メモリーズオフ-Innocent Fille- is Now Available on Steam!

「メモリーズオフ」シリーズ史上最もライトで最もシリアスな二面性のあるシナリオが、切なくて儚い、ただ一つの想いを選ぶ“二者択一の恋物語”(オルタナティブ・ラヴストーリー)へとプレイヤーを誘います。

Bethesda Softworks and id Software have canceled their annual QuakeCon fan event, announcing over Twitter that uncertainties involving the ongoing COVID-19 coronavirus pandemic make planning this year’s event too difficult to proceed with the show.

While QuakeCon is far from the first event to be called off due to the spread of COVID-19, the event makes for one of the earlier cancellations so far as QuakeCon 25 was originally scheduled for this August.

“In recent weeks we have spent a lot of time discussing how we might still move forward with QuakeCon this year, particularly given that it’s the 25th year of the event. However, with all the logistical challenges and uncertainties we currently face due to the COVID-19 pandemic, we have made the difficult decision to cancel this year’s QuakeCon,” reads a portion of the statement shared to Twitter.

“While we don’t know what the state of the pandemic will be this August, we do know it will not be possible to complete the work and planning with partners, vendors, volunteers, and others that is required to make QuakeCon a success.”

The Gamasutra Job Board is the most diverse, active and established board of its kind for the video game industry!

Here is just one of the many, many positions being advertised right now.

Location: Cambridge, Massachusetts​

Fire Hose Games is growing! The studio behind Go Home Dinosaurs, Catlateral Damage, and 20XX is now working on 2 new unannounced games for PC/console that are ambitious departures from our previous titles. We’re looking for talented, passionate developers to expand our team.

Job Description

• Develop sustainable monetization strategies and engaging progression systems for a live game
• Work with design and production to align monetization with the game’s vision and milestones
• Design economic systems within and outside the game
• Balance the game economy and progression systems
• Identify and analyze metrics for adjusting progression and economy systems
• Collaborate with engineering, art, and production

Qualifications

• 2+ years industry experience
• At least 1 shipped title
• Experience developing monetization strategies for free to play games
• Experience designing economic, metagame, and other progression systems
• Understanding of player psychology and motivation for different player segments
• Experience with visualizing and processing user data
• Knowledge of and experience with online games as a service and free to play games

Preferences

• General game design experience
• Basic programming skills
• Experience with agile/Scrum practices
• Experience with rapid prototyping
• Experience with A/B testing during live development
• Proximity to the Greater Boston area is strongly preferred, but not required

Additional Information

Fire Hose Games is located in Cambridge MA, next door to Boston. We pay good salaries and our generous benefits package includes health care, dental, unlimited vacation/sick time, 401(k) with matching, excellent maternity/paternity leave, flexible work hours, relocation costs, and free snacks at the office. Fire Hose hasn’t crunched in years and we don’t intend to start. Our company culture is inclusive, supportive, and fun. We care about diversity and strongly encourage applicants with diverse backgrounds and voices to apply. We hope to hear from you!

Interested? Apply now.

Whether you’re just starting out, looking for something new, or just seeing what’s out there, the Gamasutra Job Board is the place where game developers move ahead in their careers.

Gamasutra’s Job Board is the most diverse, most active, and most established board of its kind in the video game industry, serving companies of all sizes, from indie to triple-A.

Looking for a new job? Get started here. Are you a recruiter looking for talent? Post jobs here.

In this article, you’ll learn the ins and outs of the sorting function in Python. In particular, you’re going to learn how to sort a list of dictionaries in all possible variations. [1] So let’s get started!

How to Sort a List of Dictionaries …

… By Value?

Problem: Given a list of dictionaries. Each dictionary consists of multiple (key, value) pairs. You want to sort them by value of a particular dictionary key (attribute). How do you sort this dictionary?

Minimal Example: Consider the following example where you want to sort a list of salary dictionaries by value of the key 'Alice'.

salaries = [{'Alice': 100000, 'Bob': 24000}, {'Alice': 121000, 'Bob': 48000}, {'Alice': 12000, 'Bob': 66000}] sorted_salaries = # ... Sorting Magic Here ... print(sorted_salaries)

The output should look like this where the salary of Alice determines the order of the dictionaries:

[{'Alice': 12000, 'Bob': 66000},

{'Alice': 100000, 'Bob': 24000},

{'Alice': 121000, 'Bob': 48000}]

Solution: You have two main ways to do this—both are based on defining the key function of Python’s sorting methods. The key function maps each list element (in our case a dictionary) to a single value that can be used as the basis of comparison.

