更新列表理解中的 if 条件内的变量

Question

i tried my best to write the following code in a list comprehension but ended up getting wrong

for i in range(1,n+1):
    L.append(i+(k*sin))
    if(i % (k) == 0):
        sin*=-1


return [i+(k*sin) if(i % (k*2) == 0) sin*=-1 for i in range(1,n+1)]

Answer 1

You can do this with a list comprehension, but you should take a different approach.

Let's say k == 3 : then your sequence will look like this:

1 + k*sin
2 + k*sin
3 + k*sin
4 + -k*sin
5 + -k*sin
6 + -k*sin
7 + k*sin
8 + k*sin
9 + k*sin
 ...

Basically, you just want a list comprehension that looks like

[i + n*sin for i, n in zip(range(1, n+1), ???)]

The question is, what do we replace ??? with to get our sequence of repeating k and -k values?

The itertools module provides several tools for generating this:

• repeat - repeat a value some number of times
• chain - concatenate two or more sequences into a single sequence
• cycle - cycle through the values in a sequence repeatedly

In this case, it's

cycle(chain(repeat(k, k), repeat(-k, k)))

giving you

from itertools import repeat, chain, cycle


[i + n*sin for i, n in zip(range(1, n+1), cycle(chain(repeat(k, k), repeat(-k, k)))]

zip ensures we only take a finite number of values from the infinite sequence.

Another approach is to use another tool from itertool , islice , to take only the first n values from the cycle. Then we don't need range anymore; we can use enumerate to number those values from 1 to n instead.

from itertools import repeat, chain, cycle, islice


[i + n*sin for i, n in enumerate(islice(cycle(chain(repeat(k, k), repeat(-k, k))), n), start=1]

---

Both read better if you define ks first:

ks = cycle(chain(repeat(k, k), repeat(-k, k)))
[i + n*sin for i, n in zip(range(1, n+1), ks)]

or

ks = cycle(chain(repeat(k, k), repeat(-k, k)))
[i + n*sin for i, n in enumerate(islice(ks, n), start=1)]

---

You can also optimize this a bit by repeating the value of k*sin , rather than performing the multiplication for each element of the list.

ks = cycle(chain(repeat(k*sin, k), repeat(-k*sin, k)))
[i + n for i, n in ...] # using either zip or enumerate as before

Answer 2

What you want is to mathematically formulate your problem a little bit better, than it can easily be represented as a list comprehension.

In essence, you want the a formula for the signal of sin which is dependent on the index i and on k that will invert the signal every k . That is represented by (-1) ** (i - 1) // k .

So, if you include that into your code, you should arrive at exactly what you want:

return [i + (k * ((-1) ** ((i - 1) // k)) * sin) for i in range(1,n+1)]

Hope that helps!