Reputation: 21
I am trying to convert the below for loop to Python.
for (i = 5; i < n; i = i*5):
I am not sure how to make use of the Range function when i want the i value to be set to the multiple of 5. For example, 1st time I want the i to be 5, then followed by 25, then followed by 125 and it should go on.
The following is what i have tried:
i = 5
for i in range (i, n+1, i*5)
The problem with the above being, the value of i getting incremented by 25, making it to 30 whereas i want the i to be 25 in the second iteration. It is pretty easy when using the while loop. But I am seeing if there is a way to implement the same in the for loop. Please help. Thanks in advance.
Upvotes: 2
Views: 413
Reputation: 11330
Most Python programmers would just use a while loop:
i = 5
while i < n:
....
i = i * 5
If you really, really want a for loop:
import itertools
for i in itertools.takewhile(lambda x: x < n, (5 ** i for i in itertools.count(1))):
... whatever ...
Upvotes: 0
Reputation: 57135
There is a theorem in computer science that states that any "C-style" for
loop can be transformed into an equivalent while
loop. This is one of those cases where the transformation is desirable:
i = 5
while i < n:
# Loop body goes here
i *= 5
You can hide the loop logic behind a generator:
def multrange(start, stop, ratstep):
i = start
while i < stop:
yield i
i *= ratstep
list(multrange(5, 10000, 5))
#[5, 25, 125, 625, 3125]
Upvotes: 1
Reputation: 27577
You can define your own range
function using yield
!
def range(i, j, k):
while i * k < j:
i *= k
yield i
for i in range(5, 2000, 5):
print(i)
Output:
25
125
625
Upvotes: 0
Reputation: 61643
I am not sure how to make use of the Range function when i want the i value to be set to the multiple of 5
It will not work that way. range
can only create arithmetic sequences; multiplying every time creates a geometric sequence.
What you can do is take advantage of the fact that the i
values are successive powers of 5; so make a loop over the desired exponent values, and compute i
inside the loop:
# Computing the `limit` in terms of `n` is left as an exercise.
# Just in case you were already computing `n` in terms of an existing `limit`,
# in which case you could just use it directly ;)
for j in range(limit):
i = 5**j
Upvotes: 1