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Python numpy.repeate方法代码示例

本文整理汇总了Python中numpy.repeate方法的典型用法代码示例。如果您正苦于以下问题:Python numpy.repeate方法的具体用法?Python numpy.repeate怎么用?Python numpy.repeate使用的例子?那么恭喜您, 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的用法示例。


在下文中一共展示了numpy.repeate方法的6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: bbox_overlaps_2D

# 需要导入模块: import numpy [as 别名]
# 或者: from numpy import repeate [as 别名]
def bbox_overlaps_2D(boxes1, boxes2):
    """Computes IoU overlaps between two sets of boxes.
    boxes1, boxes2: [N, (y1, x1, y2, x2)].
    """
    # 1. Tile boxes2 and repeate boxes1. This allows us to compare
    # every boxes1 against every boxes2 without loops.
    # TF doesn't have an equivalent to np.repeate() so simulate it
    # using tf.tile() and tf.reshape.
    boxes1_repeat = boxes2.size()[0]
    boxes2_repeat = boxes1.size()[0]
    boxes1 = boxes1.repeat(1,boxes1_repeat).view(-1,4)
    boxes2 = boxes2.repeat(boxes2_repeat,1)

    # 2. Compute intersections
    b1_y1, b1_x1, b1_y2, b1_x2 = boxes1.chunk(4, dim=1)
    b2_y1, b2_x1, b2_y2, b2_x2 = boxes2.chunk(4, dim=1)
    y1 = torch.max(b1_y1, b2_y1)[:, 0]
    x1 = torch.max(b1_x1, b2_x1)[:, 0]
    y2 = torch.min(b1_y2, b2_y2)[:, 0]
    x2 = torch.min(b1_x2, b2_x2)[:, 0]
    zeros = Variable(torch.zeros(y1.size()[0]), requires_grad=False)
    if y1.is_cuda:
        zeros = zeros.cuda()
    intersection = torch.max(x2 - x1, zeros) * torch.max(y2 - y1, zeros)

    # 3. Compute unions
    b1_area = (b1_y2 - b1_y1) * (b1_x2 - b1_x1)
    b2_area = (b2_y2 - b2_y1) * (b2_x2 - b2_x1)
    union = b1_area[:,0] + b2_area[:,0] - intersection

    # 4. Compute IoU and reshape to [boxes1, boxes2]
    iou = intersection / union
    overlaps = iou.view(boxes2_repeat, boxes1_repeat)
    return overlaps 
开发者ID:MIC-DKFZ,项目名称:medicaldetectiontoolkit,代码行数:36,代码来源:model_utils.py

示例2: bbox_overlaps_3D

# 需要导入模块: import numpy [as 别名]
# 或者: from numpy import repeate [as 别名]
def bbox_overlaps_3D(boxes1, boxes2):
    """Computes IoU overlaps between two sets of boxes.
    boxes1, boxes2: [N, (y1, x1, y2, x2, z1, z2)].
    """
    # 1. Tile boxes2 and repeate boxes1. This allows us to compare
    # every boxes1 against every boxes2 without loops.
    # TF doesn't have an equivalent to np.repeate() so simulate it
    # using tf.tile() and tf.reshape.
    boxes1_repeat = boxes2.size()[0]
    boxes2_repeat = boxes1.size()[0]
    boxes1 = boxes1.repeat(1,boxes1_repeat).view(-1,6)
    boxes2 = boxes2.repeat(boxes2_repeat,1)

    # 2. Compute intersections
    b1_y1, b1_x1, b1_y2, b1_x2, b1_z1, b1_z2 = boxes1.chunk(6, dim=1)
    b2_y1, b2_x1, b2_y2, b2_x2, b2_z1, b2_z2 = boxes2.chunk(6, dim=1)
    y1 = torch.max(b1_y1, b2_y1)[:, 0]
    x1 = torch.max(b1_x1, b2_x1)[:, 0]
    y2 = torch.min(b1_y2, b2_y2)[:, 0]
    x2 = torch.min(b1_x2, b2_x2)[:, 0]
    z1 = torch.max(b1_z1, b2_z1)[:, 0]
    z2 = torch.min(b1_z2, b2_z2)[:, 0]
    zeros = Variable(torch.zeros(y1.size()[0]), requires_grad=False)
    if y1.is_cuda:
        zeros = zeros.cuda()
    intersection = torch.max(x2 - x1, zeros) * torch.max(y2 - y1, zeros) * torch.max(z2 - z1, zeros)

    # 3. Compute unions
    b1_volume = (b1_y2 - b1_y1) * (b1_x2 - b1_x1)  * (b1_z2 - b1_z1)
    b2_volume = (b2_y2 - b2_y1) * (b2_x2 - b2_x1)  * (b2_z2 - b2_z1)
    union = b1_volume[:,0] + b2_volume[:,0] - intersection

