Javascript Copies Arrays Of Unreferenced Objects

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When working with arrays in Java, you may decide to make a copy of an array. For example, if you are running a cafe and want to create a seasonal menu , you can create a copy of the original menu on which to base the new menu.

In Java there are several ways you can copy an array. This tutorial will explore four common methods of copying tables and discuss how they work row by row. After reading this tutorial, you will be a master at copying tables in Java.

Java Arrays

In Java, an array is a container that holds values ‚Äã‚Äãthat contain a single type. For example, a table can be used to store a list of books or a list of scores that players have earned in a game of darts.

Arrays are useful when you want to work with many values ‚Äã‚Äãbecause you can store them in a collection. This allows you to condense your code and simultaneously run the same methods on the same values.

The word Let we want to create an array that stores the coffees sold in our coffeeshop. We could do it using this code:

String [] = coffees {"Espresso", "Mocha", "Latte", "cappuccino", "Versare Over", "Flat white" };

In this example, we declare an array called cafes which stores string values . Our array contains six values.

Each element of an array is assigned an index number, starting with 0, which can be used to refer to individual elements in an array.

Now that we have explored the basics of Java arrays, we can discuss the methods you can use to copy the contents of your table.

Copy the array using the assignment operator

One of the most commonly used cloning methods to copy an array is to use the assignment operator.

The assignment operator is used to assign a value to an array. Using the assignment operator, we can assign the contents of an existing array to a new variable, which will create a copy of our existing array.

Let go back to the coffeeshop. Let’s say we want to create a copy of the cafes array that we’ll be basing our summer coffee menu on. We could use this code to create a copy of the array:

Our code returns:

Espresso, Moka, Latte, Cappuccino, Pour over, flat white,

Break from Descended our code. On the first line of code in our CopyAssignment class, we declare an array called cafes which stores our standard cafe menu.

On the next line, we use the assignment operator to assign the value of cafes to a new array called summer_coffees. Next, we create a "for-each‚" loop that passes each summer_coffees array element and prints to the console.

There is a downside to using this method: if you change the elements of one array, the other array will also be changed. So if we changed the value of Latte Summer Latte in our summer_coffee list, our coffee list would also be be changed

loop to copy arrays

the first approach we discussed to copy an array -. by using the assignment operator - creates what is called a copy shallow. This is the case because we have assigned an existing array object to a new one, which means that when we change any object they will both be changed - . the two objects are related

However, we often need to create a deep copy . Deep Copies copy the values ‚Äã‚Äãof an existing object and create a new array object. When creating a full copy, you can edit your new table without affecting the original.

One approach that can be used to create a full copy is to create a loop through the contents of an array and create a new array.

Suppose we want to create a full copy of our cafes table called summer_coffees. This should be a thorough copy as we intend to modify the contents of the summer_coffees table to reflect the new coffees we will be offering in the summer months.

Here is the code we would use to create a deep copy using a loop:

When we run our code, the output is as follows:

[Espresso, Mocha, Latte, Cappuccino, Pour over, Flat White]

As you can see, our code created a copy of our original array. We explain how it works step by step:

  1. We import java.util.Arrays which includes toString () method, we will use it to print our array on the console at the end of the example.
  1. We declare an array called cafes which stores the coffee list on our standard menu.
  2. We initialize an array called summer_coffees which can store six values.
  3. We use a loop to iterate through each item in the cafe list.
  4. Whenever the loop is executed, the element with the value of index i in summer_coffees will be assigned the element with the code index value <> i in cafes.
  5. We use Arrays.toString () to convert summer_coffees to a string, then print the new array with our copied items to the console.

Using Java copyOfRange () methods

The Java copyOfRange () method to copy Arrays .copyOfRange () is part of the java.util.Arrays class Here is the syntax for the copyOfRange () method:

We are going to analyze the syntax of the copyOfRange () method:

  1. DataType is the data type that the new array is will store.
  2. newArray is the name of the new array.
  3. oldArray is the array whose values ‚Äã‚Äãyou want to copy into the newArray directory.
  4. indexPos is where the copy operation should start in the oldArray.
  5. length is the number of values ‚Äã‚Äãthat should be copied from oldArray newArray.
  6. Let’s take an example to illustrate the copyOfRange () method in action. Suppose we want to make a copy of our cafes array from earlier. We could do that using this code:

    Our code returns:

    summer coffees: [Espresso, Moka, Latte, Cappuccino, Pour over, flat white]

    break down our code:

    1. We import java.util.Arrays library which stores copyOfRange () methods and toString () which we will use in our example.
    2. we declare an array called cafes which stores the coffees in our standard menu.
    3. We declare an array named summer_coffees and use the copyOfRange () method to create a copy of the cafes array. The parameters that we specify are as follows:
      1. cafes is the name of the table that we want to copy
      2. indicates that we. want to copy the values starting at the position of the index of 0 coffee table .
      3. coffees.length indicates that we want to copy each value into the. list
      4. We print " summer coffees:" . followed by the resulting array named summer_coffees on the console

      Now we’ve created a copy of our cafe list called summer_coffees .

      Using Java Arraycopy () Methods

      The arraycopy () method is used to copy data from one network to another network. The arraycopy () method is part of the system class and includes a set of options that allow you to customize the copy you create of an existing array.

      Here is the syntax of the arraycopy () method:

      System.arraycopy (sourceArray, StartingPos, newArray, newArrayStartingPos, length);

      Breaking this Let down method. copyarray () takes five parameters:

      • sourceArray is the name of the array you want to copy
      • startingPos. is the location of the index from which you want to start copying values ‚Äã‚Äãinto the source_array.
      • newArray is the name of the new array in which the values ‚Äã‚Äãwill be copied.
      • newArrayStartingPos is the index position where the copied values ‚Äã‚Äãare to be added.
      • length is the number of elements you want to copy to the new_array directory.

      Back to the coffeeshop. Suppose we want to copy each value in our cafes array into a new array called summer_coffees. We could do that using this code:

      Our code returns:

summer coffees: [espresso, mocha, latte, cappuccino, Pour over, flat white]

break down our code step by step -Step:

  1. we import the package java.util.Arrays at the start of our program, including the toString () method that we will use to print the copy of the array we create at the end of our program.
  2. We declare an array called coffe es which stores the coffees in our standard menu.
  3. We initialize an array called summer_coffees which will contain 6 values.
  4. We use arraycopy () to create a copy of our cafes array. Here are the parameters we specify.
    1. cafes is the array we want to copy
    2. is the position where we want our copy to start in the cafes array.
    3. summer_coffees is the array where we want the copied values ‚Äã‚Äãto be appended.
    4. is where we want the copied values ‚Äã‚Äãto start being added to the summer_coffees array.
    5. coffees.length is the number of array elements we want to copy. In this case, using coffees.length allows us to copy each item from the cafes list.
    6. We print a message that says " summer coffees:"., Followed by the list of summer coffees we have created
    7. Conclusion

      Copying an array is a common operation when working with lists. This tutorial explored four ways an array can be copied in Java.

      First , we discussed how to create a shallow copy depth with the assignment operator, so we proceeded to explain how to make a full copy copy using a loop. So we explored how to use the copyOfRange () method to create a copy of an array and how the arraycopy () system method is used to copy an array.

      You are now ready to start copying arrays in Java as one! pro

      Javascript Copies Arrays Of Unreferenced Objects 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 Copies Arrays Of Unreferenced Objects JavaScript: Questions

      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

      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.

      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.

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