How to create a Conda environment based on a YAML file?
Question:
I am working on a online course and the instructors have asked me to create a Conda environment, and they have provided a YAML file to provide the library and package requirements for the Conda package.
However, I am unable to find a way to create the environment using the YAML file.
Answers:
conda env create takes an optional flag --file:
-f FILE, –file FILE environment definition file (default: environment.yml)
So do:
conda env create --file=myfile.yaml
Of course, replace =myfile.yaml with your YAML file name.
Use:
conda env create --name NAME --file FILE
where FILE is the YAML file and NAME is what you want to name the environment.
(The accepted answer used to suggest conda create, but that only works on the output of conda list --explicit, not on Conda environment YAML files.)
conda env create --file=dand-env-win.yaml
worked well for me where my file name was dand-env-win.yaml for a course on the Internet.
I am working on a online course and the instructors have asked me to create a Conda environment, and they have provided a YAML file to provide the library and package requirements for the Conda package.
However, I am unable to find a way to create the environment using the YAML file.
conda env create takes an optional flag --file:
-f FILE, –file FILE environment definition file (default: environment.yml)
So do:
conda env create --file=myfile.yaml
Of course, replace =myfile.yaml with your YAML file name.
Use:
conda env create --name NAME --file FILE
where FILE is the YAML file and NAME is what you want to name the environment.
(The accepted answer used to suggest conda create, but that only works on the output of conda list --explicit, not on Conda environment YAML files.)
conda env create --file=dand-env-win.yaml
worked well for me where my file name was dand-env-win.yaml for a course on the Internet.