Lambda Expression Javascript

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How to use lambda expressions in Java

Have you ever thought that some methods could fit on one line, if only there was no Java you would need to use detailed syntax to declare a method? You are not the only one.

Methods are an extremely useful function in programming.

A method is a block of code that does something specific to your program. They are used because once a method is defined it can be called multiple times. This helps to reduce redundancy in your code, which makes it easier to maintain your code.

In this guide, we are going to talk about lambda expressions in Java. This is a special type of method that you can use called a lambda expression, which implements a functional interface. We are going to talk about how they work and why they are used.

What is a functional interface?

Before talking about lambda expressions, you need to understand functional interfaces. These are interfaces that only contain an abstract method. The only abstract method contained in a functional interface indicates the purpose of the interface.

Let’s take this example:.

We’ve created an interface called CalculateThree. This interface has only one method, multiply_by_three, which means it is a functional interface.

What is a Java lambda expression?

A lambda expression is an anonymous process. It is used to implement a method which has been defined within a functional interface. Lambda expressions are sometimes referred to as anonymous methods because they do not have a name

Lambda expressions make use of the arrow operator, which separates them into two sections:

the side left contains the parameters used by the expression; the right side contains the code that will be executed when the Lambda expression is executed. Lambda functions can accept a single parameter or multiple parameters

How to use a lambda expression

Let’s create a program that takes a number entered by a user and multiplies by three. start by writing code that accepts a number from a user:

When we run this code, our user will be asked to enter a number to multiply. Our program will read this number from the "user_number‚" variable. You can learn more about how the Scanner class works by reading our Java Scanner tutorial .

So, let’s go g to define a lambda expression to multiply the number by three. Paste the following code above your main method:

This code defines the interface that our lambda expression will refer to. Then add the following code to the bottom of the main method in the class:

We used the CalculateThree interface to declare a variable called "multiply". This stores the code for our lambda expression. The lambda expression is related to multiply_by_three in our CalculateThree interface because multiply_by_three is the only function in the interface.

We point notation therefore used to call this function:

dot notation lets you specify the name of the class or interface you want to reference, followed by a dot, the name of the method you want to access. In this case, we want to access the multiply_by_three in the "Multiply" Interface

Our last code looks like this:

Let’s run our code and insert the number 3 to multiply:

Our code multiplied with successful e the number we entered by 3. This action was carried out by using the lambda expression that we have defined

How to use an expression lambda block

expressions lambda can appear in two forms:. expressions or using blocks. The lambda block syntax is used when the code on the right side of the arrow spans multiple lines.

Block syntax is where you place code in the right side of a lambda expression with braces. ({})

Let’s take this example:

We have declared a lambda expression called send_greeting. This expression refers to the GreetUser interface that we defined at the start of our program. Our lambda expression does two things. It:

Since this requires two lines of code, we used syntax block. The code on the right side of our lambda expression is enclosed in braces.

Conclusion

Lambda expressions are anonymous methods that implement a functional interface. These functions are sometimes referred to as classes or anonymous functions because they do not have a name and do not operate on their own

Are you up for a challenge? Write a lambda expression that checks whether a number is same or not. In this case, "X is same" should be printed on the console, where X is the number being tested; otherwise, "X is odd" must be printed on the console.

You are now ready to start working with lambda expressions in Java like an expert!

Lambda Expression Javascript 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)

Lambda Expression Javascript insert: Questions

insert

How to insert newlines on argparse help text?

5 answers

I"m using argparse in Python 2.7 for parsing input options. One of my options is a multiple choice. I want to make a list in its help text, e.g.

from argparse import ArgumentParser

parser = ArgumentParser(description="test")

parser.add_argument("-g", choices=["a", "b", "g", "d", "e"], default="a",
    help="Some option, where
"
         " a = alpha
"
         " b = beta
"
         " g = gamma
"
         " d = delta
"
         " e = epsilon")

parser.parse_args()

However, argparse strips all newlines and consecutive spaces. The result looks like

~/Downloads:52$ python2.7 x.py -h
usage: x.py [-h] [-g {a,b,g,d,e}]

test

optional arguments:
  -h, --help      show this help message and exit
  -g {a,b,g,d,e}  Some option, where a = alpha b = beta g = gamma d = delta e
                  = epsilon

How to insert newlines in the help text?

406

Answer #1

Try using RawTextHelpFormatter:

from argparse import RawTextHelpFormatter
parser = ArgumentParser(description="test", formatter_class=RawTextHelpFormatter)

insert

Is a Python list guaranteed to have its elements stay in the order they are inserted in?

5 answers

If I have the following Python code

>>> x = []
>>> x = x + [1]
>>> x = x + [2]
>>> x = x + [3]
>>> x
[1, 2, 3]

Will x be guaranteed to always be [1,2,3], or are other orderings of the interim elements possible?

366

Answer #1

Yes, the order of elements in a python list is persistent.

insert

Inserting image into IPython notebook markdown

5 answers

I am starting to depend heavily on the IPython notebook app to develop and document algorithms. It is awesome; but there is something that seems like it should be possible, but I can"t figure out how to do it:

I would like to insert a local image into my (local) IPython notebook markdown to aid in documenting an algorithm. I know enough to add something like <img src="image.png"> to the markdown, but that is about as far as my knowledge goes. I assume I could put the image in the directory represented by 127.0.0.1:8888 (or some subdirectory) to be able to access it, but I can"t figure out where that directory is. (I"m working on a mac.) So, is it possible to do what I"m trying to do without too much trouble?

277

Answer #1

Most of the answers given so far go in the wrong direction, suggesting to load additional libraries and use the code instead of markup. In Ipython/Jupyter Notebooks it is very simple. Make sure the cell is indeed in markup and to display a image use:

![alt text](imagename.png "Title")

Further advantage compared to the other methods proposed is that you can display all common file formats including jpg, png, and gif (animations).

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