I am trying to run my python script on docker. I tried different ways to do it but not able to run it on docker. My python script is given below:
import os
print ('hello')
I have already installed docker on my mac. But i want to know how i can make images and then push it to docker after that i wanna pull and run my script on docker itself.
Going by question title, and if one doesn't want to create docker image but just want to run a script using standard python docker images, it can run using below command
docker run -it --rm --name my-running-script -v "$PWD":/usr/src/myapp -w /usr/src/myapp python:3.7-alpine python script_to_run.py
Alright, first create a specific project directory for your docker image. For example:
mkdir /home/pi/Desktop/teasr/capturing
Copy your dockerfile and script in there and change the current context to this directory.
cp /home/pi/Desktop/teasr/capturing.py /home/pi/Desktop/teasr/dockerfile /home/pi/Desktop/teasr/capturing/
cd /home/pi/Desktop/teasr/capturing
This is for best practice, as the first thing the docker-engine does on build, is read the whole current context.
Next we'll take a look at your dockerfile. It should look something like this now:
FROM python:latest
WORKDIR /usr/local/bin
COPY capturing.py .
CMD ["capturing.py", "-OPTIONAL_FLAG"]
The next thing you need to do is build it with a smart name. Using dots is generally discouraged.
docker build -t pulkit/capturing:1.0 .
Next thing is to just run the image like you've done.
docker run -ti --name capturing pulkit/capturing:1.0
The script now get executed inside the container and will probably exit upon completion.
Edit after finding the problem that created the following error:
standard_init_linux.go:195: exec user process caused "exec format error"
There's a different architecture beneath raspberry pi's (ARM instead of x86_64), which COULD'VE BEEN the problem, but wasn't. If that would've been the problem, a switch of the parent image to FROM armhf/python would've been enough.
Source
BUT! The error kept occurring.
So the solution to this problem is a simple missing Sha-Bang on top of the python script. The first line in the script needs to be #!/usr/bin/env python and that should solve the problem.
Source
You need to create a dockerfile in the directory your script is in.
You can take this template:
FROM python:latest
COPY scriptname.py /usr/local/share/
CMD ["scriptname.py", "-flag"]
Then simply execute docker build -t pulkit/scriptname:1.0 . and your image should be created.
Your image should be visible under docker images. If you want to execute it on your local computer, use docker run.
If you want it to upload to the DockerHub, you need to log into the DockerHub with docker login, then upload the image with docker push.
I Followed #samprog (most accepted) answer on my machine running on UBUNTU VERSION="14.04.6".
and was getting "standard_init_linux.go:195: exec user process caused "exec format error"
None of the solution worked for me mentioned above.
Fixed the error after changing my Dockerfile as follows
FROM python:latest
COPY capturing.py ./capturing.py
CMD ["python","capturing.py"]
Note: If your script import some other module then you need to modify COPY statement in your Dockerfile as follows - COPY *.py ./
Hope this will be useful for others.
Another way to run python script on docker can be:
copy the local python script to docker:
docker cp yourlocalscript.path container_id:/dst_path/
container id can be found using:
docker ps
run the python script on docker:
docker exec -it python /container_script_path.py
its very simple
1- go to your Python script directory and create a file with this title without any extension
Dockerfile
2-now open the docker file and write your script name instead of sci.py
( content of Dockerfile )
FROM python:slim #i choice slim version you can choose another tag for example python:3
WORKDIR /usr/local/bin
COPY sci.py . #replace you scrip name with sci.py
CMD [ "python", "sci.py" ] #replace you scrip name with sci.py
save it and now you should create image file from this dockerfile and script py
and next run it
3-in path address folder write CMD and press Enter key :
4-When the cmd window opens for you, type in it :
docker build -t my-python-app . #this create image in docker by this title my-python-app
5- and findly run image:
docker run -it --rm --name my-running-app my-python-app
I've encountered this problem recently, this dependency HELL between python2 and python3 got me. Here is the solution.
Bind your current working directory to a Docker container with python2 and pip2 running.
Pull the docker image.
docker pull frolvlad/alpine-python2
Add this alias into /home/user/.zshrc or /home/user/.bashrc
alias python2='docker run -it --rm --name python2 -v "$PWD":"$PWD" -w
"$PWD" frolvlad/alpine-python2'
Once you type python2 into your CMD you'll be thrown into the Docker instance.
Related
First off, I might have formulated the question inaccurately, feel free to modify if necessary.
Although I am quite new to docker and all its stuff, yet somehow managed to create an image (v2) and a container (cont) on my Win 11 laptop. And I have a demo.py which requires an .mp4 file as an arg.
