Python - Append Help - String & list

Question

Another append question... This is my code:

def s(xs,n,m):
    t = []
    while n < m:
        n += 2
        t.append(xs[n])
    return t

When I evaluate the following:

x = s('African', 0, 3)

Why does it return this?:

['r', 'c']

Answer 1

while n < m:
    n += 2 # at this point n = 2 because you've passed 0
    t.append(xs[n]) # you append r to t since xs[2] = r

but n < m still, so next iteration:

while n < m:
    n += 2 # at this point n = 4
    t.append(xs[n]) # you append c to t since xs[4] = c

now n > m, so the function returns ['r', 'c'] . Everything is correct.

Answer 2

Ok, so line-by-line...

Your call looks like this:

x = s('African', 0, 3)

so what happens is:

Step 1. - initial assignement
```
 def s(xs,n,m): 
```
xs='African' , n=0 and m=3 and then:
```
 t = [] 
```
(so, empty list t is introduced).
Step 2. - loop
1. Then the following condition is evaluated:
```
 while n < m: 
```
  to True , because 0 < 3 .
2. And then n is increased:
```
 n += 2 
```
  so it is now equal to 2 .
3. Then the appropriate element is appended to the empty t list:
```
 t.append(xs[n]) 
```
  and this element is " r ", because xs[2] == 'r' .
4. Then n < m condition is again evaluated to True (because 2 < 3 ), so the loop executes again:
```
 n += 2 
```
  and n is now equal to 4 .
5. Then appropriate char from xs string is appended to t list (which already has one element, r , as we mentioned above).
```
 t.append(xs[n]) 
```
  and this element is " c " (because xs[4] is exactly " c ").
6. Then condition for while loop is again evaluated, but this time to False (because 4 < 3 is not true), so the loop stops executing...
(Step 3. - after the loop) ...and the program flow goes to the final statement of the function, which is:
```
 return t 
```

And t returns the list we filled with two elements - as a result, the function returns list ['r', 'c'] .

Is it clear enough? Did it help?