Get Time In Seconds Javascript

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Python’s time module represents time objects in Python. Developers use the time () function to return the current time as a UNIX timestamp. The ctime () function translates to a UNIX timestamp in a standard format of year, month, day, and time.

Working with time is an essential part of many programs. For example, suppose you are creating a registration form for a website. You may want to retrieve the current time so you know when a user has signed up for your site.

is where the Python time module comes in. The time module allows you to work with time in your Python program. You can perform actions such as retrieving the time and creating a calendar break.

This tutorial will discuss, with reference to examples, the basics of the Time Python module and how to use it.

Import Python Time Module

The time module has a number of time related functions that you can implement in your Python programs. You can use the time module in all versions of Python.

Before discussing how to use the time module, we must first import it into our program. We can do this using a Python import statement :

Python time () function

the Python time () function retrieves the current time. Time is represented as the number of seconds since January 1, 1970. This is where UNIX time, also called "epoch, begins."

Suppose we want to recover the time right now. We could do it using this program:

Our code returns:

Note that our program did not return a regular time, for example 10:30 AM Our program returned a number at floating point (that is, a decimal number).

This floating point number is the number of seconds since the start of the epoch. We printed the time to the console using a declaration of printing .

in programming, the period time is a standard way of representing time. the time module does not take time zones into account.

This is because the time of the time only stores the number of seconds that have elapsed since the date and time it - it. No time zone data is stored, such as Coordinated Universal Time, with an epoch value

Python Get the current time using ctime ()

the time of age is not readable by most humans. You can accept epoch time to local time using Python ctime () .

ctime () function takes one argument argument. Is the number of seconds since the start of the epoch and returns a string with local time. This value is based on your computer time zone .

Suppose we want to convert our floating point number from the previous epoch to a timestamp. We could do it using this program:

Our code returns the time:

First of all, we declared a Python variables called epoch_time . This variable stores our timestamp.

So we used time.ctime () to convert the timestamp to local time. On the next line of our code, we printed the instruction Local time is:, followed by local time. Additionally, the comma between Local time is: and LOCAL_TIME variables adds a space between the two values.

Python time sleep function

One of the most common functions in the Python time library is the time.sleep () function. time.sleep () suspends execution of a program for a specified number of seconds.

Suppose you want to print the expression "... Disconnection ", then wait three seconds before printing "Disconnection successful!" " To the console. We could do it using this code:

This is how our program works:

The time.sleep () function accepts either floating point (decimal) or integers (integer). if you want a program to stop 3.7 seconds, you can do so by specifying 3.7 as a parameter. In our previous example, we specified 3.

Python struct_time

struct_time odatetime of the Python non-epochal time library structure timestamps. struct_time object is useful because it allows you to work with a standardized form when using timestamps.

Here are the values ‚Äã‚Äãof the struct_time object can store:

We can access all of these attributes via an struct_time object. We are using struct_time objects in our examples below

Using Python structime_object

Python time.localtime ()

Now suppose we want to convert the time from the epoch to l ’local time and having the result returns an struct_time object. We could do this using the localtime () or gmtime () functions.

localtime () function takes an epoch time and returns an struct_time object.

Suppose we want to get the day of the week and the minutes from the hour. We could do it using this program:

Our code returns:

First , we have imported the time library. So we used localtime () to convert our timestamp from previous to an struct_time object.

On the next line , we printed the Day of the week: statement on the console. We then printed the day of the week (which is stored in the tm_wday value of our current_time object).

Next, we printed the minute of the hour: at the console. We followed this instruction with the minutes of the hour (stored in the tm_min value of our current_time object).

Python time.gmtime ()

gmtime () function works the same as the localtime () function, return struct_time object of an epoch.

Suppose we want to know the month of the year and the day of the month. We could do it using this program:

Our code returns:

In our code, we have used gmtime () instead of localtime () to generate a struct_time object. We assigned the value of the gmtime () method current_time variable.

Then we printed Python string Month of the year: on the console, followed by the current month of the year. This value is stored in the tm_mon in our current_time object

So we would print day of the month: on the console, then the current day of the month This value is in the value tm_mday in the current_time object.

python time.asctime ()

asctime () convert an struct_time object into a formatted string Consider the following example:.

Our program returns:

in this example we are using the localtime ( ) at conv ep ert times in a struct_time object. Next, we use asctime () to convert our struct_time object to a fully formatted string representing the current time.

This string is assigned to the FORMATTED_TIME. Then we print the declaration The current time is:., followed by formatted time

time.strftime ()

strftime () formats the values ‚Äã‚Äãin a struct_time in a particular chain.

For example, suppose you want to transform a struct_time object into a string with the following format: MONTH / DAY / YEAR. We could do it using this code:

Our code returns:

We have imported the time library. So, we used localtime () to generate an struct_time object that reflects the current time. We then used strftime () to convert our struct_time object into a string that displays the month, day, and year.

Finally, we printed final_time array value. This table stores our formatted time.

