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Either JavaScript is one of three ways to declare a variable. JavaScript let is used to declare block scope variables and cannot be reassigned or declared again. The let variable is best used for loops.

Variables are an essential part of most programming languages. When you start to use JavaScript, you will come across the term often . "Variable" is used to describe a container that stores one or more values ‚Äã‚Äãin code

There are three ways to declare a JavaScript variable. Var, and leave const. Each of these types of variables can be used in different circumstances, which we will discuss later in this article.

That said, the focus of this tutorial is the JavaScript keyword "let it go." " " Leave " allows you to declare variables that can be reassigned. We’ll see how the "leave‚" keyword works in JavaScript and explore an example of a variable declaration using "leave‚".

Last day of JavaScript variables

JavaScript variables store values. For example, if we have a name that we want to store in our code, we might want to assign it to a variable. Variables can contain any type of data, including numbers, strings, and objects.

The most common way to declare a JavaScript variable is to use "var", largely because the original specification of the language on which JavaScript is based no longer part of declaring a variable. Here is the syntax for declaring a variable using "var:"

In the example above, we declared a variable called "name" and assigned it the value "John Appleseed." Our variable has a few components:

  • The variable is declared using the "var" . "Full_name" keyword
  • The name of our variable is
  • "=" syntax tells our program to assign our variable "full_name" with a value.
  • The value assigned to our variable is "John Appleseed".

Now that we have declared our variable, we can l ’ using all along . our code in the following code, we print our name variable:

Our program returns: "John Appleseed".

In this example, our variable is a string, but we could have assigned a number, an array, a JSON object, a boolean, or any other data type. Here is an example of a variable that stores an array:

In this example we have declared a variable called This stores four strings variables, which are the names of people who might attend our example concert

Types of JavaScript variables

In the examples above, the keyword "var" is used to declare a variable, but sometimes it is more appropriate to use another keyword. The keywords, let "and" const "come into play.

In JavaScript, we can declare variables using each of these three keywords, which are used in different circumstances. .

The following table shows the differences between these three types of variables:

Word < / td> variable scope Reassign? Reassign? Raise? var Function Yes Yes Yes const Block Yes No No quit Block No No No

This table looks complex, but when you start using the different types of variables you will start to understand better when to use each one.

The following are general rules of thumb for using these types of variables:

  1. Use const , whenever possible, unless you need to re-declare or hoist a variable .
  2. Use let if you are working with loop .
  3. Use var only if:
    1. You are working on previous code ,
    2. You need a variable that you can re-claim or
    3. you need a variable that is accessible anywhere in the program (in the world) .

Two mnemonics to help you remember the three keywords for variables in JavaScript and when to use them are:

  1. Variables I am with TAINER who > allow to keep going in your code.
    1. Const Antine could neither redeclare nor raise the
    2. Either pass through the loops, so that the

      variety of vegetables has become a world heirloom.

      Now let’s take a look at the keyword "leave" . We will explore its use in JavaScript and how it compares to other keywords for variables.

      JavaScript Let

      JavaScript "let" word - key allows you to declare a range variable block. (In contrast to global variables, block scope variables are local to the block they are declared in.) -You remember that you declare global variables with

What does this mean? scope refers to where you can access a variable in your code. "Block field" refers to the area inside a loop or in general conditional braces ({}) distinguish a block of code.

Local variables are the variables declared in a block of code. the global variables, on the other hand., are reported to the outside.

The "let" keyword is useful in programs that use variables in an "if" or "for" loop

Here is a JavaScript example & ldquo; leave " keyword in action:

Our code returns The following:

It’s a working day. Make a banana for lunch today.

It’s not a working day. Prepare an apple for lunch today

In this example, we are declaring two variables:. " Work day" and "fruit lunch" " work day " is a variable of global scope that we can access our entire program (note that the code of practice "var") and "lunchFruit" is a local variable that is reachable reassigned only in our loop (note that the code uses "let").

When our program runs the "if" statement, it sees that "business day" equals "true". Therefore, our program executes the code in the "if" block. In this case, our program reassigns "lunchFruit" to "banana" and then prints a message to the console.

