Python : How to refer itself in the list comprehension? [duplicate]

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.

Answer 1

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])

Answer 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))

Answer 3

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)]