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Python.Engineering Wiki Javascript Card To List

Javascript Card To List

exp | filter | find | JavaScript | log | repeat | sin
Michael Zippo Michael Zippo 04.11.2021

Programmers like to reduce repetition in their code. The more you repeat your code, the less manageable it is and the slower your programs will run. That is why there are different methods in JavaScript to iterate through datasets.

One of these methods is called the map () method. This method iterates through an existing array and performs a specific function on all the elements of that array.

In this guide we will talk about how to use the JavaScript map () function. We’ll take a look at three common examples of the map () function in action to get you started.

What is the map function ?

The JavaScript map () method calls a specific function on each element of an array. The result of this function is then moved into its own table.

For example, suppose you want to multiply each element in an array by two. You can do this by creating a function that multiplies each element in the array by two and moving that function into a map () method.

The syntax of the map () function is:

function is the callback function that will be executed on each element of the array. To learn more about functions, you can read our ultimate guide to JavaScript functions. thisValue is the default value that will be stored in the function’s this variable. By default, this is not defined.

The map method creates a new array based on the results of the callback function.

The () card has several uses. The most common is to call a function on a list of array elements. An example would be to multiply each number in a list of numbers or find the length of each string in a list of strings.

You will also find the function used to display lists in JavaScript libraries like React or Vue.js.

Calling a function with Map

The map () method allows you to perform a repetitive task on each item in a list, which makes it useful in many cases.

Let’s say you have a store cookie and you will increase the price of each cookie by 5%. Instead of calculating all the new prices individually, you can use the map () method.

Here is the code you would use:

Our code returns:

In this example, we have increased the price of each cookie by 5%. First, we have declared a cookie price list in the "cookiePrices" variable. So we used the map () method with a function that calculates a 5% price increase for each cookie.

"cookie * 1.05" calculated the percentage increase, then we returned that increase, rounded to the nearest two decimal places. Finally, we printed the new table to the console using the console.log () method.

We could also move our function to its function to create our own code. readable:

Our code returns the same values ‚Äã‚Äãas before, but is more readable:

Modify elements to an array

The map () method is often used to modify the elements of an array. Suppose our cookie store is developing a loyalty program for its customers. Each cookie you buy earns you 10 points, and if you earn 100 points you will earn a free cookie.

The following program would allow us to do this:

Our code returns:

In this example, we’ve calculated the total number of points each customer should receive based on the number of cookies they’ve purchased.

First, we have declared a series of objects called " clients ". This list stores the names of our customers and the number of cookies they have purchased using key: value pairs .

So we have declared a function called calculateLoyaltyPoints (). This function adds a new item to each customer item called "points" which is calculated by multiplying the number of cookies each the customer has purchased before 10am.

We use the map () method to scroll through our list of customers and apply our calculateLoyaltyPoints () function Finally, we print the revised list of clients. This list now reflects our " point ‚" value that we added to each customer’s entry.

Render a list using a library

The map () method is commonly used in JavaScript libraries such as React. The purpose of this method is to render the items in a list. Let’s see an example of map () in React.

Say you want to view a list of cookies that our store sells on our website. We could do it using this code:

); const root = document.getElementById ("root"); ReactDOM.render (, root);

This code creates a component in React that renders a list. Each item in the list is contained in a

  • tag which is displayed using map (). map () creates an individual
  • tag for each item in the list, and the "i" tag variable assigns each item in the list a unique key.

    Map Methods vs. Iterator

    Map is an example of an iterator method in JavaScript. These methods allow you to cycle through all the items in a list and perform certain actions.

    When deciding to use the map () function, it is a good idea to first ask yourself if another iterator method would be better. Make sure you choose the right one tool for the job.

    Here are the other iterator methods that exist in JavaScript:

    You might be thinking to yourself, "The map looks a lot like the forEach () method. Are they equal? ‚Äã‚Äã" That’s a good question. There is a subtle difference between these two methods.

    The map () function walks through a list, modifying each element of the list and returns a new list, while the forEach () function iterates through a list and, as a side effect, performs certain operations on the list.

    The map () function is best used if you need to call a function on every element of a list, declare a list component in a framework like React, or modify the contents of a list.



    Conclusion

    The map () method is useful for performing repetitive tasks several times. It is e also useful for declaring components in React such as lists.

    In this tutorial, we have analyzed three main uses of the map () method. You are now ready to start using the map () JavaScript method like an expert developer!

    Javascript Card To List 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 Card To List filter: Questions

    filter

    List comprehension vs. lambda + filter

    5 answers

    I happened to find myself having a basic filtering need: I have a list and I have to filter it by an attribute of the items.

    My code looked like this:

    my_list = [x for x in my_list if x.attribute == value]

    But then I thought, wouldn"t it be better to write it like this?

    my_list = filter(lambda x: x.attribute == value, my_list)

    It"s more readable, and if needed for performance the lambda could be taken out to gain something.

    Question is: are there any caveats in using the second way? Any performance difference? Am I missing the Pythonic Way‚Ñ¢ entirely and should do it in yet another way (such as using itemgetter instead of the lambda)?

    957

    Answer #1

    It is strange how much beauty varies for different people. I find the list comprehension much clearer than filter+lambda, but use whichever you find easier.

    There are two things that may slow down your use of filter.

    The first is the function call overhead: as soon as you use a Python function (whether created by def or lambda) it is likely that filter will be slower than the list comprehension. It almost certainly is not enough to matter, and you shouldn"t think much about performance until you"ve timed your code and found it to be a bottleneck, but the difference will be there.

    The other overhead that might apply is that the lambda is being forced to access a scoped variable (value). That is slower than accessing a local variable and in Python 2.x the list comprehension only accesses local variables. If you are using Python 3.x the list comprehension runs in a separate function so it will also be accessing value through a closure and this difference won"t apply.

    The other option to consider is to use a generator instead of a list comprehension:

    def filterbyvalue(seq, value):
   for el in seq:
       if el.attribute==value: yield el

    Then in your main code (which is where readability really matters) you"ve replaced both list comprehension and filter with a hopefully meaningful function name.

    957

    Answer #2

    This is a somewhat religious issue in Python. Even though Guido considered removing map, filter and reduce from Python 3, there was enough of a backlash that in the end only reduce was moved from built-ins to functools.reduce.

    Personally I find list comprehensions easier to read. It is more explicit what is happening from the expression [i for i in list if i.attribute == value] as all the behaviour is on the surface not inside the filter function.

    I would not worry too much about the performance difference between the two approaches as it is marginal. I would really only optimise this if it proved to be the bottleneck in your application which is unlikely.

    Also since the BDFL wanted filter gone from the language then surely that automatically makes list comprehensions more Pythonic ;-)

    filter

    How do I do a not equal in Django queryset filtering?

    5 answers

    MikeN By MikeN

    In Django model QuerySets, I see that there is a __gt and __lt for comparative values, but is there a __ne or != (not equals)? I want to filter out using a not equals. For example, for

    Model:
    bool a;
    int x;

    I want to do

    results = Model.objects.exclude(a=True, x!=5)

    The != is not correct syntax. I also tried __ne.

    I ended up using:

    results = Model.objects.exclude(a=True, x__lt=5).exclude(a=True, x__gt=5)
    784

    Answer #1

    You can use Q objects for this. They can be negated with the ~ operator and combined much like normal Python expressions:

    from myapp.models import Entry
from django.db.models import Q

Entry.objects.filter(~Q(id=3))

    will return all entries except the one(s) with 3 as their ID:

    [<Entry: Entry object>, <Entry: Entry object>, <Entry: Entry object>, ...]

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