Expression that returns mutated list


I’m looking for a single expression, mutates an element and returns the modified list

The following is a bit verbose

# key=0; value=3; rng=[1,2]
[(v if i != key else value) for i, v in enumerate(rng)]


I’m looking for a way to inline the following function in a single expression

def replace(rng: List, key: int, value):
    a = list(rng) 
    a[key] = value
    return a

Edit 2: the code that actually motivated this question

class TextDecoder(nn.Module):
    def forward(self, x: Tensor, kv_cache: Tensor):
        kv_cache_write = torch.zeros((_:=list(kv_cache.shape))).__setitem__(-2, x.shape[-1]) or _)
Asked By: Tom Huntington



# key=0; value=3; rng=[1,2]
print(rng.__setitem__(key, value) or rng)

If you don’t want to modify the original list rng, you can do

print((_:=list(rng)).__setitem__(key, value) or _)

unfortunately this leaks the variable _.

Answered By: Tom Huntington

Try list concatenation:

key = 0
value = 3
rng = [1, 2]

out = rng[:key] + [value] + rng[key+1:]

rng[:key] is a copy of the list up to the key (exclusive), [value] is a new list where the only element is value, and rng[key+1] is a copy of the list from the key on (exclusive). Concatenate these together, and you get a copy where the key is replaced.

Answered By: Michael M.

Maybe better than list concatenation:

[*rng[:key], value, *rng[key+1:]]


[a for a, a[key] in [(rng[:], value)]][0]

Or if you are just assigning the result to a variable (as discussed) you can do:

a, a[key] = rng[:], value
Answered By: Kelly Bundy