    # 4. Compute IoU and reshape to [boxes1, boxes2]
    iou = intersection / union
    overlaps = iou.view(boxes2_repeat, boxes1_repeat)
    return overlaps 
开发者ID:MIC-DKFZ,项目名称:medicaldetectiontoolkit,代码行数:38,代码来源:model_utils.py

示例3: overlaps_graph

# 需要导入模块: import numpy [as 别名]
# 或者: from numpy import repeate [as 别名]
def overlaps_graph(boxes1, boxes2):
    """Computes IoU overlaps between two sets of boxes.
    boxes1, boxes2: [N, (y1, x1, y2, x2)].
    """
    # 1. Tile boxes2 and repeate boxes1. This allows us to compare
    # every boxes1 against every boxes2 without loops.
    # TF doesn't have an equivalent to np.repeate() so simulate it
    # using tf.tile() and tf.reshape.
    b1 = tf.reshape(tf.tile(tf.expand_dims(boxes1, 1),
                            [1, 1, tf.shape(boxes2)[0]]), [-1, 4])
    b2 = tf.tile(boxes2, [tf.shape(boxes1)[0], 1])
    # 2. Compute intersections
    b1_y1, b1_x1, b1_y2, b1_x2 = tf.split(b1, 4, axis=1)
    b2_y1, b2_x1, b2_y2, b2_x2 = tf.split(b2, 4, axis=1)
    y1 = tf.maximum(b1_y1, b2_y1)
    x1 = tf.maximum(b1_x1, b2_x1)
    y2 = tf.minimum(b1_y2, b2_y2)
    x2 = tf.minimum(b1_x2, b2_x2)
    intersection = tf.maximum(x2 - x1, 0) * tf.maximum(y2 - y1, 0)
    # 3. Compute unions
    b1_area = (b1_y2 - b1_y1) * (b1_x2 - b1_x1)
    b2_area = (b2_y2 - b2_y1) * (b2_x2 - b2_x1)
    union = b1_area + b2_area - intersection
    # 4. Compute IoU and reshape to [boxes1, boxes2]
    iou = intersection / union
    overlaps = tf.reshape(iou, [tf.shape(boxes1)[0], tf.shape(boxes2)[0]])
    return overlaps 
开发者ID:olgaliak,项目名称:segmentation-unet-maskrcnn,代码行数:29,代码来源:model.py

示例4: bbox_overlaps_2D

# 需要导入模块: import numpy [as 别名]
# 或者: from numpy import repeate [as 别名]
def bbox_overlaps_2D(boxes1, boxes2):
    """Computes IoU overlaps between two sets of boxes.
    boxes1, boxes2: [N, (y1, x1, y2, x2)].
    """
    # 1. Tile boxes2 and repeate boxes1. This allows us to compare
    # every boxes1 against every boxes2 without loops.
    # TF doesn't have an equivalent to np.repeate() so simulate it
    # using tf.tile() and tf.reshape.

    boxes1_repeat = boxes2.size()[0]
    boxes2_repeat = boxes1.size()[0]

    boxes1 = boxes1.repeat(1,boxes1_repeat).view(-1,4)
    boxes2 = boxes2.repeat(boxes2_repeat,1)

    # 2. Compute intersections
    b1_y1, b1_x1, b1_y2, b1_x2 = boxes1.chunk(4, dim=1)
    b2_y1, b2_x1, b2_y2, b2_x2 = boxes2.chunk(4, dim=1)
    y1 = torch.max(b1_y1, b2_y1)[:, 0]
    x1 = torch.max(b1_x1, b2_x1)[:, 0]
    y2 = torch.min(b1_y2, b2_y2)[:, 0]
    x2 = torch.min(b1_x2, b2_x2)[:, 0]
    #--> expects x1<x2 & y1<y2
    zeros = torch.zeros(y1.size()[0], requires_grad=False)
    if y1.is_cuda:
        zeros = zeros.cuda()
    intersection = torch.max(x2 - x1, zeros) * torch.max(y2 - y1, zeros)

    # 3. Compute unions
    b1_area = (b1_y2 - b1_y1) * (b1_x2 - b1_x1)
    b2_area = (b2_y2 - b2_y1) * (b2_x2 - b2_x1)
    union = b1_area[:,0] + b2_area[:,0] - intersection

