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Not an answer to your question but just FYI `something.property.apply(lambda name: name)` is equivalent to `something.property` . The lambda function is just the identity function so no transformation is performed.

looks like you need
```clusterName: rancher_ci_cluster.name```

you are wrapping it in a `{}` but it looks like that property probably just wants a `string`

yeah just want a string but in an yaml format and i'm passing the value dynamically from previous resource

ohh right. does Pulumi Python have the `interpolate` function?

doing string interpolation with an `Output&lt;string&gt;` doesn't work

<https://www.pulumi.com/docs/intro/concepts/inputs-outputs/#outputs-and-strings>

oh so need to use `all()` and `apply()` together?

yeah i think soemthing like this will work. might be an even better way to format it but i forget if python supports array param deconstruction
```    values=Output.all(config.require("rancherAwsCiRegion"), rancher_ci_cluster.name).apply(lambda s: f"""
awsRegion: {s[0]}
autoDiscovery:
  clusterName: {s[1]}
extraArgs:
  balance-similar-node-groups: true
priorityClassName: system-cluster-critical
""",```

might have to make that function a named function if python doesn't allow it for being multi-line

it might fail having `config` (as this is `pulumi.Config` ) in `Output` as its not a resource

`Output.all` just takes Outputs. it doesn't need to be from a cloud resource

perfect , thanks for help. appreciate it :clap:

on side note you mentioned this is `catalog.name.apply(lambda name: name)` same as `catalog.name` right, but `catalog.name` was returning object, after using `apply()`  only it returned raw output, any thoughts on it? btw the catalog here is `rancher2.CatalogV2` resource

A Pulumi `Output&lt;string&gt;` is an object. It is not the same as a `string`. So you can't use it in normal string interpolation. The reason why you got an object reference earlier is because you used it python's string interpolation, and the string value of an object is that string you got with the memory address.

What `apply` does is it allows you to access the value wrapped by `Output`. Within the callback function, you can treat an `Output&lt;string&gt;` as a normal string. This allows you to use `s[0]` and `s[1]` in the code block above as regular strings.

`catalog.name.apply(lambda name: name)` is the same as `catalog.name` because the callback function you passed in is the identity function, not for any other reason.

understood, thanks for sharing out in detail. this helps.