Kemal - 7 months ago 13

Python Question

Everybody,

a =[0, 0, 2, 4, 6]

x=5

This a list (a) and a fix value (x).

I need a loop codes which must add with x every element of list and add this value with previous list elements in every loop ( loop must continue as x value). Other words result should like below:

`0`

0

2

4

6

0

0

7

9

11

0

0

12

14

16

0

0

15

19

21

0

0

21

24

26

I prepared codes as below but it doesn’t work. Other words produce something as below (incorrect)

`i=0`

counter=0

while counter < x:

for i in a:

if i >0:

i=i+x

elif i ==0:

i=0

print i

counter=counter+1

0

0

7

9

11

0

0

7

9

11

0

0

7

9

11

0

0

7

9

11

0

0

7

9

11

So, I need to help for this…

Thank you.

Answer

I think this does mostly what you want (at least, as I understand the question)...

```
def make_it_so(a, x):
i = 0
counter=0
while counter < x:
for i in a:
if i == 0:
yield 0
else:
yield i + counter * x
counter = counter + 1
# Demo
for item in make_it_so([0, 0, 2, 4, 6], 5):
print item
```

Note that I've made it a generator function. You could easily turn it into a regular function that returns a list if you created an output list at the top of the function and swapped `yield ...`

for `output_list.append(...)`

and then `return output_list`

at the end of the function...

The key here is to understand that after the first loop, you are adding `0`

to all of the (non-zero) items. After the second loop, you are adding `x`

. After the third loop, you're adding the `x + x`

(since the first loop added `x`

and now you're adding `x`

more). In general, for the Nth loop, you'll be adding `(N-1) * x`

to all of the non-zero items. So, you just need to keep track of `N`

, (or `N-1`

). In fact, your original code was already doing this (with `counter`

), so we just re-purpose that and it's all good.