load_chembl1871_ki

skfp.datasets.moleculeace.load_chembl1871_ki(data_dir: str | PathLike | None = None, as_frame: bool = False, verbose: bool = False) → DataFrame | tuple[list[str], ndarray]

Load the ChEMBL1871 Ki dataset.

The task is to predict the inhibitor constant (Ki) of molecules against the Androgen receptor target [1] [2].

Tasks	1
Task type	regression
Total samples	659
Recommended split	activity_cliff
Recommended metric	RMSE

Parameters:

• data_dir ({None, str, path-like}, default=None) – Path to the root data directory. If None, currently set scikit-learn directory is used, by default $HOME/scikit_learn_data.

• as_frame (bool, default=False) – If True, returns the raw DataFrame with columns: “SMILES”, “label”. Otherwise, returns SMILES as list of strings, and labels as a NumPy array (1D integer binary vector).

• verbose (bool, default=False) – If True, progress bar will be shown for downloading or loading files.

Returns:

data – Depending on the as_frame argument, one of: - Pandas DataFrame with columns: “SMILES”, “label” - tuple of: list of strings (SMILES), NumPy array (labels)

Return type:

pd.DataFrame or tuple(list[str], np.ndarray)

References

[1]

D. van Tilborg, A. Alenicheva, and F. Grisoni “Exposing the Limitations of Molecular Machine Learning with Activity Cliffs” Journal of Chemical Information and Modeling, vol. 62, no. 23, pp. 5938–5951, Dec. 2022

[2]

B. Zdrazil et al. “The ChEMBL Database in 2023: a drug discovery platform spanning multiple bioactivity data types and time periods” Nucleic Acids Research, vol. 52, no. D1, Nov. 2023

Examples

>>> from skfp.datasets.moleculenet import load_chembl1871_ki
>>> dataset = load_chembl1871_ki()
>>> dataset  
(['CC1=CC(C)(C)Nc2ccc3c(c21)/C(=C/c1ccsc1)Oc1ccc(F)cc1-3, ..., 'CN(C[C@](C)(O)C(=O)Nc1ccc(C#N)c(C(F)(F)F)c1)c1ccc(C#N)c(-c2ccccc2)c1'], \
array([-2.825, ..., -1.892]))

>>> dataset = load_chembl1871_ki(as_frame=True)
>>> dataset.head() 
                                                                            SMILES        Ki
0                            CC1=CC(C)(C)Nc2ccc3c(c21)/C(=C/c1ccsc1)Oc1ccc(F)cc1-3 -2.825426
1                         CCc1ccccc1/C=C1\Oc2ccc(F)cc2-c2ccc3c(c21)C(C)=CC(C)(C)N3 -3.201124
2                      CC1=CC(C)(C)Nc2ccc3c(c21)/C(=C/c1ccccc1N(C)C)Oc1ccc(F)cc1-3 -2.913284
3                           CC1=CC(C)(C)Nc2ccc3c(c21)/C(=C/c1ccccc1)Oc1c(F)cccc1-3 -3.163161
4  CC(=O)O[C@]1(C(C)=O)CC[C@H]2[C@@H]3C[C@H](C)C4=CC(=O)CC[C@]4(C)[C@H]3CC[C@@]21C -0.462398