Ever since Banjo and Kazooie were officially confirmed as Super Smash Bros. Ultimate fighters last summer, rumours and speculation surrounding new games and remasters for Rare’s legendary series have been doing the rounds online. One such rumour suggests that Playtonic, a studio housing plenty of ex-Rare staff, is helping Rare, Microsoft, or both to create a new title.

Today, and not for the first time, Playtonic Games has taken to social media to gently let down any fans believing such rumours to be true.

Playtonic says, “it’s never fun to come out and say these things, but we’re trying to avoid hopes being pinned on us”.

What the Yooka-Laylee developer really is currently up to is being kept under wraps for now, but it might be best to avoid dreaming about any brand new Banjo-Kazooie games guh-huh-ing onto your Switch. Is guh-huh-ing a word? It is now.

Not a stability update, then?

• 
• by Ryan Craddock 7 hours ago

We’ve seen quite a lot of rogue-lite games in the past few years, but HyperParasite is quite a unique one. Taking place within a typical cyberpunk cityscape, an alien parasite has invaded the planet and is looking to leech onto every unfortunate soul that happens to get in its way. The twist here is that you get to play as the parasite, and all the heroic characters defending the city are your enemies. It’s a neat little spin on a game that otherwise borrows its aesthetics and gameplay structure from other examples of the genre.

With solid twin-stick controls, HyperParasite boasts a wide range of characters that you can play as. At first, though, the game can be a bit of a grind. You’ll only have access to a few characters, and in order to unlock more, you’ll need to kill them during your run and transport its brain to the in-game store. That’s not all though; once the characters’ brains are safely stored away, you’ll need to invest precious currency in order to actually unlock the character. Once you’ve got a few under your belt, the game becomes a lot more manageable, as you can always infect a new host if your current one gets killed. When playing as the parasite in its pure form, however, caution is highly advised, as one hit will immediately kill you and end your run.

A mix of offence and defence is required if you’re to successfully navigate the stages within HyperParasite. Whether playing as the parasite in its pure form or as an infected host, you’re granted the ability to dash, which makes you invulnerable for a split second as you execute it. Each character also has an enhanced version of their main attack, so, for example, the hobo can blast his trolley towards enemies, ploughing through several at once, and the police officer can power up his handgun so it shoots out gigantic bullets. It’s worth saving these abilities for boss battles, or for when you’re feeling completely overwhelmed by enemies.

Overall, the gameplay is pretty solid, with the unlockable characters proving to be a great incentive to dive in again and again. It takes a great deal of patience during your initial sessions to keep playing, however, because without having a decent range of hosts to choose from, you’re at risk of dying over and over again if you’re stuck within the parasite’s pure form. Thankfully, the enemies do tend to telegraph their attacks from a mile off, so you can mitigate the danger significantly once you know how to look out for this.

Unfortunately, the graphics and overall presentation don’t hold up quite as well at the gameplay itself. The environments are nice enough and display an ’80s aesthetic that works rather well, but the character models themselves lack detail, resulting in a bit of difficulty telling them apart during the more hectic moments. We also noticed some slowdown here and there, but thankfully this is minimal. If you’re not too fussed about detailed visuals, then HyperParasite will no doubt keep you hooked with its unique parasite gameplay mechanic and exciting gun-play.

Yesterday, Sony revealed the upcoming PlayStation 5’s brand new controller, the DualSense. Our chums over at Push Square have details on everything you need to know about if you’re thinking of grabbing a PS5 later this year, but we couldn’t help but notice one small – and very silly – detail ourselves.

Actually brought to our attention thanks to the tweet below, we’ve realised that the controller’s shiny new ‘Create’ button looks strangely familiar. We’re not sure what the design is really supposed to represent exactly, but it sure does look an awful lot like the honk of everyone’s favourite indie goose.

Since launching on Switch last September, Untitled Goose Game has also waddled on over to the PS4. If a sequel should be released later down the line on PS5, wouldn’t that just make for the perfect honk button?

Do you like the new PS5 controller design? Think it’ll be any nicer than the Switch’s Pro Controller? Give us a honk in the comments.

I have spent my morning hours on an important mission. What is the cleanest, fastest, and most concise answer to the following question: How do you find the key with the minimum value in a Python dictionary?  Most answers on the web say you need to use a library but this is not true!

Simply use the min function with the key argument set to dict.get:

income = {'Anne' : 1111, 'Bert' : 2222, 'Cara' : 9999999} print(min(income, key=income.get))

# Anne

The min function goes over all keys, k, in the dictionary income and takes the one that has minimum value after applying the income.get(k) method. The get() method returns the value specified for key, k, in the dictionary.

