For loop with custom steps in python

Question

I can make simple for loops in python like:

for i in range(10):

However, I couldn't figure out how to make more complex ones, which are really easy in c++.

How would you implement a for loop like this in python:

for(w = n; w > 1; w = w / 2)

The closest one I made so far is:

for w in reversed(range(len(list)))

Answer 1

for i in range(0, 10, 2):
    print(i)

>>> 0
>>> 2
>>> 4
>>> 6
>>> 8

http://docs.python.org/2/library/functions.html

>>> range(10)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> range(1, 11)
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> range(0, 30, 5)
[0, 5, 10, 15, 20, 25]
>>> range(0, 10, 3)
[0, 3, 6, 9]

Answer 2

First and foremost: Python for loops are not really the same thing as a C for loop. They are For Each loops instead. You iterate over the elements of an iterable. range() generates an iterable sequence of integers, letting you emulate the most common C for loop use case.

However, most of the time you do not want to use range(). You would loop over the list itself:

for elem in reversed(some_list):
    # elem is a list value

If you have to have a index, you usually use enumerate() to add it to the loop:

for i, elem in reversed(enumerate(some_list)):
    # elem is a list value, i is it's index in the list

For really 'funky' loops, use while or create your own generator function:

def halved_loop(n):
    while n > 1:
        yield n
        n //= 2

for i in halved_loop(10):
    print i

to print 10, 5, 2. You can extend that to sequences too:

def halved_loop(sequence):
    n = -1
    while True:
        try:
            yield sequence[n]
        except IndexError:
            return
        n *= 2

for elem in halved_loop(['foo', 'bar', 'baz', 'quu', 'spam', 'ham', 'monty', 'python']):
    print elem

which prints:

python
monty
spam
foo

Answer 3

For your exact example, you probably wouldn't use a for loop at all, but a while loop:

w = n
while w > 1:
    do stuff
    w = w / 2

Answer 4

Something like for i in [n/(2**j) for j in range(int(math.log(n))+1)]

Answer 5

You need to use a generator. You could implement this as follows:

def stepDown(n):
    while n>1:
        yield n
        n = n/2

for i in stepDown(n):
    print i # or do whatever else you wish.

Note that this generalizes easily to other complicated patterns you may have in mind.

Answer 6

For the more general case, you could create a custom generator function, that takes a start, stop, and a function for generating the next step from the last:

def my_range(start, stop, f):
    x = start
    while x < stop if stop > start else x > stop:
        yield x
        x = f(x)

>>> list(my_range(1, 1024, lambda x: x*2))
[1, 2, 4, 8, 16, 32, 64, 128, 256, 512]

>>> list(my_range(1000, 1, lambda x: x/2))
[1000, 500.0, 250.0, 125.0, 62.5, 31.25, 15.625, 7.8125, 3.90625, 1.953125]