(x.items() + y.items())
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python version 3.9.0a4 or greater, then you can directly use:

x = {"a":1, "b": 2}
y = {"b":10, "c": 11}
z = x | y
print(z)
{"a": 1, "c": 11, "b": 10}

5839

Answer #3

An alternative:

z = x.copy()
z.update(y)

We hope this article has helped you to resolve the problem. Apart from How to expire session due to inactivity in Django?, check other around-related topics.

Want to excel in Python? See our review of the best Python online courses 2022. If you are interested in Data Science, check also how to learn programming in R.

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Davies Wu

Munchen | 2022-12-03

Thanks for explaining! I was stuck with How to expire session due to inactivity in Django? for some hours, finally got it done 🤗. Checked yesterday, it works!

Carlo Robinson

Shanghai | 2022-12-03

I was preparing for my coding interview, thanks for clarifying this - How to expire session due to inactivity in Django? in Python is not the simplest one. I am just not quite sure it is the best method

Walter Richtgofen

Abu Dhabi | 2022-12-03

exp is always a bit confusing 😭 How to expire session due to inactivity in Django? is not the only problem I encountered. Will get back tomorrow with feedback

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