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I need a Matlab coder.

Job Description:

Image processing using deep learning for bone fracture detection.

It should compromise training, testing and prediction

Skills: Matlab and Mathematica, Algorithm, Engineering, MATLAB, Coding

About the Client:
( 0 reviews ) India

Project ID: #35862036

8 freelancers are bidding on average ₹1081 for this job

vikram2web

Hi there, I have 8 years of experience in Matlab. Let's get into chat when you are available. Thanks

₹1050 INR in 7 days
(35 Reviews)
4.7
scriptwriterx

Hi there I have read your project discription and I'm confident I can do this project for you perfectly . I still have few questions , come inbox for further details . Thankyou

₹600 INR in 2 days
(4 Reviews)
2.7
lalitdeveloper7

I can do the task Image processing using deep learning for bone fracture detection. Let me know more about it I am a freelancer with having 6 years of experience. I've gone through your project. I have read your requir More

₹1500 INR in 1 day
(3 Reviews)
1.5
DianaBlueCrystal

I am familiar with deep learning on MATLAB and made a lot of projects with it. From preprocessing to classification with a file of performance metrics results. You can contact me

₹1050 INR in 3 days
(1 Review)
0.6
kashishvaishnav2

“Honestly, I possess all the skills and experience that you're looking for. I'm pretty confident that I am the best candidate for this job role. It's not just my background in the past projects, but also my people skil More

₹1050 INR in 7 days
(0 Reviews)
0.0
faridaashraff

I just finished my 5 year engineering journey and I have been studying matlab since my first year in uni. I have also done a lot of similar projects and I can finish the whole task in less than 5 days!

₹1500 INR in 5 days
(0 Reviews)
0.0
macesco

I am biomedical engineer with great experience in Matlab and C/C++ programming, using also Excel since many years. I have worked a lot with processing of biomedical images at university. I am new here in [login to view URL] More

₹850 INR in 4 days
(0 Reviews)
0.0