# 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 _)
...
```

## Answers:

```
# 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 `_`

.

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.

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
```