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How to learn Swift

Are you interested in creating your own iO S or Mac application? If so, you will need to learn a programming language: Swift .

This guide will explore the best way to learn Swift online. In addition, it will give you a number of tips to help you learn to program in Swift.

What is Swift ?

Swift is a programming language that lets you create apps for Mac, iPhone, Apple Watch, and iPad. Swift was released in 2014 by Apple.

Swift is the foundation for most of the major apps you’ll use on Apple devices. Spotify, Instagram, and YouTube are three of the many names that use Swift to power their iOS apps.

Why should you learn Swift ?

Swift is easy to learn

The Swift programming language was created to be efficient and easy to use.

Swift uses simple and expressive syntax that is easy to understand, even if you have no previous experience with coding. In fact, Swift was designed to be the first programming language to learn, according to Apple. .

Swift developers are in high demand

According to Apple, the company created 2 000 000 jobs in all 50 US states. It is estimated that 1,530,000 are attributable to the App Store ecosystem. This means that the jobs were created thanks to companies hiring developers to build apps for Apple devices.

In terms of the job market for Swift developers, the outlook is bright. According to Glassdoor, there are more than 3,400 open jobs available in iOS development in the US. This work only includes part of the mobile application development work.

This shows that Swift is, in fact, a highly sought after skill and that many employers are actively seeking .

Swift is the future of iOS

Swift may not have been released until 2014, but since its launch it has become one of the most popular programming languages ‚Äã‚Äãin the world.

According to TIOBE Index , Swift is the 12th most sought after programming language. The TIOBE index tracks the popularity of major programming languages ‚Äã‚Äãusing search engine traffic.

What is Swift used for ?

Swift is a programming language used to develop applications for Apple devices such as the iPhone and the Apple Watch . It is a modern replacement for the C-based languages ‚Äã‚Äãpreviously used in Apple development.

Most of the apps you see on Apple’s App Store today were written in Swift. In fact, most apps for Apple devices today are written in Swift, with the exception of the old ones. This is because Swift is the recommended language to use for Apple development.

Is Swift difficult to learn?

Swift is not a difficult programming language to learn as long as you invest the right time. Many programmers find it easy to get started with Swift due to its syntax and beginner’s resources.

In fact, Apple has detailed tutorials on how to code in Swift on their website and Swift documentation site . These tutorials give you all the knowledge you need to get started with Swift. Finding resources is therefore not a problem.

Swift uses clear and expressive syntax. The architects of the language wanted Swift to be easy to read and write. Therefore, Swift is a great place to start if you want to learn programming. You shouldn’t have a hard time reading Swift code once you have mastered the basics.

How long does it take to learn Swift ?

It takes about a month or two to develop a basic understanding of Swift, assuming you spend about an hour through day to study. Se studi part-time o full-time, puoi imparare i fondamenti di Swift in un periodo di tempo pi√π breve.

Puoi iniziare a lavorare alla tua prima applicazione con Swift abbastanza rapidamente grazie a tutti i tutorial dettagliati there outside. To start building apps on your own, you will need a few months of practice.

If you want to become a professional developer of iOS or Apple devices, you have a long way to go. It will take at least six months, maybe even a year, to build the skills to become a professional Swift developer.

As with any programming language, your learning journey will never really end. Swift is always updating and there are so many frameworks out there that you will never run out of material to learn. The time it takes to learn Swift depends on how long it takes you and your willingness to learn the language.

How to learn to program Swift

But how do you really learn programming in Swift? To answer this question, we are going to discuss the two things you need to know. We’ll talk about the basic skills Swift developers need and how you can learn them.

Develop your skills

Before pursuing a career as a Swift developer, you must first master the basics of the language of Swift programming. Although Swift is a relatively modern language, there are already a number of great learning resources available to help beginners learn the language.

To start your journey, you need to focus on learning the basics. But before you even do that, you need to download Xcode , which is Apple’s application for creating software. You should take a look at the Apple Swift documentation . This documentation includes a wide variety of guides that can be helpful as you begin to get started with Swift.

We’ll cover the main topics you should explore when getting started with Swift.

Syntax and variables

The first step in learning a programming language is to know how it works and what rules govern the language. This is where the syntax comes in.

So, once you have learned the basic Swift syntax, you can start exploring variables. Variables are used to store data in a program. Here are the main topics you should explore in this regard. which concerns syntax and variables:

Conditionals and Loops

Conditionals allow programmers to execute a block of code only if and when a certain condition or set of conditions is met. This allows them to control how their program works in different contexts. Loops, on the other hand, allow you to perform repetitive tasks multiple times.

Here are the main topics you should be familiar with when it comes to conditions and loops:

arrays and dictionaries

an array is a type of data that allows you to store a list of values ‚Äã‚Äãin a program. A dictionary is a type of data that allows you to store a list of labels along with a list of values. Data is stored in key-value pairs.

When working in Swift it is essential that you can use these types of data. Here are the main topics you should learn:

Functions

Functions are blocks of code that perform a particular task in a program. For example, a function can add two numbers in a program or force a user to sign in to a mobile application. Here are the main topics you should know about functions:

Classes and Structures

Swift is an object-oriented programming language, which means that it uses classes to define the structure of data in a program. The goal of the lessons is to make your code more readable and to reduce repetitions. Additionally, Swift also uses structures to allow developers to create their own data types.

Here are the main topics you should read when it comes to classes and structures:

These topics form the basis of the Swift programming language. There are frameworks like SwiftUI which allow you to create more detailed user interfaces. None of these frameworks are out of reach once you’ve learned the basics of Swift.