In this example:

Here are some other values you can use with strftime function:

Conclusion

The Python’s time module allows you to work with time in your Python code. The module is equipped with a number of different functions.

You can use them to retrieve the current time, convert the time between epoch time and more readable values, pause a program, format time, and perform other tasks.

Read our guide How to Program in Python for tips and tricks on how to improve your language skills of Python programming.

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Get Time In Seconds Javascript exp: Questions

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)

JavaScript

JSON datetime between Python and JavaScript

4 answers

kevin By kevin

I want to send a datetime.datetime object in serialized form from Python using JSON and de-serialize in JavaScript using JSON. What is the best way to do this?

403

Answer #1

You can add the "default" parameter to json.dumps to handle this:

date_handler = lambda obj: (
    obj.isoformat()
    if isinstance(obj, (datetime.datetime, datetime.date))
    else None
)
json.dumps(datetime.datetime.now(), default=date_handler)
""2010-04-20T20:08:21.634121""

Which is ISO 8601 format.

A more comprehensive default handler function:

def handler(obj):
    if hasattr(obj, "isoformat"):
        return obj.isoformat()
    elif isinstance(obj, ...):
        return ...
    else:
        raise TypeError, "Object of type %s with value of %s is not JSON serializable" % (type(obj), repr(obj))

Update: Added output of type as well as value.
Update: Also handle date

JavaScript

What blocks Ruby, Python to get Javascript V8 speed?

4 answers

Are there any Ruby / Python features that are blocking implementation of optimizations (e.g. inline caching) V8 engine has?

Python is co-developed by Google guys so it shouldn"t be blocked by software patents.

Or this is rather matter of resources put into the V8 project by Google.

260

Answer #1

What blocks Ruby, Python to get Javascript V8 speed?

Nothing.

Well, okay: money. (And time, people, resources, but if you have money, you can buy those.)

V8 has a team of brilliant, highly-specialized, highly-experienced (and thus highly-paid) engineers working on it, that have decades of experience (I"m talking individually – collectively it"s more like centuries) in creating high-performance execution engines for dynamic OO languages. They are basically the same people who also created the Sun HotSpot JVM (among many others).

Lars Bak, the lead developer, has been literally working on VMs for 25 years (and all of those VMs have lead up to V8), which is basically his entire (professional) life. Some of the people writing Ruby VMs aren"t even 25 years old.

Are there any Ruby / Python features that are blocking implementation of optimizations (e.g. inline caching) V8 engine has?

Given that at least IronRuby, JRuby, MagLev, MacRuby and Rubinius have either monomorphic (IronRuby) or polymorphic inline caching, the answer is obviously no.

Modern Ruby implementations already do a great deal of optimizations. For example, for certain operations, Rubinius"s Hash class is faster than YARV"s. Now, this doesn"t sound terribly exciting until you realize that Rubinius"s Hash class is implemented in 100% pure Ruby, while YARV"s is implemented in 100% hand-optimized C.

So, at least in some cases, Rubinius can generate better code than GCC!

Or this is rather matter of resources put into the V8 project by Google.

Yes. Not just Google. The lineage of V8"s source code is 25 years old now. The people who are working on V8 also created the Self VM (to this day one of the fastest dynamic OO language execution engines ever created), the Animorphic Smalltalk VM (to this day one of the fastest Smalltalk execution engines ever created), the HotSpot JVM (the fastest JVM ever created, probably the fastest VM period) and OOVM (one of the most efficient Smalltalk VMs ever created).

In fact, Lars Bak, the lead developer of V8, worked on every single one of those, plus a few others.

JavaScript

Django Template Variables and Javascript

4 answers

When I render a page using the Django template renderer, I can pass in a dictionary variable containing various values to manipulate them in the page using {{ myVar }}.

Is there a way to access the same variable in Javascript (perhaps using the DOM, I don"t know how Django makes the variables accessible)? I want to be able to lookup details using an AJAX lookup based on the values contained in the variables passed in.

256

Answer #1

The {{variable}} is substituted directly into the HTML. Do a view source; it isn"t a "variable" or anything like it. It"s just rendered text.

Having said that, you can put this kind of substitution into your JavaScript.

<script type="text/javascript"> 
   var a = "{{someDjangoVariable}}";
</script>

This gives you "dynamic" javascript.

We hope this article has helped you to resolve the problem. Apart from Get Time In Seconds Javascript, check other exp-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.

By the way, this material is also available in other languages:



Jan Emmerson

California | 2022-12-06

Thanks for explaining! I was stuck with Get Time In Seconds Javascript for some hours, finally got it done 🤗. I just hope that will not emerge anymore

Cornwall Innsbruck

Abu Dhabi | 2022-12-06

Thanks for explaining! I was stuck with Get Time In Seconds Javascript for some hours, finally got it done 🤗. I just hope that will not emerge anymore

Carlo Zelotti

Vigrinia | 2022-12-06

Simply put and clear. Thank you for sharing. Get Time In Seconds Javascript and other issues with sin was always my weak point 😁. Will use it in my bachelor thesis

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