Javascript State

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Have you ever heard of State in React JavaScript library but don’t know how to use it? State refers to "the way every object appears on a web page." " The concept is simple - it is a collection of data about the current state of the component, which can be changed by modifying the data described. Whenever the data changes , it is displayed again on the computer screen. The hardest part of the condition is knowing how and when to use it. But don’t worry because this tutorial explains how state is used.

To give you a quick overview, to set the initial state you use this.state in a JavaScript constructor. To update the state, use the setState method. We will talk more about state and ways of use state in this tutorial.

What is React State?

React state is an object with a set of properties designated for a component and this helps with the way the component is rendered. A component with state is called stateful. You usually use if a component need to track information about it between makes.

An example would be a "like" button component that needs to keep track of the number of times it clicks. This should reappear on the screen each time the button is pressed as if to display the updated number of likes. And for that, we have to make sure that the button like component follows the information, instead of repeating the information that we have coded with over and over. After all, you don’t want to miss these people.

The state data of a component can be accessed via this.state and edited using the .setState method. The state is private because it cannot be changed by the component that contains the state

Many people confuse state with accessories, another object of the reaction, but there are important differences between the two. One of them is that the state is dynamic, which means that the data it contains can change and thus affect the behavior of the component. Props, on the other hand, are immutable, which means that "its state cannot change after it has been built. "

React State Syntax

the state is initially defined in the function named constructor. the state may change, we want to assure you that there is an initial current value. The general format for setting the status is:

For example, if you are creating a timer component , you should set the start state as this.state = {0} seconds:. You know a timer will change its state every second, but it has to start counting somewhere. The component that you code in is called "this". This statement changes the state of the component by setting one of its attributes ("seconds") to be equal to the value you choose (in this case "0") to set.

Here is a component named "person" record of a sample. We will add four values ‚Äã‚Äã- name, occupation, hair and age - to the state object in the component:

// basic React statement import

// components that we named "Person", with properties of the state defined in the constructor:

// A render function that defines what will be displayed on the screen:

person

); }} ReactDOM.render (, document.getElementById (’root’));

The last line indicates the code will go through your HTML document and look for the element that has the id "root", and does what you instructed it to do in the render () method above. This is why you should have the following added to the document body of your index.html file:

This HTML tag has the entire ID attribute at "root". This is where all you asked the component to render will go.

You can reference the state object by typing this.state.attributeName. This will replace "attributename" with the defined attributes.

Here is an example that uses the properties we defined earlier to create the state of our "Person" component . We will use the form {} this.state.attribute to call the attribute value of the component in which we are writing our code.

In this case, we will copy the following method into our "person" component, which will be this that "that" refers. The screen displays the value belonging to the attribute we wrote. Replace the render () declaration of our "Person" component in the last example with this:

{this.state.name}

{{this.state.name} is that. state.occupation} and it is {} this.state.age years Currently her hair is {this.state.hair}

) .. }

The screen displays a header with the value of the "name" attribute, which is Tom Hiddleston. Below , this will be a paragraph that states, "Tom Hiddleston is an actor and he is 40 years old." Her hair is black. "

The

.setState () uses this.setState () to update the values of a status object. This method takes one argument, which will be a new attribute and its value: {attribute: "new value" }. methods are used to do this because React will call the render method behind the scenes so that your new content is always displayed.

Here is an example using our "person" component from before. Add the following code to the "person" code component we saw at the beginning of the article.

  1. Add an occupation modification function to "person" in the body class . This function will call the .setState method. the attribute written here is the occupancy attribute and we write the value we want to change. reminder "person" component code that job is currently
    job = () => {this.setState ({occupation "model"});} "actor".
  2. Insert the following keycode inside the .render () main method a>

    multiple .setState () are not processed in order, so this may present more about these - conflict you should be aware of . This is important because your code is getting over big and there are more lines and more React written states, you have to think about how they work together.This can be difficult to do, especially with React being asynchronous in nature, where you can’t predict who is controlling your setstate methods going to be processed.

    Do you have to use state?

    The answer is simple: no, you shouldn’t. There are two main types of components: React components functional and class components. Stateless components are functional components. By mistake there was no state in a component - you have to create it yourself

    Stateless, however, your application will not be very interactive .. Even if you opt for setting state work , it is advisable not to be overzealous in using it. This is because excessive use of state in components can make it difficult to predict possible conflicts due to the complexity.

    Conclusion

    State and Props are two important objects to keep in mind when using React. State tracks the current state of a component as it changes based on updated data that defines a state. The accessories, on the other hand, don’t change. A simple rule of thumb which one to use is to ask, "Are the attributes of this component being changed at any time?" ". If the answer is "yes" then you want to use the status category, as Props is immutable.

    Before you fully immerse yourself in React , make sure you are familiar with JavaScript ES6, like React VERSATILE of its functions, such as direction functions, classes and types of variables. In addition, changes were made to the React use what concerns the state and other functions without having to write a class in the recent release day React, reaction hooks .

    Javascript State __del__: Questions

    How can I make a time delay in Python?

    5 answers

    I would like to know how to put a time delay in a Python script.

    2973

    Answer #1

    import time
    time.sleep(5)   # Delays for 5 seconds. You can also use a float value.
    

    Here is another example where something is run approximately once a minute:

    import time
    while True:
        print("This prints once a minute.")
        time.sleep(60) # Delay for 1 minute (60 seconds).
    

    2973

    Answer #2

    You can use the sleep() function in the time module. It can take a float argument for sub-second resolution.

    from time import sleep
    sleep(0.1) # Time in seconds
    

    How to delete a file or folder in Python?

    5 answers

    How do I delete a file or folder in Python?

    2639

    Answer #1


    Path objects from the Python 3.4+ pathlib module also expose these instance methods:

    Javascript State __dict__: Questions

    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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