How to use filter, map, and reduce in Python 3

Question

 In python 2, the filter, map, and reduce work perfectly. Here is an example:

>>> def f(x):

        return x % 2 != 0 and x % 3 != 0

>>> filter(f, range(2, 25))

[5, 7, 11, 13, 17, 19, 23]

>>> def cube(x):

        return x*x*x

>>> map(cube, range(1, 11))

[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]

>>> def add(x,y):

        return x+y

>>> reduce(add, range(1, 11))

55

But when I am trying in Python 3, I receive the following outputs:

>>> filter(f, range(2, 25))

<filter object at 0x0000000002C14908>

>>> map(cube, range(1, 11))

<map object at 0x0000000002C82B70>

>>> reduce(add, range(1, 11))

Traceback (most recent call last):

  File "<pyshell#8>", line 1, in <module>

    reduce(add, range(1, 11))

NameError: name 'reduce' is not defined

Answer 1

Please be informed that the map() and filter() was purposely changed in newer version of Python so that it can return iterators, and reduce was eliminated from being a built-in and placed in functools.reduce.

You can use the below code:

>>> def f(x): return x % 2 != 0 and x % 3 != 0

...

>>> list(filter(f, range(2, 25)))

[5, 7, 11, 13, 17, 19, 23]

>>> def cube(x): return x*x*x

...

>>> list(map(cube, range(1, 11)))

[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]

>>> import functools

>>> def add(x,y): return x+y

...

>>> functools.reduce(add, range(1, 11))

55

>>>

The below code will replace your usage of map and filter with generators expressions or list comprehensions. Example:

>>> def f(x): return x % 2 != 0 and x % 3 != 0

...

>>> [i for i in range(2, 25) if f(i)]

[5, 7, 11, 13, 17, 19, 23]

>>> def cube(x): return x*x*x

...

>>> [cube(i) for i in range(1, 11)]

[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]

>>>