scvelo.tl.velocity_graph throws ValueError

[alexblaessle]

Hi

I was trying to reproduce the intro tutorial, however, I am getting a ValueError when running scv.tl.velocity_graph(adata).

import scvelo as scv

adata=scv.datasets.pancreas()

scv.pp.filter_genes(adata, min_shared_counts=20)
scv.pp.normalize_per_cell(adata)
scv.pp.filter_genes_dispersion(adata, n_top_genes=2000)
scv.pp.log1p(adata)

scv.pp.filter_and_normalize(adata, min_shared_counts=20, n_top_genes=2000)
scv.pp.moments(adata, n_pcs=30, n_neighbors=30)

scv.tl.velocity(adata)

scv.tl.velocity_graph(adata)

Error output

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
Cell In[10], line 1
----> 1 scv.tl.velocity_graph(adata)

File ~/venvs/scvelo/lib/python3.11/site-packages/scvelo/tools/velocity_graph.py:363, in velocity_graph(data, vkey, xkey, tkey, basis, n_neighbors, n_recurse_neighbors, random_neighbors_at_max, sqrt_transform, variance_stabilization, gene_subset, compute_uncertainties, approx, mode_neighbors, copy, n_jobs, backend)
    359 n_jobs = get_n_jobs(n_jobs=n_jobs)
    360 logg.info(
    361     f"computing velocity graph (using {n_jobs}/{os.cpu_count()} cores)", r=True
    362 )
--> 363 vgraph.compute_cosines(n_jobs=n_jobs, backend=backend)
    365 adata.uns[f"{vkey}_graph"] = vgraph.graph
    366 adata.uns[f"{vkey}_graph_neg"] = vgraph.graph_neg

File ~/venvs/scvelo/lib/python3.11/site-packages/scvelo/tools/velocity_graph.py:175, in VelocityGraph.compute_cosines(self, n_jobs, backend)
    172 n_obs = self.X.shape[0]
    174 # TODO: Use batches and vectorize calculation of dX in self._calculate_cosines
--> 175 res = parallelize(
    176     self._compute_cosines,
    177     range(self.X.shape[0]),
    178     n_jobs=n_jobs,
    179     unit="cells",
    180     backend=backend,
    181 )()
    182 uncertainties, vals, rows, cols = map(_flatten, zip(*res))
    184 vals = np.hstack(vals)

File ~/venvs/scvelo/lib/python3.11/site-packages/scvelo/core/_parallelize.py:138, in parallelize.<locals>.wrapper(*args, **kwargs)
    126     pbar, queue, thread = None, None, None
    128 res = Parallel(n_jobs=n_jobs, backend=backend)(
    129     delayed(callback)(
    130         *((i, cs) if use_ixs else (cs,)),
   (...)
    135     for i, cs in enumerate(collections)
    136 )
--> 138 res = np.array(res) if as_array else res
    139 if thread is not None:
    140     thread.join()

ValueError: setting an array element with a sequence. The requested array has an inhomogeneous shape after 2 dimensions. The detected shape was (1, 4) + inhomogeneous part.

Versions

python 3.11.1

scvelo==0.2.5  scanpy==1.9.3  anndata==0.9.1  loompy==3.0.7  numpy==1.24.3  scipy==1.10.1  matplotlib==3.7.1  sklearn==1.2.2  pandas==2.0.1  

[alexblaessle]

Ok, I did some more digging. Downgrading from numpy-1.24.3 to numpy-1.23.5 fixes the problem. However, not working with the latest numpy version.

[WeilerP]

Closing as duplicate of #1058.

[tobylanser]

@WeilerP I wrestled with this too. Tried downgrading to numpy==1.21.1 which didn't work, but 1.23.5 makes it run smoothly. Thanks!

[lzygenomics]

@alexblaessle I occur this bug too. Downgrading from numpy-1.24.3 to numpy-1.23.5 is working! Thank you! @WeilerP And I think maybe it is not totally duplicate of #1058, following your solution for #1058, the bug on scv.tl.recover_dynamics is fixed perfectly~ but the bug on scv.tl.velocity_graph still remain until downgrading numpy. Thanks!

[Tomzhangmx]

I occur this bug too. Manipulating numpy versions didn't work.

[Zodiac111111]

I changed numpy to 1.23.5, but it still not work. then I updated the TensorFlow 2, it suddenly worked. I don't know if this can fix the problem, just for your reference.

[liyiyuian]

I'm not reporting new errors; I just want to let people know changing to 1.23.5 makes it work for me.

[pyhell]

Hello,
For those that might have the same problem. I tried downgrading to numpy 1.23.5 and it did not work. I had Python 3.11 and had to go down to 3.8 for it to work.