Python : How to refer itself in the list comprehension?

Question

lsit_a = [2, 4, 3, 6, 3, 8, 5]

list comprehension is very useful.

list_b = [a**2 for a in list_a]

I want to know how to write a self-reference in the python list comprehension.

for example -

list_c = [a**2 if i == 0 else a*2 + (itself[i-1]) for i, a in enumurate(list_a)]

how to write the part of itself[i-1]?

ADD

Sorry, my explain is bad...

I want to get the list by a high-speed way.

input list

list_a = [0,0,1,0,0,0,1,0,0,1,0]

output list

list_c = [0, 0, 1, 0.9, 0.8, 0.7, 1, 0.9, 0.8, 1, 0.9]

like a saw tooth list.

What is `itself` supposed to be a reference to? The as-yet uncompleted list comprehension…?! — deceze, Jan 11 '18 at 10:20
You cannot refer to things that **do not exist** yet. And `list_c` does not exist inside the list-comprehension that *creates it*. You have to go with a for-loop here or find a mathematical way to re-write your comprehension that only references objects that exist — Ma0, Jan 11 '18 at 10:23
People are mocking this question but it is perfectly reasonable, and I would go so far as to say that Python *should* allow self-referential list comprehensions-I'm sure there are reasons why this isn't supported, but a priori this seems like a perfectly reasonable feature to have in a language (Haskell, for example, supports this...). The snarky comments are unwarranted. — Ben Kushigian, Jun 10 '18 at 18:32

score 9 · Accepted Answer · answered Jan 11 '18 at 10:24

List comprehension is meant to be a compact expression for the most straightforward use cases. You cannot reference the list itself in list comprehension, for that just use a regular for-loop:

list_c = []
for i, a in enumerate(list_a):
    if i == 0:
        list_c.append(a ** 2)
    else:
        list_c.append(a * 2 + list_c[i-1])

You can also rewrite that loop in a more efficient way:

list_a_iterator = iter(list_a)
list_c = [next(list_a_iterator) ** 2]  # May raise StopIteration if list_a is empty
for item in list_a_iterator:
    list_c.append(item * 2 + list_c[-1])

score 2 · Answer 2 · answered Jan 11 '18 at 10:34

2

You cannot reference a list before it is created. However, you can use this hacky reduce approach if you desperately want a one-liner:

list_c = reduce(lambda lst, a: lst + [lst[-1] + a**2], list_a, [0])[1:]

And in Python > 3.2, you can use itertools.accumulate:

from itertools import accumulate
list_c = list(accumulate(a**2 for a in list_a))

answered Jan 11 '18 at 10:34

user2390182

67,685
6
55
77

Thank you! I don't know itertools. I will try to learn this library. – Hiroyuki Taniichi Jan 11 '18 at 11:01
Well, you can't reference it before it is created, but it is created at the *start* of the list comprehension. I'm even able to actually reference it and use it *during* the list comprehension. Implementing the OP's desired list comp: https://repl.it/repls/NoxiousWordyBracket#main.py (Yes, of course I know it's absolutely awful, just having some fun and making some point :-) – Stefan Pochmann Oct 02 '20 at 16:46

zipa · Answer 3 · 2018-01-11T10:41:34.813

1

You can access list_a from your comprehension:

[a**2 if i == 0 else a*2 + list_a[i-1]**2 if i == 1 else a*2 + list_a[i-1]*2 for i, a in enumerate(list_a)]

edited Jan 11 '18 at 10:41

answered Jan 11 '18 at 10:36

zipa

26,044
6
38
55

Python : How to refer itself in the list comprehension?

ADD

input list

output list

3 Answers3