load_chembl1862_ki#

skfp.datasets.moleculeace.load_chembl1862_ki(data_dir: str | PathLike | None = None, as_frame: bool = False, verbose: bool = False, force_update: bool = False) → DataFrame | tuple[list[str], ndarray]#

Load the ChEMBL1862 Ki dataset.

The task is to predict the inhibitor constant (Ki) of molecules against the Tyrosine-protein kinase abl1 target [1] [2].

Tasks

1

Task type

regression

Total samples

794

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.

  • force_update (bool, default=False) – If True, always re-download the dataset from HuggingFace Hub, even if it is already present locally. If False, the dataset is downloaded only if it is not yet available locally.

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

Examples

>>> from skfp.datasets.moleculeace import load_chembl1862_ki
>>> dataset = load_chembl1862_ki()
>>> dataset  
(['Nc1[nH]cnc2nnc(-c3ccc(Cl)cc3)c1-2, ..., 'CCCCNc1ncnc2c1cnn2CC(Cl)c1ccccc1'], \
array([-2.699, ..., -3.3]))

>>> dataset = load_chembl1862_ki(as_frame=True)
>>> dataset.head() 
                                              SMILES       Ki
0                  Nc1[nH]cnc2nnc(-c3ccc(Cl)cc3)c1-2 -2.69897
1  Cc1ccc(N2NC(=O)/C(=C/c3ccc(-c4ccc(C)c(Cl)c4)o3... -3.69897
2  O=C1NN(c2ccc(Cl)c(Cl)c2)C(=O)/C1=C\c1cccc(OCc2... -3.00000
3    O=C1NN(c2ccc(I)cc2)C(=O)/C1=C\c1cc2c(cc1Br)OCO2 -3.39794
4   O=C1NN(c2ccc(I)cc2)C(=O)/C1=C\c1ccc(N2CCOCC2)cc1 -4.30103