gaut's questions

我执行此操作是为了使用 terraform 在本地构建 docker 镜像。第一次运行成功，但每次尝试都会失败，Error: failed to read dockerfile: unexpected EOF除非我完全手动删除该terraform.tfstate文件。

terraform {
  backend "local" {}
    required_providers {
      docker = {
        source  = "kreuzwerker/docker"
        version = "~>3.0.2"
      }
    # other providers
}
provider "docker" {

}

resource "docker_image" "dbx_kedro_img" {
  name         = "docker-build-test:latest" 
  build {
    context = "."
    dockerfile = "./Dockerfile"
  }
}

我尝试将版本修改为docker-build-test:latest2或docker-build-test2:latest，运行terraform state rm docker_image.dbx_kedro_img，删除本地 docker 镜像和容器，但没有成功。唯一有效的方法是手动删除terraform.tfstate。欢迎提出任何想法。

docker rm $(docker ps -aq) # not useful
docker rmi $(docker images -q) -f # idem
terraform state rm docker_image.dbx_kedro_img # idem

我的docker文件：

# Use a lightweight base image
FROM alpine:latest
#
# Set the command to run when the container starts
CMD ["echo", "Hello, World3!"]

编辑：添加这样的触发器后，行为相同，第一次尝试后失败，即使修改了图像名称。

# not useful
resource "docker_image" "dbx_kedro_img" {
  name         = "docker-build-test5:latest" 
  build {
    context = "."
    dockerfile = "./Dockerfile"
  }
    triggers = {
    always_rebuild = timestamp()
  }
}

所以我正在开发这个功能

#' Disambiguated Equality operator between two vectors that may contain NA's
#'
#' Forces the equality result to be TRUE if both corresponding vector elements are either equal, or both NA.
#' @param x vector of any type compatible with is.na and ==.
#' @param y vector of any type compatible with is.na and ==.
#' @export
#' @examples 
#' c(NA, 1:5) %==% c(NA, 1:3,"nope", NA) #[1]  TRUE  TRUE  TRUE  TRUE FALSE FALSE

gp.is.equal.force <- `%==%` <- function(x, y, vect = T) {
  res <- (is.na(x) & is.na(y)) | (!is.na(x) & !is.na(y) & x == y)
  if (!vect) res <- all(res)
  res
}

我注意到 R 版本 4.3.2 (2023-10-31) 上的 NaN 值出现了我没有预料到的行为——“Eye Holes”。

> is.na(NaN) # [1] TRUE
> is.na(c(NaN, 5)) # [1]  TRUE FALSE
> is.na(c(NaN, NA)) # [1] TRUE TRUE
> is.nan(c(NaN, NA)) # [1]  TRUE FALSE
> is.na(c(NaN, as.factor("abc"))) # [1]  TRUE FALSE
> is.nan(c(NaN, 5)) # [1]  TRUE FALSE

到目前为止一切都很好，但是：

> is.na(c(NA, NaN, "abc")) # [1]  TRUE FALSE FALSE
> is.nan(c(NA, NaN, "abc")) # [1] FALSE FALSE FALSE
> is.nan(c(NaN, "abc")) #[1] FALSE FALSE

似乎将characters 添加到混合中会以某种方式破坏值的逻辑NaN。是不是很奇特？