CODEWITHSUNDEEP
python-3.xlambdareduce
Prateek B
Prateek B

Reputation: 31

How does lambda function work with reduce function in python?

I'm trying to understand the following code:

>>> print(reduce(lambda x, y: x + y, [1,2,3,4,5]))
15

What is x and y? Which is the accumulator here? Does the accumulator always has to be the first argument (i.e. x here) What happens when no initialization is done? What is the default value picked?

Upvotes: 2

Views: 1744

Answers (2)

JRajan
JRajan

Reputation: 702

An extended form of this code which would might be easier to understand

lst = [1, 2, 3, 4, 5]

x = 0  # Default value
for i in lst:
    y = i
    x = x + y
print(x)

Or, if you consider the lambda function:

lst = [1, 2, 3, 4, 5]
x = 0

def func(x,y):
    x = x + y
    return x

for i in lst:
    x = func(x, i)

print(x)

As you can see, x is the accumulator here. You won't be able to use the reduce method if you don't specify the accumulator as the first argument.

Hope that clears how your code works.

Update: As @Wolf points out, we can specify an initial value for the args for reduce method. Updated x to be 0, as that is the default value.

Upvotes: 2

Wolf
Wolf

Reputation: 10238

I think the documentation of functools.reduce(function, iterable[, initializer]) is quite readable. The iteration is carried out, applying the operation from left to right to consecutive elements and accumulating the result in the left operand. To be sure about the aspects in question, it helps to create variations of above code.

In the following example, we get the sum of the elements; we see that there is no implicit 1 as "default" value at work:

>>> print(reduce(lambda x, y: x + y, [2,3]))
5

The reduction stops if there are no elements left in the iterable:

>>> print(reduce(lambda x, y: x + y, [1]))
1

The reduction fails if there are no elements at all:

>>> print(reduce(lambda x, y: x + y, []))
Traceback (most recent call last):
  File "<pyshell#11>", line 1, in <module>
    print(reduce(lambda x, y: x + y, []))
TypeError: reduce() of empty sequence with no initial value

The optional initializer value is used to start the reduction (and immediately stops it if there are no elements):

>>> print(reduce(lambda x, y: x + y, [], 0))
0

So x is indeed the accumulator that is initialized with the optional initializer value before entering the iterable. And y is the next value from the iterable, if there are no elements (left), the function is ignored.

Here another example to better understand the roles of x and y:

>>> print(reduce(lambda x, y: x - y, [1,2,3], 10))
4

In single-step mode, above reduction would look like this:

x = 10   # initializer
y = 1    # from iterable
x = x-y  # x==9
y = 2    # from iterable
x = x-y  # x==7
y = 3    # from iterable
x = x-y  # x==4
4        # result

Upvotes: 2

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