But when our runtime code finishes in the "if" statement field , prints the message "It’s not a business day." Prepare an apple today ‚". This is because our "fruit lunch" variable only changes in our "if" statement. The name of the fruit does not change because our final console.log () is outside the "if" statement

The code above shows the two main attributes of the "let". Variable:

  1. "let" has block scope, which means it cannot be accessed in a particular block.
  2. "let" can be reassigned if you want, but only in our block. field
  3. If we wanted our variable to be global we would use ’var’ instead, but since the previous example uses an ’if’ we just had to use the ’let & rdquo; variable.

    JavaScript Let Alternatives

    To illustrate the differences between "var" and "leave" in more detail, here is an example of the same code above, but using "var" instead. instead of "leaving" to declare our "lunchFruit" variable:

    Our code returns:

    It’s a working day . Make a banana for lunch today.

    It’s not a working day. Prepare a banana for lunch today.

    As you can see, when our code changes the "lunchFruit" variable value "banana" the change is made in the world, so when our code executes the last "console.log ()" statement, prints the variable of "lunchFruit" which stores the value "banana".

    , you can also use "const" to declare a variable. However, the variable contrast declared using the "let" word - key command , once the variable is declared, it can not be redeclared.

    Here’s an example of declaring a variable using ’const’ ’the word - key:

    The variable "full_name" now stores "John Appleseed." But if we wanted to change the name, we need to declare a new variable, like this:

    If we tried to reassign the variable "full_name," our program would return an error, "full_name has already been declared." So "const" can be useful for storing values ‚Äã‚Äãthat won’t change, but if you plan to re-declare a variable, you can use "var.

    Conclusion

    Variables are an essential part of programming. In this tutorial, we have discussed the basics of variables and how to work with "var" and "leave" in JavaScript. We also briefly discussed the differences between "var" "leave" and "cost" and how each of these variable keywords should be used in JavaScript.

    You are now ready to declare the JavaScript variable like a professional developer!

    👻 Read also: what is the best laptop for engineering students in 2022?

    Javascript Sheets cos: Questions

    cos

    How do I install pip on macOS or OS X?

    5 answers

    I spent most of the day yesterday searching for a clear answer for installing pip (package manager for Python). I can"t find a good solution.

    How do I install it?

    1672

    Answer #1

    UPDATE (Jan 2019):

    easy_install has been deprecated. Please use get-pip.py instead.


    Old answer:

    easy_install pip
    

    If you need admin privileges to run this, try:

    sudo easy_install pip
    

    1672

    Answer #2

    ⚡️ TL;DR — One line solution.

    All you have to do is:

    sudo easy_install pip
    

    2019: ⚠️easy_install has been deprecated. Check Method #2 below for preferred installation!

    Details:

    ⚡️ OK, I read the solutions given above, but here"s an EASY solution to install pip.

    MacOS comes with Python installed. But to make sure that you have Python installed open the terminal and run the following command.

    python --version
    

    If this command returns a version number that means Python exists. Which also means that you already have access to easy_install considering you are using macOS/OSX.

    ℹ️ Now, all you have to do is run the following command.

    sudo easy_install pip
    

    After that, pip will be installed and you"ll be able to use it for installing other packages.

    Let me know if you have any problems installing pip this way.

    Cheers!

    P.S. I ended up blogging a post about it. QuickTip: How Do I Install pip on macOS or OS X?


    ✅ UPDATE (Jan 2019): METHOD #2: Two line solution —

    easy_install has been deprecated. Please use get-pip.py instead.

    First of all download the get-pip file

    curl https://bootstrap.pypa.io/get-pip.py -o get-pip.py
    

    Now run this file to install pip

    python get-pip.py
    

    That should do it.

    Another gif you said? Here ya go!

    1672

    Answer #3

    You can install it through Homebrew on OS X. Why would you install Python with Homebrew?

    The version of Python that ships with OS X is great for learning but it’s not good for development. The version shipped with OS X may be out of date from the official current Python release, which is considered the stable production version. (source)

    Homebrew is something of a package manager for OS X. Find more details on the Homebrew page. Once Homebrew is installed, run the following to install the latest Python, Pip & Setuptools:

    brew install python
    

    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)
    

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