Adding integers progressively to a list

Question

It's a while loop and I need to add the log variable (most recent variable) to a list. I have no idea how to do it.

I'm going to add every number that is a multiple of both 5 and 7 (i.e. if x % 7==0 and x % 5==0) to a list which I am going to state at the end.

But how?

--

"This program is about the find every number between 1500 and 2700 (inclusive) which is divisible by both 5 and 7"

x=1500
for x in range(1500,2701):
    if x % 7==0 and x % 5==0:
        print("\n", x,"IS DIVISIBLE\n")
        x=x+1
        #I THINK THE LIST STUFF GOES HERE

    else:
        print(x,"is not a common multiple")
        x=x+1

input()

Basically, I just want the x variable which is divisible by 7 and 5 (each time the loop runs) to be added to a list. E.g. 1505, 1540 etc.

`values = [x for x in range(1500, 2701) if not x % 7 and not x % 5]` — han solo, Feb 12 '19 at 15:47
Don't say "adding" to a list when you mean "appending". And as you'll find out in Python, generally the more Pythonic alternative to iteratively appending elements to a list is to use a **list comprehension**. — smci, Jun 26 '21 at 11:26

Ma0 · Answer 1 · 2019-02-12T16:19:08.990

1

If you are looking for efficiency, do some math first:

5 and 7 are primes to themselves¹ so a number is divisible from both if and only if it is divisible from their product.
After finding the first number that satisfies point 1, simply keep adding 35 until you reach the end of your range. Note that you know how many such numbers will be in your range!

Now the code:

first = next(x for x in range(1500, 2701, 5) if x % 35 == 0)
res = [first + 35*i for i in range((2701-1-1500)//35 + 1)]

which produces:

[1505, 1540, 1575, 1610, 1645, 1680, 1715, 1750, 1785, 1820, 1855, 1890, 1925, 1960, 1995, 2030, 
 2065, 2100, 2135, 2170, 2205, 2240, 2275, 2310, 2345, 2380, 2415, 2450, 2485, 2520, 2555, 2590, 
 2625, 2660, 2695]

This will be faster than any if-based approach.

As far as the problem with your code goes, it has been thoroughly discussed by the other answers and comments, so I will not go into that.

¹and in general but that is irrelevant here

edited Feb 12 '19 at 16:19

answered Feb 12 '19 at 15:56

Ma0

15,057
4
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65

I am confused by the use of `next` here, as I never use it myself. How is that different from this code except for the fact that it's a one liner? `for x in range(1500, 2701, 5):` `if x%35 == 0:` `first = x` `break` – Guimoute Feb 12 '19 at 16:06
it is not; it is just a more compact way to write it. – Ma0 Feb 12 '19 at 16:13
Oh I get it. `next()` is simply the first element of the generator. Smart. – Guimoute Feb 12 '19 at 16:17

Marisa · Answer 2 · 2019-02-12T15:50:04.740

Create a list variable:

foo_list = []

After that, all you need to do is at the end of your loop append the value to your list

foo_list.append(<your variable>)

So it would look something like this on your code:

my_list = []
x=1500
for x in range(1500,2701):
    if x % 7==0 and x % 5==0:
        print("\n", x,"IS DIVISIBLE\n")
        my_list.append(x)
    else:
        print(x,"is not a common factor")

As you can see, the x+=1 was deleted, the loop is doing it for you!

As some people pointed out, if you are just interested in getting the list and not printing wether if the number is a common factor or not you could use list comprehensions, as follows:

my_list= [x for x in range(1500, 2701) if x%5==0 and x%7==0]

You can remove the `x=x+1` too. – Guimoute Feb 12 '19 at 15:49 — Guimoute, Feb 12 '19 at 15:49

score 0 · Answer 3 · answered Feb 12 '19 at 15:48

You don't need to increment x when you have met a condition, as that's handled for you by the range generator.

You are close:

somelist = []

for x in range(1500, 2701):
    if x%5==0 and x%7==0:
        somelist.append(x)
    # You don't need an else block as the loop will just continue

You could alternatively do this in a list comprehension for more speed:

somelist = [x for x in range(1500, 2701) if x%5==0 and x%7==0]

score 0 · Answer 4 · answered Feb 12 '19 at 15:49

0

result =[]    
for x in range(1500,2701):
        if x % 7==0 and x % 5==0:
            result.append(x)    
        else:
            print(x,"is not a common factor")
            x=x+1

answered Feb 12 '19 at 15:49

John Willson

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score 0 · Answer 5 · answered Feb 12 '19 at 15:52

0

my_list = []
for x in range(1500,2701):
    if x % 7==0 and x % 5==0:
        print("\n", x,"IS DIVISIBLE\n")
        my_list.append(x)
            #I THINK THE LIST STUFF GOES HERE


    else:
        print(x,"is not a common factor")
print(my_list)

answered Feb 12 '19 at 15:52

kederrac

16,819
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Adding integers progressively to a list

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