• Use a lambda function as key function to sort the list of dictionaries.
• Use the itemgetter function as key function to sort the list of dictionaries.

Here’s the code of the first option using a lambda function that returns the value of the key 'Alice' from each dictionary:

# Create the dictionary of Bob's and Alice's salary data

salaries = [{'Alice': 100000, 'Bob': 24000}, {'Alice': 121000, 'Bob': 48000}, {'Alice': 12000, 'Bob': 66000}] # Use the sorted() function with key argument to create a new dic.

# Each dictionary list element is "reduced" to the value stored for key 'Alice'

sorted_salaries = sorted(salaries, key=lambda d: d['Alice']) # Print everything to the shell

print(sorted_salaries)

The output is the sorted dictionary. Note that the first dictionary has the smallest salary of Alice and the third dictionary has the largest salary of Alice.

[{'Alice': 12000, 'Bob': 66000}, {'Alice': 100000, 'Bob': 24000}, {'Alice': 121000, 'Bob': 48000}]

Try It Yourself:

You’ll learn about the second way below (where you use the itemgetter function from the operator module).

Related articles on the Finxter blog:

• Sorting Lists in Python
• Lambda Functions
• Dictionaries

… Using Itemgetter?

Same Problem: Given a list of dictionaries. Each dictionary consists of multiple (key, value) pairs. How to sort them by value of a particular dictionary key (attribute)?

Minimal Example: Consider again the following example where you want to sort a list of salary dictionaries by value of the key 'Alice'.

salaries = [{'Alice': 100000, 'Bob': 24000}, {'Alice': 121000, 'Bob': 48000}, {'Alice': 12000, 'Bob': 66000}] sorted_salaries = # ... Sorting Magic Here ... print(sorted_salaries)

The output should look like this where the salary of Alice determines the order of the dictionaries:

[{'Alice': 12000, 'Bob': 66000},

{'Alice': 100000, 'Bob': 24000},

{'Alice': 121000, 'Bob': 48000}]

Solution: Again, you’re going to define a key function. But instead of creating it yourself with the lambda keyword, you’re going to use an existing one. In particular, you’ll use the itemgetter function from the operator module to sort the list of dictionaries.

Here’s the code of the first option using a lambda function that returns the value of the key 'Alice' from each dictionary:

# Import the itemgetter function from the operator module

from operator import itemgetter # Create the dictionary of Bob's and Alice's salary data

salaries = [{'Alice': 100000, 'Bob': 24000}, {'Alice': 121000, 'Bob': 48000}, {'Alice': 12000, 'Bob': 66000}] # Use the sorted() function with key argument to create a new dic.

# Each dictionary list element is "reduced" to the value stored for key 'Alice'

sorted_salaries = sorted(salaries, key=itemgetter('Alice')) # Print everything to the shell

print(sorted_salaries)

The output is the sorted dictionary. Note that the first dictionary has the smallest salary of Alice and the third dictionary has the largest salary of Alice.

[{'Alice': 12000, 'Bob': 66000}, {'Alice': 100000, 'Bob': 24000}, {'Alice': 121000, 'Bob': 48000}]

Now, you know how to sort a list of dictionaries by value. But what if you want to sort by key?

… By Key?

Problem: Given a list of dictionaries. Each dictionary consists of multiple (key, value) pairs. How to sort them by a particular key (attribute)?

Solution: It doesn’t make sense. If you assume that all dictionaries have that same particular key, you cannot really sort because all dictionaries have the same key. If there’s no tie-breaker, it’s impossible to do. But even if there’s a tie-breaker (e.g. the value associated to the key), it doesn’t make sense because you could have sorted by value in the first place.

… By Multiple Keys?

Problem: Given a list of dictionaries. How do you sort by multiple key value pairs?

Minimal Example: Consider the following example where you want to sort a list of database entries (e.g. each stored as a dictionary) by value of the key 'username'. If the 'username' is the same, you use the 'joined' value as a tiebreaker. If the 'joined' date is the same, you use the 'age' as a tie breaker.

db = [{'username': 'Alice', 'joined': 2020, 'age': 23}, {'username': 'Bob', 'joined': 2018, 'age': 19}, {'username': 'Alice', 'joined': 2020, 'age': 31}] sorted_salaries = # ... Sorting Magic Here ... print(sorted_salaries)

The output should look like this where the salary of Alice determines the order of the dictionaries:

[{'username': 'Alice', 'joined': 2020, 'age': 23}, {'username': 'Alice', 'joined': 2020, 'age': 31}, {'username': 'Bob', 'joined': 2018, 'age': 19}]

Solution: You’re going to define a key function with the lambda keyword. But instead of returning a single value to a given key, you return a tuple—one entry per value that should be considered. For example, the database entry {'username': 'Alice', 'joined': 2020, 'age': 23} is mapped to ('Alice', 2020, 23). This ensures that two tuples that have the same first tuple value will still be sorted correctly by using the second tuple value as a tiebreaker.

