Over the last few weeks, the world at large has taken on many new challenges in daily life. Adopting new ways to work from home, often while parenting and balancing other priorities. It takes time, patience and problem solving. It’s like learning a new language.

We have received questions from across the disability community looking for tools, resources and best practices. Below you’ll see some of our key learnings since transitioning to working from home in early March. We have identified ways to accelerate the learning curve by leaning into our employees expertise and by continuing to prioritize accessibility to ensure that what we deliver is accessible to all our colleagues in these new and shifting circumstances.

Consider this the start of a series with more chapters ahead. Our hope is that by sharing these learnings we can accelerate your new your journey of being accessible, inclusive and productive – no matter where you’re working.

Learn from others with disabilities

The biggest source of knowledge right now are your employees, especially those in your disability employee communities. The insight and learning here will accelerate the learning curve. Three of our employees have already posted articles with details of how they are working day to day pragmatics:

• Leah Katz-Hernandez, a member of our CEO communications team, shares her experience as a profoundly deaf, visual-only American Sign Language (ASL) user – with tips on how to conduct great productive meetings in the virtual meeting space.
• Alyson Boote, a staff ASL interpreter, lists her recommendations for at-home remote ASL interpretation including downloadable reference guides.
• Megan Lawrence, member of our mental health employee resource group, details key tools and imperatives for maintaining emotional wellbeing including use of the toolset, MyAnalytics.

Leverage assistive technology

Now more than ever, accessibility isn’t option – it’s imperative. Online content, conference calls and virtual forums have replaced in person meetings and events for immediate future, and ‘in person’ accommodations that empower people with disabilities to consume that content have understandably reduced or stopped. However, if you embed accessibility into design of virtual or online content, you remove or reduce the potential of exclusion. You have the power to include and accessibility is the key. Here are a few tools at your disposal to assist you:

• Use Accessibility Insights to check your website, Windows or Android app for accessibility with quick easy guides on how to make them more accessible. Do this before you post content.
• Caption your videos. There are lots of ways to do this, I upload videos into Microsoft Stream (available as part of Microsoft 365) and auto captioning/editing feature prior to sharing within my organization.
• Use Accessibility Checker on any Microsoft 365 document to catch simple gotchas. Add alt-text to all images and ensure the format is screen reader friendly.
• Microsoft Teams is a one stop shop for conference calls, meetings, collaboration. If you’re looking for the simple answer to ‘is it accessible’ – yes – we’ve worked hard to making this an accessible platform for online meetings. Turn on live captions in any call or webinar, use ‘pin’ feature to keep one speaker (or in my case, my ASL interpreter) on the screen to avoid distractions, use Immersive Reader in the chat window or my favorite feature ‘background blur’ which was specifically designed by one of our deaf engineers to power up lipreading, great example of how an accessibility feature has powered up millions.

Contact us any time

The Microsoft Disability Answer Desk is available 24/7 by phone, chat or ASL Video. Please reach out if you need any advice or assistance as you learn your ‘new language.’

More to come. If you have any feedback or topics you want me to cover in next blog, pop ideas into the comments field below. Stay safe and wash your hands.

Yesterday, PlatinumGames revealed the fourth and final part of its much-hyped ‘Platinum4’ series of announcements, and the whole thing turned out to be an April Fools’ joke. It didn’t seem to sit particularly well with fans who were hoping for something more, but Hideki Kamiya has taken to social media to explain the thought process behind the whole thing.

In case you missed it, the prank teased an arcade machine for a new game called ‘Sol Cresta’, a made-up sequel to ’80s side-scrolling shoot em ups Moon Cresta and Terra Cresta. On Twitter, Kamiya says that he hopes fans “were able to get a good laugh”. His full statement can be found below (translation via Siliconera):

“We, PlatinumGames, went all-out for this year’s April Fools’ again. While I’m aware that we’re living in troubled times, as an entertainer there’s a part of me that just wants to cheer you all up, even if it’s just a little.

And a big thank you goes out to the folks at Hamster for their cooperation.

I’ve always had my own rules about April Fools’ jokes such as ‘make it obvious that it’s a lie’ and ‘make sure it doesn’t disappoint even if they know it’s a lie’ and so on. However, this time it was ‘make something that will get us (me?) excited’ and we went full stupid with it. I hope that you were able to get a good laugh and remind yourself that this is PlatinumGames.”

Judging by the reactions of fans in the prank’s YouTube comments, and indeed in our very own comment section yesterday, it sadly didn’t seem to get viewers very excited at all. Instead, people were simply left wondering whether we’ll still be getting a ‘proper’ fourth announcement in the future.

That still remains to be seen, but we sure hope there’s something else up PlatinumGames’ sleeve – for many, hyping up a joke won’t quite cut it.

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?”