# Partition error: mesh has 1 parts

**URL:** <https://pyfr.discourse.group/t/partition-error-mesh-has-1-parts/1273>\
**Category:** Errors\
**Created:** [28 June 2025 17:32 UTC](https://pyfr.discourse.group/t/partition-error-mesh-has-1-parts/1273 "2025-06-28T17:32:27Z")\
**Posts on this page:** 2\
**Page:** 1

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**Author:** ![Amirshau](https://avatars.discourse-cdn.com/v4/letter/a/c89c15/32.png) [@Amirshau](https://pyfr.discourse.group/u/Amirshau)\
**Post date:** [28 June 2025 17:32 UTC](https://pyfr.discourse.group/t/partition-error-mesh-has-1-parts/1273/1 "2025-06-28T17:32:27Z")

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Hello evryone,

I’m trying to partion a mesh file, but as I run the line I’m getting the following error:  
line: `$ pyfr partition -p metis 2 -e balanced try.pyfrm .`

```bash
Traceback (most recent call last):
  File "/pyfr-venv3/bin/pyfr", line 8, in <module>
    sys.exit(main())
  File "/pyfr-venv3/lib/python3.10/site-packages/pyfr/ __main__.py", line 124, in main
    args.process(args)
  File "/pyfr-venv3/lib/python3.10/site-packages/pyfr/ __main__.py", line 190, in process_partition
    mesh, rnum, part_soln_fn = part.partition(mesh, args.progress)
  File "/envs/pyfr-venv3/lib/python3.10/site-packages/pyfr/partitioners/base.py", line 396, in partition
    raise RuntimeError(f'Partitioner error: mesh has {n} '
RuntimeError: Partitioner error: mesh has 1 parts versus goal of 2

```

Any suggestions?

---

<div class="post-metadata">

**Author:** ![fdw](https://avatars.discourse-cdn.com/v4/letter/f/a587f6/32.png) [@fdw](https://pyfr.discourse.group/u/fdw)\
**Post date:** [29 June 2025 03:53 UTC](https://pyfr.discourse.group/t/partition-error-mesh-has-1-parts/1273/2 "2025-06-29T03:53:30Z")

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How many elements are in your mesh? Are you able to post a picture of your mesh?

This error typically occurs when you have a similar number of elements to partitions. With that said there are some known bugs in METIS which can trigger it for larger element counts. If this is the case then please consider using a different partitioner.

Regards, Freddie.