Play with it yourself in our interactive code shell:

Now, read the 4-min article or watch the short video to fully understand this concept.

What’s the Min Function in Python?

Most likely, you already know Python’s min(…) function. You can use it to find the minimum value of any iterable or any number of values. Here are a few examples using the min function without specifying any optional arguments.

income = {'Anne' : 1111, 'Bert' : 2222, 'Cara' : 9999999} print(min(income, key=income.get))

# Anne # Key that starts with 'smallest' letter of the alphabet

print(min(income))

# Anne # Smallest value in the dictionary income

print(min(income.values()))

# 1111 # Smallest value in the given list

print(min([1,4,7,5,3,99,3]))

# 1 # Compare lists element-wise, min is first list to have a larger

# element print(min([1,2,3],[5,6,4]))

# [1, 2, 3] # Smallest value in the given sequence of numbers

print(min(5,7,99,88,123))

# 5

So far so good. The min function is very flexible. It works not only for numbers but also for strings, lists, and any other object you can compare against other objects.

Now, let’s look at the optional arguments of the min function. One of them is 'key'. Let’s find out what it does.

How Does the Key Argument of Python’s min() Function Work?

The last examples show the intuitive workings of the min function: you pass one or more iterables as positional arguments.

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Intermezzo: What are iterables? An iterable is an object from which you can get an iterator. An iterator is an object on which you can call the next() method. Each time you call next(), you get the ‘next’ element until you’ve got all the elements from the iterator. For example, Python uses iterators in for loops to go over all elements of a list, all characters of a string, or all keys in a dictionary.

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When you specify the key argument, define a function that returns a value for each element of the iterable. Then each element is compared based on the return value of this function, not the iterable element (the default behavior).

Here is an example:

lst = [2, 4, 8, 16] def inverse(val): return -val print(min(lst))

# 2 print(min(lst, key=inverse))

# 16

We define a function inverse() that returns the value multiplied by -1. Now, we print two executions of the min() function.

• The first is the default execution: the minimum of the list [2, 4, 8, 16] is 2.
• The second uses key. We specify inverse as the key function. Python applies this function to all values of [2, 4, 8, 16]. It compares these new values with each other and returns the min. Using the inverse function Python does the following mappings:

Original Value	Value after inverse() (basis for min())
2	-2
4	-4
8	-8
16	-16

Python calculates the minimum based on these mappings. In this case, the value 16 (with mapping -16) is the minimum value because -2 > -4 > -8 > -16. 

Now let’s come back to the initial question:

How to Get the Key with the Minimum Value in a Dictionary?

We use the same example as above. The dictionary stores the income of three persons John, Mary, and Alice. Suppose you want to find the person with the smallest income. In other words, what is the key with the minimum value in the dictionary?

Now don’t confuse the dictionary key with the optional key argument of the min() function. They have nothing in common – it’s just an unfortunate coincidence that they have the same name!

From the problem, we know the result is a dictionary key. So, we call min() on the keys of the dictionary. Note that min(income.keys()) is the same as min(income).

To learn more about dictionaries, check out our article Python Dictionary – The Ultimate Guide.

However, we want to compare dictionary values, not keys. We’ll use the key argument of min() to do this. We must pass it a function but which?

To get the value of 'Anne', we can use bracket notation – income['Anne']. But bracket notation is not a function, so that doesn’t work. Fortunately, income.get('Anne') does (almost) the same as income['Anne'] and it is a function! The only difference is that it returns None if they key is not in the dictionary. So we’ll pass that to the key argument of min().

income = {'Anne' : 1111, 'Bert' : 2222, 'Cara' : 9999999} print(min(income, key=income.get))

# Anne

How to Get the Key with the Maximum Value in a Dictionary?

If you understood the previous code snippet, this one will be easy. To find the key with maximum value in the dictionary, you use the max() function.

income = {'Anne' : 1111, 'Bert' : 2222, 'Cara' : 9999999} print(max(income, key=income.get))

# Cara

The only difference is that you use the built-in max() function instead of the built-in min() function. That’s it.

Related article:

• How to Get the Key with the Maximum Value in a Dictionary?

Find the Key with the Min Value in a Dictionary – Alternative Methods

There are lots of different ways to solve this problem. They are not as beautiful or clean as the above method. But, for completeness, let’s explore some more ways of achieving the same thing.