Now, if I want to run the demo.py file, 1) I go to the project's folder (where demo.py lives), 2) open cmd and 3) run: docker start -i cont. This starts the container as:
root:834b2342e24c:/project#
Then, I should copy 4) my_video.mp4 from local project folder to container's project/data folder (with another cmd) as follows:
docker cp data/my_video.mp4 cont:project/data/.
Then I run: 5) python demo.py data/my_video.mp4. After a while, it makes two files: my_video_demo.mp4 and my_video.json in the data folder in the container. Similarly, I should copy them back to my local project folder: 6)
docker cp cont:project/data/my_video_demo.mp4 data/, docker cp cont:project/data/my_video_demo.json data/
Only then I can go to my local project/data folder and inspect the files.
I want to be able to just run a particular command that does 4) - 6) all in one.
I have read about -v option where, in my case, it would be(?) -v /data:project/data, but I don't know how to proceed.
Is it possible to do what I want? If everything is clear, I hope to get your support. Thank you.
You should indeed use Docker volumes. The following command should do it.
docker run -it -v /data:project/data v2
Well, I think I've come up with a way of dealing with it.
I have learned that with -v one can create a place that is shared between the local host and the container. All you need to do is that run the docker and provide -v as follows:
docker run --gpus all -it -v C:/Workspace/project/data:/project/data v2:latest python demo.py data/my_video_demo.mp4
--gpus - GPU devices to add to the container ('all' to pass all GPUs);
-it - tells the docker that it should open an interactive container instance.
Note that every time you run this, it will create a new container because of -it flag.
Partial credit to #bill27
I am struggling with running the latest changes. Below are the details.
Dockerfile
FROM python:3.7.3
RUN mkdir -p /usr/apps
COPY test.py /usr/apps
RUN pip install mindsdb
CMD [ "python","test.py" ]
Build
docker build -t py37:custom .
Run
docker run -it -v /Development/PetProjects/mindsdb:/usr/apps/ py37:custom
But it shows only the changes at the time of build.
First of all while starting your container you are not using volumes but bind mounts. So you mount directory /Development/PetProjects/mindsdb on your host machine to /usr/apps/ directory. Every change made to files on your host machine in this directory, will be visible in the container, and the other way round.
If you wanted to use volumes, you could create one using docker volume create command and then running container with this volume : docker container run -v volume_name:path_in_container image_name. Then you would be able to stop container and run it again by passing this volume to run command and changes to path_in_container directory could be stored across container creations.
Another thing is that you are trying to mount /usr/apps/ in your container and you copied a python script there using Dockerfile. Note that in you current docker run command contents of /Development/PetProjects/mindsdb will replace content of /usr/apps/ in your container and if you do not have your script in /Development/PetProjects/mindsdb - script will not be visible in the container.
Moreover your CMD seems not to work because of path relativeness. You should change your CMD to CMD [ "python","/usr/apps/test.py" ] or use WORKDIR option - WORKDIR /usr/apps/ so your python command could be executed from this directory and script could be visible there.
More information about differences between volumes and bind mounts can be found in docker documentation.
I have set up Docker Toolbox on a Win 10 machine. I have some simple single file Python scripts that I want to run in Docker, just for learning purpose.
Started learning Docker today, and Python 3 days ago.
I assume I have set up Docker correctly, I can run the example hello-world image. No error messages during setup.
I am following an instruction from here https://runnable.com/docker/python/dockerize-your-python-application,
which says:
If you only need to run a simple script (with a single file), you can avoid writing a complete Dockerfile. In the examples below, assume you store my_script.py in /usr/src/widget_app/, and you want to name the container my-first-python-script:
docker run -it --rm --name my-first-python-script -v "$PWD":/usr/src/widget_app python:3 python my_script.py
If I type pwd, it shows:
/c/Program Files/Docker Toolbox
And the script I want to run is located here:
C:\Docker\Python\my_script.py
This is what I think should work:
docker run -it --rm --name my-first-python-script -v "$PWD":/c/Docker/Python python:3 python my_script.py
No matter how I try to specify the file directory, I get an error:
python: can't open file 'my_script.py': [Errno 2] No such file or directory
When you run -v "$PWD":/c/Docker/Python, you are saying you want to link your current working directory to the path /c/Docker/Python in the container, which isn't what you want to do. What you are trying to do is link C:\Docker\Python\ on your host to the container folder /usr/src/widget_app.
This command will put your script inside the container path /usr/src/widget_app, then run it:
docker run -it --rm --name my-first-python-script -v /c/Docker/Python:/usr/src/widget_app python:3 python /usr/src/widget_app/my_script.py
I have a python code and to convert it to docker image, I can use below command:
sudo docker build -t customdocker .
