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Python numerictypes.issubdtype方法代码示例

本文整理汇总了Python中numpy.core.numerictypes.issubdtype方法的典型用法代码示例。如果您正苦于以下问题:Python numerictypes.issubdtype方法的具体用法?Python numerictypes.issubdtype怎么用?Python numerictypes.issubdtype使用的例子?那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在numpy.core.numerictypes的用法示例。


在下文中一共展示了numerictypes.issubdtype方法的1个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的Python代码示例。

示例1: ix_

# 需要导入模块: from numpy.core import numerictypes [as 别名]
# 或者: from numpy.core.numerictypes import issubdtype [as 别名]
def ix_(*args):
    """
    Construct an open mesh from multiple sequences.

    This function takes N 1-D sequences and returns N outputs with N
    dimensions each, such that the shape is 1 in all but one dimension
    and the dimension with the non-unit shape value cycles through all
    N dimensions.

    Using `ix_` one can quickly construct index arrays that will index
    the cross product. ``a[np.ix_([1,3],[2,5])]`` returns the array
    ``[[a[1,2] a[1,5]], [a[3,2] a[3,5]]]``.

    Parameters
    ----------
    args : 1-D sequences

    Returns
    -------
    out : tuple of ndarrays
        N arrays with N dimensions each, with N the number of input
        sequences. Together these arrays form an open mesh.

    See Also
    --------
    ogrid, mgrid, meshgrid

    Examples
    --------
    >>> a = np.arange(10).reshape(2, 5)
    >>> a
    array([[0, 1, 2, 3, 4],
           [5, 6, 7, 8, 9]])
    >>> ixgrid = np.ix_([0,1], [2,4])
    >>> ixgrid
    (array([[0],
           [1]]), array([[2, 4]]))
    >>> ixgrid[0].shape, ixgrid[1].shape
    ((2, 1), (1, 2))
    >>> a[ixgrid]
    array([[2, 4],
           [7, 9]])

    """
    out = []
    nd = len(args)
    for k, new in enumerate(args):
        new = asarray(new)
        if new.ndim != 1:
            raise ValueError("Cross index must be 1 dimensional")
        if new.size == 0:
            # Explicitly type empty arrays to avoid float default
            new = new.astype(_nx.intp)
        if issubdtype(new.dtype, _nx.bool_):
            new, = new.nonzero()
        new = new.reshape((1,)*k + (new.size,) + (1,)*(nd-k-1))
        out.append(new)
    return tuple(out) 
开发者ID:ryfeus,项目名称:lambda-packs,代码行数:60,代码来源:index_tricks.py


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