lambda is an anonymous function, it is equivalent to:
def func(p):
return p.totalScore
Now max becomes:
max(players, key=func)
But as def statements are compound statements they can't be used where an expression is required, that's why sometimes lambda's are used.
Note that lambda is equivalent to what you'd put in a return statement of a def. Thus, you can't use statements inside a lambda, only expressions are allowed.
What does max do?
max(a, b, c, ...[, key=func]) -> value
With a single iterable argument, return its largest item. With two or more arguments, return the largest argument.
So, it simply returns the object that is the largest.
How does key work?
By default in Python 2 key compares items based on a set of rules based on the type of the objects (for example a string is always greater than an integer).
To modify the object before comparison, or to compare based on a particular attribute/index, you've to use the key argument.
Example 1:
A simple example, suppose you have a list of numbers in string form, but you want to compare those items by their integer value.
>>> lis = ['1', '100', '111', '2']
Here max compares the items using their original values (strings are compared lexicographically so you'd get '2' as output) :
>>> max(lis)
'2'
To compare the items by their integer value use key with a simple lambda:
>>> max(lis, key=lambda x:int(x)) # compare `int` version of each item
'111'
Example 2: Applying max to a list of tuples.
>>> lis = [(1,'a'), (3,'c'), (4,'e'), (-1,'z')]
By default max will compare the items by the first index. If the first index is the same then it'll compare the second index. As in my example, all items have a unique first index, so you'd get this as the answer:
>>> max(lis)
(4, 'e')
But, what if you wanted to compare each item by the value at index 1? Simple: use lambda:
>>> max(lis, key = lambda x: x[1])
(-1, 'z')
Comparing items in an iterable that contains objects of different type:
List with mixed items:
lis = ['1','100','111','2', 2, 2.57]
In Python 2 it is possible to compare items of two different types:
>>> max(lis) # works in Python 2
'2'
>>> max(lis, key=lambda x: int(x)) # compare integer version of each item
'111'
But in Python 3 you can't do that any more:
>>> lis = ['1', '100', '111', '2', 2, 2.57]
>>> max(lis)
Traceback (most recent call last):
File "<ipython-input-2-0ce0a02693e4>", line 1, in <module>
max(lis)
TypeError: unorderable types: int() > str()
But this works, as we are comparing integer version of each object:
>>> max(lis, key=lambda x: int(x)) # or simply `max(lis, key=int)`
'111'
lambda is an anonymous function, it is equivalent to:
def func(p):
return p.totalScore
Now max becomes:
max(players, key=func)
But as def statements are compound statements they can't be used where an expression is required, that's why sometimes lambda's are used.
Note that lambda is equivalent to what you'd put in a return statement of a def. Thus, you can't use statements inside a lambda, only expressions are allowed.
What does max do?
max(a, b, c, ...[, key=func]) -> value
With a single iterable argument, return its largest item. With two or more arguments, return the largest argument.
So, it simply returns the object that is the largest.
How does key work?
By default in Python 2 key compares items based on a set of rules based on the type of the objects (for example a string is always greater than an integer).
To modify the object before comparison, or to compare based on a particular attribute/index, you've to use the key argument.
Example 1:
A simple example, suppose you have a list of numbers in string form, but you want to compare those items by their integer value.
>>> lis = ['1', '100', '111', '2']
Here max compares the items using their original values (strings are compared lexicographically so you'd get '2' as output) :
>>> max(lis)
'2'
To compare the items by their integer value use key with a simple lambda:
>>> max(lis, key=lambda x:int(x)) # compare `int` version of each item
'111'
Example 2: Applying max to a list of tuples.
>>> lis = [(1,'a'), (3,'c'), (4,'e'), (-1,'z')]
By default max will compare the items by the first index. If the first index is the same then it'll compare the second index. As in my example, all items have a unique first index, so you'd get this as the answer:
>>> max(lis)
(4, 'e')
But, what if you wanted to compare each item by the value at index 1? Simple: use lambda:
>>> max(lis, key = lambda x: x[1])
(-1, 'z')
Comparing items in an iterable that contains objects of different type:
List with mixed items:
lis = ['1','100','111','2', 2, 2.57]
In Python 2 it is possible to compare items of two different types:
>>> max(lis) # works in Python 2
'2'
>>> max(lis, key=lambda x: int(x)) # compare integer version of each item
'111'
But in Python 3 you can't do that any more:
>>> lis = ['1', '100', '111', '2', 2, 2.57]
>>> max(lis)
Traceback (most recent call last):
File "<ipython-input-2-0ce0a02693e4>", line 1, in <module>
max(lis)
TypeError: unorderable types: int() > str()
But this works, as we are comparing integer version of each object:
>>> max(lis, key=lambda x: int(x)) # or simply `max(lis, key=int)`
'111'
Strongly simplified version of max:
def max(items, key=lambda x: x):
current = item[0]
for item in items:
if key(item) > key(current):
current = item
return current
Regarding lambda:
>>> ident = lambda x: x
>>> ident(3)
3
>>> ident(5)
5
>>> times_two = lambda x: 2*x
>>> times_two(2)
4
max(a,...) is always going to return an element of a. So the result will be either 1,2,3, or 4.
For each value w in a, the key value is b[w]. The largest key value is 10, and that corresponds with w equalling 1. So max(a,key=lambda w: b[w]) returns 1.
Try:
a = [1,2,3,4]
b = {}
b[1] = 10
b[2] = 8
b[3] = 7
b[4] = 5
c = a + b.values()
print max(*c)
