python numpy np.arctan2()函数(批量计算反正切?)

def arctan2(x1, x2, *args, **kwargs): # real signature unknown; NOTE: unreliably restored from __doc__ """ arctan2(x1, x2, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj]) Element-wise arc tangent of ``x1/x2`` choosing the quadrant