# Python reduce and accumulate total guide

## Python reduce function

Python’s reduce() implements a mathematical technique commonly known as folding or reduction. You’re doing a fold or reduction when you reduce a list of items to a single cumulative value. Python’s reduce() operates on any iterable (not just lists) and performs the following steps:

• `Apply a function` (or callable) to the first two items (default) in an iterable and generate a partial result.
• `Use that partial result`, together with the third item in the iterable, to generate another partial result.
• `Repeat the process` until the iterable is exhausted and then return a single cumulative value.

``````reduce(f, [x1, x2, x3, x4]) = f(f(f(x1, x2), x3), x4)
``````

The idea behind Python’s reduce() is to take an existing function, apply it cumulatively to all the items in an iterable, and generate a single final value. In general, Python’s reduce() is handy for processing iterables without writing explicit for loops. Since reduce() is written in C, its internal loop can be faster than an explicit Python for loop.

Python’s reduce() was originally a built-in function (and still is in Python 2.X ), but it was moved to functools.reduce() in Python 3.0 . This decision was based on some possible performance and readability issues.

In Python 3.X, if you need to use reduce(), then you first have to import the function into your current scope using an import statement in one of the following ways:

• import functools and then use fully-qualified names like functools.reduce().
• from functools import reduce and then call reduce() directly.
PYTHON REDUCE FUNCTION
``````functools.reduce(function, iterable[, initializer])
``````

#### find sum of iterable int with reduce

``````reduce(lambda x, y: x+y, [1, 2, 3, 4, 5])

# output
# 15
``````

#### Find the largest number in the iterable using reduce()

``````from functools import reduce
num1 = [15,12,30,4,5]
num2 = reduce(lambda x,y: x if x>y else y, num1)
print(num2)

# output
# 30
``````

#### Using User-defined function in reduce()

``````from functools import reduce

def my_add(a, b):
result = a + b
print(f"{a} + {b} = {result}")
return result

numbers = [0, 1, 2, 3, 4]
reduce(my_add, numbers)

# output:
# 0 + 1 = 1
# 1 + 2 = 3
# 3 + 3 = 6
# 6 + 4 = 10
# 10
``````

#### Iterable contains only one item, reduce() will return that item.

``````from functools import reduce

def sum1(x, y):
return x + y

num1 = [5]
num2 = reduce(sum1,num1)
print(num2)

# output
# 5

from functools import reduce
num1 = [15]
num2 = reduce(lambda x,y: x if x>y else y, num1)
print(num2)

# output
# 15
``````

### python reduce initializer

PYTHON
``````functools.reduce(function, iterable[, initializer])
``````

``````from functools import reduce

def my_add(a, b):
result = a + b
print(f"{a} + {b} = {result}")
return result

numbers = [0, 1, 2, 3, 4]

reduce(my_add, numbers, 100)

# output
# 100 + 0 = 100
# 100 + 1 = 101
# 101 + 2 = 103
# 103 + 3 = 106
# 106 + 4 = 110
# 110
``````

## Python accumulate function

PYTHON ACCUMULATE SYNTAX
``````itertools.accumulate(iterable[, func, *, initial=None])
``````

itertools.accumulate
``````def accumulate(inputs, func):
itr = iter(inputs)
prev = next(itr)
for cur in itr:
yield prev
prev = func(prev, cur)
``````

#### Multiple list items

The function argument is given as operator.mul .

PYTHON operator.mul
``````operator.mul(a, b)
operator.__mul__(a, b)
Return a * b, for a and b numbers.
``````

It will return an iterator that yields all intermediate values. We can convert to list by using a list() constructor.

``````from itertools import accumulate
import operator

num = accumulate([1,2,3,4,5], operator.mul)
print (list(num))

# output
# [1, 2, 6, 24, 120]
``````

accumulate pictorial representation:

#### default func parameter

It will return an iterator that yields all intermediate values. We can convert to a list by using list() constructor.

``````import itertools

num = itertools.accumulate([1,2,3,4,5])
print(num)

# output
# <itertools.accumulate object at 0x7f218a1e6bc0>

# convert iterator to list object
print(list(num))

# output
# [1, 3, 6, 10, 15]

# using reduce() for same function
from functools import reduce
import operator

r = reduce(operator.add, [1,2,3,4,5])
print(r)

# output:
# 15
``````

#### own defined function

``````import itertools

def sum_numbers( *args):
return sum(args)

sum_numbers(1, 2, 3, 4)

# output:
# 10

list(itertools.accumulate([1,2,3,4,5], sum_numbers))

# output:
# [1, 3, 6, 10, 15]
``````

### accumulate initial value

``````from itertools import accumulate
import operator

# it will contain more than one element in the ouptut iterable.
num = accumulate([1,2,3,4,5], operator.add, initial=10)
print(list(num))

# output:
# [10, 11, 13, 16, 20, 25]
``````

If the iterable is empty and the initial parameter is mentioned, it will return the initial value.

``````from itertools import accumulate
import operator

num = accumulate([], operator.add, initial=10)
print(list(num))

# output:
# [10]
``````

If iterable contains one element and the initial parameter is not mentioned, it will return that element.

``````from itertools import accumulate

num = accumulate([5], lambda x,y: x+y)
print(list(num))

# output:
# [5]
``````

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