Reduce
Syntax#
- reduce(function, iterable[, initializer])
Parameters#
Parameter | Details |
---|---|
function | function that is used for reducing the iterable (must take two arguments). (positional-only) |
iterable | iterable that’s going to be reduced. (positional-only) |
initializer | start-value of the reduction. (optional, positional-only) |
Remarks#
reduce
might be not always the most efficient function. For some types there are equivalent functions or methods:
-
sum()
for the sum of a sequence containing addable elements (not strings):sum([1,2,3]) # = 6
-
str.join
for the concatenation of strings:''.join(['Hello', ',', ' World']) # = 'Hello, World'
-
next
together with a generator could be a short-circuit variant compared toreduce
:# First falsy item: next((i for i in [100, [], 20, 0] if not i)) # = []
Overview
# No import needed
# No import required...
from functools import reduce # ... but it can be loaded from the functools module
from functools import reduce # mandatory
reduce
reduces an iterable by applying a function repeatedly on the next element of an iterable
and the cumulative result so far.
def add(s1, s2):
return s1 + s2
asequence = [1, 2, 3]
reduce(add, asequence) # equivalent to: add(add(1,2),3)
# Out: 6
In this example, we defined our own add
function. However, Python comes with a standard equivalent function in the operator
module:
import operator
reduce(operator.add, asequence)
# Out: 6
reduce
can also be passed a starting value:
reduce(add, asequence, 10)
# Out: 16
Using reduce
def multiply(s1, s2):
print('{arg1} * {arg2} = {res}'.format(arg1=s1,
arg2=s2,
res=s1*s2))
return s1 * s2
asequence = [1, 2, 3]
Given an initializer
the function is started by applying it to the initializer and the first iterable element:
cumprod = reduce(multiply, asequence, 5)
# Out: 5 * 1 = 5
# 5 * 2 = 10
# 10 * 3 = 30
print(cumprod)
# Out: 30
Without initializer
parameter the reduce
starts by applying the function to the first two list elements:
cumprod = reduce(multiply, asequence)
# Out: 1 * 2 = 2
# 2 * 3 = 6
print(cumprod)
# Out: 6
Cumulative product
import operator
reduce(operator.mul, [10, 5, -3])
# Out: -150
Non short-circuit variant of any/all
reduce
will not terminate the iteration before the iterable
has been completly iterated over so it can be used to create a non short-circuit any()
or all()
function:
import operator
# non short-circuit "all"
reduce(operator.and_, [False, True, True, True]) # = False
# non short-circuit "any"
reduce(operator.or_, [True, False, False, False]) # = True
First truthy/falsy element of a sequence (or last element if there is none)
# First falsy element or last element if all are truthy:
reduce(lambda i, j: i and j, [100, [], 20, 10]) # = []
reduce(lambda i, j: i and j, [100, 50, 20, 10]) # = 10
# First truthy element or last element if all falsy:
reduce(lambda i, j: i or j, [100, [], 20, 0]) # = 100
reduce(lambda i, j: i or j, ['', {}, [], None]) # = None
Instead of creating a lambda
-function it is generally recommended to create a named function:
def do_or(i, j):
return i or j
def do_and(i, j):
return i and j
reduce(do_or, [100, [], 20, 0]) # = 100
reduce(do_and, [100, [], 20, 0]) # = []