Expression that returns mutated list

Question

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)]

Edit:

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

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Source

Answer 1

# 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

Answer 2

Try list concatenation:

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

out = rng[:key] + [value] + rng[key+1:]
print(out)

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.

Answer 3

Maybe better than list concatenation:

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

Another:

[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

Expression that returns mutated list

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