Resources to Learn Swift for Free

There is a large community of resources that can help you learn Swift. But which resources are worth your time? Excellent question.

The answer to this question depends on your learning style. Do you prefer how-to video tutorials or do you enjoy reading guides ? Do you like mentoring-led environments or do you prefer to learn at your own pace ?

Before you start looking for resources, you need to think about the best way to learn. Try to find resources that match your learning approach.

Swift Online Course

Quick tutorials

One of the best ways to learn programming with programming languages ‚Äã‚Äãlike Swift are done via tutorials. You can access Swift courses for beginners, intermediate and advanced, depending on your experience. The following article covers the best Swift tutorials to help you learn quickly and effectively.

The best Swift tutorials for beginners

Tutorials SwiftUI - Apple Developer

Learning Swift through Apple’s development platform is ideal for understanding this programming language as created by Apple. SwiftUI Tutorials is a free course that takes about four to four. five hours of apprentices wise. You will learn the basics of Swift development even if you are new to iOS and Swift.

It contains illustrated videos and tutorials to help you learn while following the steps. This course begins with a quick introduction followed by lessons on the essentials of Swift UI, drawing, animation, app design, layout and framework integration. Apple’s developer website also includes a forum section that includes a strong developer community.

iOS 12 Swift 4.2 - Full iOS Development with Swift Bootcamp - Udemy

Swift is used to write code for Apple’s technology platforms including iOS, macOS, watchOS, tvOS, and iPadOS. The quick bootcamps and courses offer a quick way to fully learn this programming language. This paid course from Udemy is ideal for novice software developers who want to learn iOS development through tutorials.

There is no technical requirement to take this course. It takes around 55 hours and includes videos, articles, tutorials, and other resources. You will learn about building apps, mastering augmented reality apps, and starting an app-based business during this time. This will prepare you to start working as a junior developer.

Complete Beginner’s Quick Tutorial Course - freeCodeCamp

freeCodeCamp is a non-profit organization dedicated to providing interactive online training for aspiring developers. This organization offers a three-hour video tutorial covering variables, constants, data types, functions, properties, classes, IF statements, and parameters. This course is ideal for beginners and available for free on YouTube.

This course will teach you the basics of Swift, including the essential steps of the development process. ’applications, through tutorials. You should consider this course because of its high mark from the hundreds of students who have taken it. It is one of the best learning paths for beginners to learn this powerful programming language.

As a budding app developer, you will learn how Swift works with Swift Data and Playgrounds. You will also adopt a protocol oriented approach. programming mindset while you work on app development on iPhone, Mac or iPad. It takes three to four hours of learning to complete this course.

Free Swift Tutorial - Code Conquest

Code Conquest has created this free course that includes illustrated tutorials to help future developers get started with Swift. The only requirement is a MacBook with the latest version of the operating system to work with the Swift language through this tutorial. However, you will need to further your education to become a professional application developer.

This program will help you on your journey to becoming a Swift developer. You will work alone on several projects using illustrated tutorials. and you will gain practical experience. You will learn how to set up a Swift environment and start your first project as an absolute beginner.

The best advanced Swift tutorials

Setting up AppSync with Amplify - Ray Wenderlich

Tutorials that include videos are a great way to learn Swift, even as a seasoned programmer. This course is designed for experienced programmers interested in learning the integration of BaaS technology with Apple platforms, especially iOS.

The introductory tutorial is free and provides a great opportunity to preview what you’ll be covering in the coming weeks. This course includes many video tutorials that collectively cover concepts such as AWS AppSync, AWS Amplify, and GraphQL.

iOS app development with Swift specialization - Coursera

He is a graduate of the courses offered by the University of Toronto by Coursera. It covers the fundamentals of developing iOS apps using Swift. It takes about five months to complete and includes dozens of tutorials to help you learn design patterns, protocol extensions, and more.

Students who complete the program receive a shareable certificate of completion. During this time, you will learn all about iOS app design and development, how to use XCode and other development tools. You will also be working on projects such as creating your own iOS app or game.

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Learn Fast Or Javascript __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)

How to get an absolute file path in Python

3 answers

izb By izb

Given a path such as "mydir/myfile.txt", how do I find the file"s absolute path relative to the current working directory in Python? E.g. on Windows, I might end up with:

"C:/example/cwd/mydir/myfile.txt"
877

Answer #1

>>> import os
>>> os.path.abspath("mydir/myfile.txt")
"C:/example/cwd/mydir/myfile.txt"

Also works if it is already an absolute path:

>>> import os
>>> os.path.abspath("C:/example/cwd/mydir/myfile.txt")
"C:/example/cwd/mydir/myfile.txt"

How to check if a path is absolute path or relative path in a cross-platform way with Python?

3 answers

UNIX absolute path starts with "/", whereas Windows starts with alphabet "C:" or "". Does python have a standard function to check if a path is absolute or relative?

175

Answer #1

os.path.isabs returns True if the path is absolute, False if not. The documentation says it works in windows (I can confirm it works in Linux personally).

os.path.isabs(my_path)

How to join absolute and relative urls?

3 answers

I have two urls:

url1 = "http://127.0.0.1/test1/test2/test3/test5.xml"
url2 = "../../test4/test6.xml"

How can I get an absolute url for url2?

139

Answer #1

You should use urlparse.urljoin :

>>> import urlparse
>>> urlparse.urljoin(url1, url2)
"http://127.0.0.1/test1/test4/test6.xml"

With Python 3 (where urlparse is renamed to urllib.parse) you could use it as follow:

>>> import urllib.parse
>>> urllib.parse.urljoin(url1, url2)
"http://127.0.0.1/test1/test4/test6.xml"

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