faridrashidi / scphylo-tools Goto Github PK

Description

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Minimal reproducible example

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Traceback

Version

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Hi Farid,

I am currently making heavy use of your implementation of the HUNTRESS algorithm.
tl.huntress

Do you have some guidance on the resources to allocate to HUNTRESS?

I am having some trouble running HUNTRESS.
When using a single thread HUNTRESS runs fine for small cell-mutation matrices (200 cells 10 mutations).

I tried upping the threads to 8 and ram to 2 GB in the light of larger mutation matrices (1000 cells, 50 mutations), yet it appears that even for the small runs (200 cells 10 mutations) multiple threads bring trouble - that I don't understand.

HUNTRESS seems to start up fine with the usual first line of output:
running HUNTRESS with alpha=1e-06, beta=0.1, n_threads=8

but just does not progress after that for more than 3 hrs, which is way more than the runtime the supplementary paper of HUNTRESS suggests.
Surprisingly, I cannot find a high CPU usage related to it. When multiple threads are run no significant resources are used. Tried multiple machines and setups.

Is there a way to check if HUNTRESS is making progress - A debug mode/log ?

Thanks in advance! Best Regards,

Gordon

Description

Running the scphylo.tl.ad() returns wrong values and division by zero errors.
Confident about the usage of input, see examples.

Minimal reproducible example

Example 1: wrong output

In this example scphylo.tl.ad() returns a wrong value, compared to the attached by-hand calculation for these small trees. See input arrays and tree topology attached.

tree1 / df1:
         0  3  2  1
Cell-1  0  0  0  0
Cell-2  0  0  1  0
Cell-3  0  0  1  1
Cell-4  1  0  1  1
Cell-5  1  1  1  1 

tree2 / df2:
         0  3  2  1
Cell-1  0  0  0  0
Cell-2  0  0  0  1
Cell-3  0  1  0  1
Cell-4  1  1  0  1
Cell-5  1  1  1  1

tree1:
4
└── 2
    └── 1
        └── 0
            └── 3
tree2:
4
└── 1
    └── 3
        └── 0
            └── 2
scphylo AD: 0.33333333333333337
correct AD: 0.42857142857142855

By hand calculation:

Example 2: Division by Zero

In this example the scphylo.tl.ad() fails.

tree 1 / df1:
         2  0  1  3
Cell-1  0  0  0  0
Cell-2  0  1  0  0
Cell-3  0  0  1  0
Cell-4  1  0  0  0
Cell-5  0  0  0  1 

tree 2 / df2:
         2  0  1  3
Cell-1  0  0  0  0
Cell-2  1  0  0  0
Cell-3  1  1  0  0
Cell-4  1  0  1  0
Cell-5  0  0  0  1

tree 1: 
4
├── 0
├── 1
├── 2
└── 3
tree 2: 
4
├── 2
│   ├── 0
│   └── 1
└── 3

These inputs lead to a Division by Zero error.

See Traceback.

Traceback

for Example 2, this resulted in a ZeroDivisionError in the very last line of the ad function.

Version

0.0.2

Unofficial fork at
https://github.com/pawel-czyz/scphylo-tools

Description

schphylo might be having some small special case not handled as such. For example, it breaks if the input consists of all ones. When I changed just one of the 1s to zero in the input, everything was fine.

Minimal reproducible example

mport scphylo
import trisicell as tsc
df_input = scphylo.io.read("temp_scistree.tsv")
alpha = 0.001
beta  = 0.2
conflict_free_matrix = scphylo.tl.scistree(df_input, alpha, beta, experiment=False)
tree = tsc.ul.to_tree(conflict_free_matrix)
tsc.pl.dendro_tree(tree)

Traceback

</details>

### Version
<!-- Output of scphylo.__version__ -->
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Hi Farid - just a minor suggestion about the MLTD.

scphylo.tl.mltd(df1, df2)

Compared to the other distances you provide, it does not say what the range of the output is,
I notice most of the other distances state "Similarity out of one."

As I understand the MLTD paper, it may or may not be normalized to be in the range [0,1].

Which do you implement? - Perhaps worth adding to the docstring.

Kindest Regards,

Gordon