Here’s the code:

# Create the dictionary of user entries in your database

db = [{'username': 'Alice', 'joined': 2020, 'age': 23}, {'username': 'Bob', 'joined': 2018, 'age': 19}, {'username': 'Alice', 'joined': 2020, 'age': 31}] # Use the sorted() function with key argument to create a new list.

# Each dictionary list element is "reduced" to the tuple of values.

db_sorted = sorted(db, key=lambda row: (row['username'], row['joined'], row['age'])) # Print everything to the shell

print(db_sorted)

The output is the sorted dictionary. Note that the first dictionary has the smallest salary of Alice and the third dictionary has the largest salary of Alice.

[{'username': 'Alice', 'joined': 2020, 'age': 23}, {'username': 'Alice', 'joined': 2020, 'age': 31}, {'username': 'Bob', 'joined': 2018, 'age': 19}]

In this example, the dictionary value for the key ‘joined’ was an integer number. But what if it’s a date?

… By Date?

Problem: Given a list of dictionaries. How do you sort the list of dictionaries by date?

Minimal Example: Consider the following example where you want to sort a list of database entries (e.g. each stored as a dictionary) by value of the key 'joined' that is from type date or timedate.

db = [{'username': 'Alice', 'joined': '2019-03-02', 'age': 23}, {'username': 'Bob', 'joined': '2020-08-08', 'age': 19}, {'username': 'Alice', 'joined': '2019-03-04', 'age': 31}] db_sorted = # ... sorting magic here ... print(db_sorted)

The output should look like this where the salary of Alice determines the order of the dictionaries:

[{'username': 'Alice', 'joined': '2019-03-02', 'age': 23}, {'username': 'Alice', 'joined': '2019-03-04', 'age': 31}, {'username': 'Bob', 'joined': '2020-08-08', 'age': 19}]

Solution: Define a key function with the lambda keyword. Simply return the string value for the key 'joined' for a given dictionary. This return value is then used to sort the dictionaries in the list.

As the join dates are given as strings in the form year-month-day (e.g. '2020-08-08'), the default alphabetical sort will also lead to a sorted overall list of dictionaries: the dictionary row with the earliest join date will be the first in the resulting list.

Here’s the code:

# Create the dictionary of user entries in your database

db = [{'username': 'Alice', 'joined': '2019-03-02', 'age': 23}, {'username': 'Bob', 'joined': '2020-08-08', 'age': 19}, {'username': 'Alice', 'joined': '2019-03-04', 'age': 31}] # Use the sorted() function with key argument to create a new list.

# Each dictionary list element is "reduced" to the join date.

db_sorted = sorted(db, key=lambda row: row['joined']) # Print everything to the shell

print(db_sorted)

The output is the sorted dictionary. Note that the first dictionary has the earliest and the third dictionary has the latest join date.

[{'username': 'Alice', 'joined': '2019-03-02', 'age': 23}, {'username': 'Alice', 'joined': '2019-03-04', 'age': 31}, {'username': 'Bob', 'joined': '2020-08-08', 'age': 19}]

You can use a similar method if the dictionary values are of format datetime as they are also comparable with default comparison operators >, <, >=, <=, and ==. [2]

… In Descending Order?

Problem: The default ordering of any Python list sort routine (sorted() and list.sort()) is in ascending order. This also holds if you sort a list of dictionaries. How can you sort in descending order?

Solution: If you want to sort in descending order, you can do one of the following:

• Use slicing to reverse the sorted list.
• Use the reverse=True argument of the sorted() or list.sort() methods.

Both ways are equivalent. Have a look at the previous example first using the reverse argument:

# Create the dictionary of user entries in your database

db = [{'username': 'Alice', 'joined': '2019-03-02', 'age': 23}, {'username': 'Bob', 'joined': '2020-08-08', 'age': 19}, {'username': 'Alice', 'joined': '2019-03-04', 'age': 31}] # Use the sorted() function with key argument to create a new list.