In a StackOverflow answer, a user compared nine (!) different methods to find the key with the minimum value in a dictionary. Here they are:

income = {'Anne' : 11111, 'Bert' : 2222, 'Cara' : 9999999} # Convert to lists and use .index(max())

def f1(): v=list(income.values()) k=list(income.keys()) return k[v.index(min(v))] # Dictionary comprehension to swap keys and values

def f2(): d3={v:k for k,v in income.items()} return d3[min(d3)] # Use filter() and a lambda function

def f3(): return list(filter(lambda t: t[1]==min(income.values()), income.items()))[0][0] # Same as f3() but more explicit

def f4(): m=min(income.values()) return list(filter(lambda t: t[1]==m, income.items()))[0][0] # List comprehension

def f5(): return [k for k,v in income.items() if v==min(income.values())][0] # same as f5 but remove the max from the comprehension

def f6(): m=min(income.values()) return [k for k,v in income.items() if v==m][0] # Method used in this article

def f7(): return min(income,key=income.get) # Similar to f1() but shortened to 2 lines

def f8(): v=list(income.values()) return list(income.keys())[v.index(min(v))] # Similar to f7() but use a lambda function

def f9(): return min(income, key=lambda k: income[k]) print(f1())

print(f2())

print(f3())

print(f4())

print(f5())

print(f6())

print(f7())

print(f8())

print(f9())

# Bert (all outputs)

In a benchmark performed on a large dictionary by the StackOverflow user, f1() turned out to be the fastest one.

So the second best way to get the key with the minimum value from a dictionary is:

income = {'Anne' : 11111, 'Bert' : 2222, 'Cara' : 9999999} v=list(income.values())

k=list(income.keys())

print(k[v.index(min(v))])

# Bert

Find Key with Shortest Value in Dictionary

We know how to find the minimum value if the values are numbers. What about if they are lists or strings?

Let’s say we have a dictionary that records the number of days each person worked this month. If they worked a day, we append 1 to that person’s list. If they didn’t work, we don’t do anything.  At the end of the month, our dictionary looks like this.

days_worked = {'Anne': [1, 1, 1, 1], 'Bert': [1, 1, 1, 1, 1, 1], 'Cara': [1, 1, 1, 1, 1, 1, 1, 1]}

The total number of days worked each month is the length of each list. If all elements of two lists are the same (as is the case here), they are compared based on their length.

# Length 2 is less than length 4

>>> [1, 1] < [1, 1, 1, 1]

True

So we can use the same code we’ve been using in the article to find the key with the minimum value.

>>> min(days_worked, key=days_worked.get) 'Anne'

If we update our dictionary so that Bert has worked the most days and apply min() again, Python returns 'Anne'.

>>> days_worked = {'Anne': [1, 1, 1, 1], 'Bert': [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], 'Cara': [1, 1, 1, 1, 1, 1, 1, 1]} # Anne has now worked the least

>>> min(days_worked, key=days_worked.get)

Find Key With Min Value in a List of Dictionaries

Let’s say we have 3 dictionaries containing income information. We want to find the key with the min value from all 3 dictionaries. 

income1 = {'Anne': 1111, 'Bert': 2222, 'Cara': 3333} income2 = {'Dani': 4444, 'Ella': 5555, 'Fred': 6666} income3 = {'Greg': 7777, 'Hope': 8888, 'Igor': 999999999999} list_of_dicts = [income1, income2, income3]

We can see that 'Anne' has the lowest income so we expect that to be returned.

There are several ways to do this. The simplest is to put all key-value pairs into one dictionary using a for loop. Then we call min() as usual.

# Initialise empty dict

>>> big_dict = {} # Use for loop and .update() method to add the key-value pairs

>>> for dic in list_of_dicts: big_dict.update(dic) # Check the result is as expected

>>> big_dict

{'Anne': 1111, 'Bert': 2222, 'Cara': 3333, 'Dani': 4444, 'Ella': 5555, 'Fred': 6666, 'Greg': 7777, 'Hope': 8888, 'Igor': 999999999999} # Call min() and specify key argument

>>> min(big_dict, key=big_dict.get) 'Anne' 

Where to Go From Here?

Every Python master must know the basics. Improving your basic code understanding skills by 20% will improve your productivity by much more than anything else. Why? Because everything else builds upon the basics.

But most material online is tedious and boring. That’s why I’ve written a new and exciting way of learning Python, while measuring and comparing your skills against other coders. Check out the book “Coffee Break Python”. It’s LeanPub 2019 bestseller in the category Python!