    # 4. Compute IoU and reshape to [boxes1, boxes2]
    iou = intersection / union
    assert torch.all(iou<=1), "iou score>1 produced in bbox_overlaps_2D"
    overlaps = iou.view(boxes2_repeat, boxes1_repeat) #--> per gt box: ious of all proposal boxes with that gt box

    return overlaps 
开发者ID:MIC-DKFZ,项目名称:RegRCNN,代码行数:41,代码来源:model_utils.py

示例5: bbox_overlaps_3D

# 需要导入模块: import numpy [as 别名]
# 或者: from numpy import repeate [as 别名]
def bbox_overlaps_3D(boxes1, boxes2):
    """Computes IoU overlaps between two sets of boxes.
    boxes1, boxes2: [N, (y1, x1, y2, x2, z1, z2)].
    """
    # 1. Tile boxes2 and repeate boxes1. This allows us to compare
    # every boxes1 against every boxes2 without loops.
    # TF doesn't have an equivalent to np.repeate() so simulate it
    # using tf.tile() and tf.reshape.
    boxes1_repeat = boxes2.size()[0]
    boxes2_repeat = boxes1.size()[0]
    boxes1 = boxes1.repeat(1,boxes1_repeat).view(-1,6)
    boxes2 = boxes2.repeat(boxes2_repeat,1)

    # 2. Compute intersections
    b1_y1, b1_x1, b1_y2, b1_x2, b1_z1, b1_z2 = boxes1.chunk(6, dim=1)
    b2_y1, b2_x1, b2_y2, b2_x2, b2_z1, b2_z2 = boxes2.chunk(6, dim=1)
    y1 = torch.max(b1_y1, b2_y1)[:, 0]
    x1 = torch.max(b1_x1, b2_x1)[:, 0]
    y2 = torch.min(b1_y2, b2_y2)[:, 0]
    x2 = torch.min(b1_x2, b2_x2)[:, 0]
    z1 = torch.max(b1_z1, b2_z1)[:, 0]
    z2 = torch.min(b1_z2, b2_z2)[:, 0]
    zeros = torch.zeros(y1.size()[0], requires_grad=False)
    if y1.is_cuda:
        zeros = zeros.cuda()
    intersection = torch.max(x2 - x1, zeros) * torch.max(y2 - y1, zeros) * torch.max(z2 - z1, zeros)

    # 3. Compute unions
    b1_volume = (b1_y2 - b1_y1) * (b1_x2 - b1_x1)  * (b1_z2 - b1_z1)
    b2_volume = (b2_y2 - b2_y1) * (b2_x2 - b2_x1)  * (b2_z2 - b2_z1)
    union = b1_volume[:,0] + b2_volume[:,0] - intersection

    # 4. Compute IoU and reshape to [boxes1, boxes2]
    iou = intersection / union
    overlaps = iou.view(boxes2_repeat, boxes1_repeat)
    return overlaps 
开发者ID:MIC-DKFZ,项目名称:RegRCNN,代码行数:38,代码来源:model_utils.py

示例6: bbox_overlaps

# 需要导入模块: import numpy [as 别名]
# 或者: from numpy import repeate [as 别名]
def bbox_overlaps(boxes1, boxes2):
    """Computes IoU overlaps between two sets of boxes.
    boxes1, boxes2: [N, (y1, x1, y2, x2)].
    """
    # 1. Tile boxes2 and repeate boxes1. This allows us to compare
    # every boxes1 against every boxes2 without loops.
    # TF doesn't have an equivalent to np.repeate() so simulate it
    # using tf.tile() and tf.reshape.
    boxes1_repeat = boxes2.size()[0]
    boxes2_repeat = boxes1.size()[0]
    boxes1 = boxes1.repeat(1,boxes1_repeat).view(-1,4)
    boxes2 = boxes2.repeat(boxes2_repeat,1)

    # 2. Compute intersections
    b1_y1, b1_x1, b1_y2, b1_x2 = boxes1.chunk(4, dim=1)
    b2_y1, b2_x1, b2_y2, b2_x2 = boxes2.chunk(4, dim=1)
    y1 = torch.max(b1_y1, b2_y1)[:, 0]
    x1 = torch.max(b1_x1, b2_x1)[:, 0]
    y2 = torch.min(b1_y2, b2_y2)[:, 0]
    x2 = torch.min(b1_x2, b2_x2)[:, 0]
    zeros = Variable(torch.zeros(y1.size()[0]), requires_grad=False)
    if y1.is_cuda:
        zeros = zeros.cuda()
    intersection = torch.max(x2 - x1, zeros) * torch.max(y2 - y1, zeros)

    # 3. Compute unions
    b1_area = (b1_y2 - b1_y1) * (b1_x2 - b1_x1)
    b2_area = (b2_y2 - b2_y1) * (b2_x2 - b2_x1)
    union = b1_area[:,0] + b2_area[:,0] - intersection

    # 4. Compute IoU and reshape to [boxes1, boxes2]
    iou = intersection / union
    overlaps = iou.view(boxes2_repeat, boxes1_repeat)

    return overlaps 
开发者ID:trancept,项目名称:decentralized_AI,代码行数:37,代码来源:model.py


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