# Each dictionary list element is "reduced" to the join date.

db_sorted = sorted(db, key=lambda row: row['joined'], reverse=True) # Print everything to the shell

print(db_sorted)

Output:

[{'username': 'Bob', 'joined': '2020-08-08', 'age': 19}, {'username': 'Alice', 'joined': '2019-03-04', 'age': 31}, {'username': 'Alice', 'joined': '2019-03-02', 'age': 23}]

And second using slicing with negative step size (without the reverse argument):

# Create the dictionary of user entries in your database

db = [{'username': 'Alice', 'joined': '2019-03-02', 'age': 23}, {'username': 'Bob', 'joined': '2020-08-08', 'age': 19}, {'username': 'Alice', 'joined': '2019-03-04', 'age': 31}] # Use the sorted() function with key argument to create a new list.

# Each dictionary list element is "reduced" to the join date.

db_sorted = sorted(db, key=lambda row: row['joined'])[::-1] # Print everything to the shell

print(db_sorted)

Same output:

[{'username': 'Bob', 'joined': '2020-08-08', 'age': 19}, {'username': 'Alice', 'joined': '2019-03-04', 'age': 31}, {'username': 'Alice', 'joined': '2019-03-02', 'age': 23}]

… in Ascending Order?

The default ordering of any Python list sort routine (sorted() and list.sort()) is in ascending order. This also holds if you sort a list of dictionaries.

Where to Go From Here

In this article, you’ve learned how to sort a list of dictionaries. In summary, use the sorted() method with a key function as argument. This gives you the flexibility to customize how exactly you want to sort the dictionary—just map each dictionary to a comparable value.

I’ve realized that professional coders tend to use dictionaries more often than beginners due to their superior understanding of the benefits of dictionaries. If you want to learn about those, check out my in-depth tutorial of Python dictionaries.

If you want to stop learning and start earning with Python, check out my free webinar “How to Become a Python Freelance Developer?”. It’s a great way of starting your thriving coding business online.

“[Webinar] How to Become a Python Freelance Developer?”

Raylib 3.0 was released today.  Raylib is an open source, cross platform C based framework of tools and libraries for making games.  Components of raylib can be used on their own, however you can also download a preconfigured development environment making it among the easiest ways to get started doing C or C++ game programming. 

Details of the 3.0 release:

• All global variables from the multiple raylib modules have been moved to a global context state, it has several benefits, first, better code readability with more comprehensive variables naming and categorization (organized by types, i.e. CORE.Window.display.width, CORE.Input.Keyboard.currentKeyState or RLGL.State.modelview). Second, it allows better memory management to load global context state dynamically when required (not at the moment), making it easy to implement a hot-reloading mechanism if desired.

• All memory allocations on raylib and its dependencies now use RL_MALLOC, RL_FREE and similar macros. Now users can easely hook their own memory allocations mechanism if desired, having more control on how memory is allocated internally by the library. Additionally, it makes it easier to port the library to embedded devices where memory control is critical. For more info check raylib issue #1074.

• All I/O file accesses from raylib are being moved to memory data access, now all I/O file access is centralized into just four functions: LoadFileData(), SaveFileData(), LoadFileText(), SaveFileText(). Users can just update those functions to any I/O file system. This change makes it easier to integrate raylib with Virtual File Systems or custom I/O file implementations.

• All raylib data structures have been reviewed and optimized for pass-by-value usage. One of raylib distinctive design decisions is that most of its functions receive and return data by value. This design makes raylib really simple for newcomers, avoiding pointers and allowing complete access to all structures data in a simple way. The downside is that data is copied on stack every function call and that copy could be costly so, all raylib data structures have been optimized to stay under 64 bytes for fast copy and retrieve.

• All raylib tracelog messages have been reviewed and categorized for a more comprehensive output information when developing raylib applications, now all display, input, timer, platform, auxiliar libraries, file-accesses, data loading/unloading issues are properly reported with more detailed and visual messages.

• raudio module has been internally reviewed to accomodate the new Music structure (converted from previous pointer format) and the module has been adapted to the highly improved miniaudio v0.10.

• text module reviewed to improve fonts generation and text management functions, Font structure has been redesigned to better accomodate characters data, decoupling individual characters as Image glyphs from the font atlas parameters. Several improvements have been made to better support Unicode strings with UTF-8 encoding.

• Multiple new examples added (most of them contributed by raylib users) and all examples reviewed for correct execution on most of the supported platforms, specially Web and Raspberry Pi. A detailed categorized table has been created on github for easy examples navigation and code access.

• New GitHub Actions CI system has been implemented for Windows, Linux and macOS code and examples compilation on every new commit or PR to make sure library keeps stable and usable with no breaking bugs.

Raylib is available on https://www.raylib.com/, with the source code hosted on GitHub under the liberal ZLib license.  The raylib cheatsheet is a one stop shop for learning the commands from the various libraries.  Raylib is also well documented on the Wiki.

Learn more about raylib in the video below.

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