Loop Javascript

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do… while loops can be used to execute a block of code once and will continue to execute if an instruction continues to evaluate as true. The while and do ... while loops are useful if you want to execute a block of code when a certain condition is met.

Loops are a fundamental feature of programming and are used to automate similar tasks. For example, suppose you want to print each name in a list of students. You can create a loop to automate the task.

JavaScript provides two main types of loops that can be used to perform a task that is repetitive multiple times. The first type, called a for loop, executes a statement a number of times. The second type is the while loop, which is executed depending on whether a condition is true or not.

In this tutorial, we’ll cover the basics of the while loop in JavaScript. We’ll explore the syntax of while loops, how to create do‚Ķ while loops, and explore some examples of these loops in action.

JavaScript While Loops

A JavaScript while loop executes a block of code while a condition evaluates to true. While loops stop executing when their condition is false. A while loop allows you to repeat a block of code multiple times without copying and pasting the code.

while loops are often used if you want to execute unspecified code multiple times. On the other hand, JavaScript for loop are used if you already know or can calculate how many times your loop should run.

Here is the syntax of a while loop in JavaScript:

The while statement is very similar to a JavaScript for .

You can create an infinite loop by specifying a condition that will always evaluate as true. Make sure your condition can be evaluated as false unless you want the loop to continue until you manually abort planning.

While JavaScript Loop Example

Let’s say we’re having an exclusive party. We want to invite only 40 people to our party at one time.

There are already 35 people at our party. Every time we invite someone, we want to calculate how many more people we can invite before we reach the limit. Here is an example of a program that uses an while loop to perform this calculation:

Our code returns the following:

In the first two lines of our code, we declare two variables. The limit variable is used to specify the maximum number of people that can join our group. The current_attendees JavaScript variable keeps track of how many people are currently attending the party.

Let’s create a while loop that runs until the number of current participants is less than the limit we set.

Next, our program adds 1 to the number of current participants and calculates the number of spaces remaining by subtracting current_attendees from limit. Finally, our program prints a statement to the console telling our group leader how many spaces are left.

Our cycle then lasted as long as there were less than 40 people in our group. Each time the loop ran, our current_attendees increased by 1. As soon as our group reached 40 participants, our loop stopped working.

while loops: Another example

Let’s use another example to illustrate the while loop in action. In the following example, we create a while loop that prints each name in our VIP party table:

Our code returns:

Our loop iterates through the vips array until our counter variable is less than the length of the JavaScript table .

JavaScript do… while Loop

A JavaScript do… while loop executes a statement once, then checks if a condition is true. If the condition is true, the loop will execute again. Otherwise, the code no longer works.

There is a big difference between a while loop and a do… while loop. Do ... while loops are executed at least once even if the specified condition never evaluates to true. In contrast, while loops are only executed if their condition is true.

Here is the syntax of a do… while loop in JavaScript:

The code in our do… while loop will always be executed at least once.

JavaScript do… while loop example

Let’s use an example to illustrate how the do‚Ķ while loop works. Let’s say we create a puzzle. We want our program to keep asking a user to guess a number until they guess the correct number.

Here is an example of a program that we could use to create this puzzle:

Our program will ask the user to enter a number between 1 and 10 as long as guessed_number is not equal to < em> number_to_guess. But our program will execute the contents of our do statement before evaluating the conditions for the first time.

When our condition is false, our loop will stop executing.

Conclusion

While loops can be used to execute a repetitive block of code while a statement evaluates to true. Do… while loops can be used to execute a block of code once. They will continue to function if an instruction continues to return true.

The while and do… while loops are useful if you want to execute a block of code when a certain condition is met.

In this tutorial, we have explained how to create a while and do‚Ķ while loop in JavaScript. We’ve also explored a few examples of these loops in action to illustrate where they can be useful. Now you have the information you need to use the while loop like a JavaScript expert !

For tips on the best JavaScript learning resources and courses, see our Article on how to learn JavaScript .

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

Loop Javascript abort: Questions

How do I abort the execution of a Python script?

1 answers

Ray Vega By Ray Vega

I have a simple Python script that I want to stop executing if a condition is met.

For example:

done = True
if done:
    # quit/stop/exit
else:
    # do other stuff

Essentially, I am looking for something that behaves equivalently to the "return" keyword in the body of a function which allows the flow of the code to exit the function and not execute the remaining code.

180

Answer #1

To exit a script you can use,

import sys
sys.exit()

You can also provide an exit status value, usually an integer.

import sys
sys.exit(0)

Exits with zero, which is generally interpreted as success. Non-zero codes are usually treated as errors. The default is to exit with zero.

import sys
sys.exit("aa! errors!")

Prints "aa! errors!" and exits with a status code of 1.

There is also an _exit() function in the os module. The sys.exit() function raises a SystemExit exception to exit the program, so try statements and cleanup code can execute. The os._exit() version doesn"t do this. It just ends the program without doing any cleanup or flushing output buffers, so it shouldn"t normally be used.

The Python docs indicate that os._exit() is the normal way to end a child process created with a call to os.fork(), so it does have a use in certain circumstances.

exp

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

5 answers

Carl Meyer By Carl Meyer

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

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

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

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

5839

Answer #1

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

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

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

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

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

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

    and now:

    z = merge_two_dicts(x, y)
    

Explanation

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

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

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

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

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

z = {**x, **y}

And it is indeed a single expression.

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

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

and now:

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

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

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

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

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

Not yet on Python 3.5, but want a single expression

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

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

and then you have a single expression:

z = merge_two_dicts(x, y)

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

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

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

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

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

Critiques of Other Answers

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

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

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

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

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

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

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

This example demonstrates what happens when values are unhashable:

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

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

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

Another hack you should not use:

z = dict(x, **y)

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

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

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

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

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

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

and

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

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

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

instead of

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

Response to comments

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

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

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

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

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

More comments:

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

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

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

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

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

from copy import deepcopy

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

Usage:

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

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

Less Performant But Correct Ad-hocs

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

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

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

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

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

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

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

Performance Analysis

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

from timeit import repeat
from itertools import chain

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

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

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

In Python 3.8.1, NixOS:

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

Resources on Dictionaries

5839

Answer #2

In your case, what you can do is:

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

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

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

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

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

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

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

5839

Answer #3

An alternative:

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

We hope this article has helped you to resolve the problem. Apart from Loop Javascript, check other abort-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:



Manuel Porretti

Berlin | 2022-11-30

Thanks for explaining! I was stuck with Loop Javascript for some hours, finally got it done 🤗. Will get back tomorrow with feedback

Ken Jackson

San Francisco | 2022-11-30

JavaScript is always a bit confusing 😭 Loop Javascript is not the only problem I encountered. Will use it in my bachelor thesis

Davies Robinson

Munchen | 2022-11-30

Simply put and clear. Thank you for sharing. Loop Javascript and other issues with join was always my weak point 😁. Checked yesterday, it works!

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