This converts python code to docker image. To convert it I use a Dockerfile with below commands:
FROM python:3
ADD my_script.py /
ADD user.conf /srv/config/conf.d/
RUN pip3 install <some-package>
CMD [ "python3", "./my_script.py" ]
In this, we have RUN command which install required packages. Lets say if we have deleted the image for some reason and want to build it again, so is there any way we can skip this RUN step to save some time because I think this is already installed.
Also in my code I am using a file user.conf which is in other directory. So for that I am including this in DOckerfile and also saving a copy of it in current directory. Is there a way in docker where I can define my working directory so that docker image searches for the file inside those directories.
Thanks
Yes you cannot remove the RUN or other statements in dockerfile, if you want to build the docker image again after deleteing.
You use the command WORKDIR in your dockerfile but its scope will be within the docker images, i.e when you create the container from the image workdir will be set to that metioned in WORKDIR
For ex :
WORKDIR /srv/config/conf.d/
This /srv/config/conf.d/ will set as workingdir, but you have to use below in dockerfile while building in-order to copy that file in specified location
ADD user.conf /srv/config/conf.d/
Answering your first question: A docker image holds everything related to your python environment including the packages you install. When you delete the image then the packages are also deleted from the image. Therefore, no you cannot skip that step.
Now on to your second question, you can bind a direectory while starting the container by:
docker run -v /directory-you-want-to-mount:/src/config/ customdocker
You can also set the working directory with -w flag.
docker run -w /path/to/dir/ -i -t customdocker
https://docs.docker.com/v1.10/engine/reference/commandline/run/
I've been searching a lot for the past days reagarding Dockerfile. I'm using cx_Oracle in python 2.7. Here's how my Dockerfile looks like:
FROM sbanal/python-oracle-xe12.1-latest
WORKDIR /code/app
COPY generate_distance.py /code/app/app.py
COPY generate_values.py /code/app/app2.py
To make it easier to explain, I've made a method to print out the name of the file. In generate_distance.py:
def test():
print "Generate distance"
test()
In generate_values.py:
def test():
print "Generate values"
test()
Then I'm running docker build with a tag:
docker build -t gen .
Sending build context to Docker daemon 13.82kB
Step 1/4 : FROM sbanal/python-oracle-xe12.1-latest
---> 723335924016
Step 2/4 : WORKDIR /code/app
---> Using cache
---> 9fde6fb3ac02
Step 3/4 : COPY generate_distance.py /code/app/app.py
---> 1dbf7ef85ee3
Removing intermediate container ae626dcef48c
Step 4/4 : COPY generate_values.py /code/app/app2.py
---> 7a54500b88a3
Removing intermediate container f496edfc237d
Successfully built 7a54500b88a3
Successfully tagged gen:latest
When running 'docker images', I can see the 'gen' image. But when I run the 'gen' image, only app.py is working:
>docker run -p 5500:5000 gen
>Generate distance
I can't see what mistake I've done. I also don't know why it has to be called app.py. If I use different file name during COPY in Dockerfile, I get 'No such file or directory' error. That is:
FROM sbanal/python-oracle-xe12.1-latest
WORKDIR /code/app
COPY generate_relation_distance.py /code/app/generate_relation_distance.py
COPY generate_ten_values.py /code/app/generate_ten_values.py
Build and run like the part over:
docker run -p 5500:5000 gen
python: can't open file 'app.py': [Errno 2] No such file or directory
Hope someone can help me :)
Your image is based on sbanal/python-oracle-xe12.1-latest (first line of your Dockerfile).
In this Dockerfile is a "CMD" defined, which specifies the first command of your container. Here, that is
CMD python app.py
(see last line of your base image).
The command will be executed as sh -c "python app.py".
This is why your Dockerfile starts on container creation python app.py
You need to override the "CMD" part in your Dockerfile, e.g.
CMD ["python", "app2.py"]
See the official docker docs to understand CMD.
You should only have one CMD in your Dockerfile containing the first command, which is automatically executed by the container.
If you want to start multiple services, you should consider, if this should really be packed into one image. Or you follow the official docs and consider using a supervisor or a script, which starts your desired services.
If anyone wondering how I solved the problem, I used bash script instead.
script.sh:
#Build the image
docker build -t image_name .
#Run image image_name in container
docker run -d -p 550:500 image_name tail -f /dev/null
#Get the container id
container_id="$(docker ps | grep $IMAGE_NAME | grep -Eo '^[^ ]+')"
#Run your program in container
docker exec -it container_id /bin/sh -c "python generate_distance.py"
docker exec -it container_id /bin/sh -c "python generate_values.py"
My goal was to store the output in text files and copy those from docker container to localhost.