📜  计算骰子分数的代码 - Python 代码示例

📅  最后修改于: 2022-03-11 14:45:59.942000             🧑  作者: Mango

代码示例1
import numpy as np
np.random.seed(0)
true = np.random.rand(10, 5, 5, 4)>0.5
pred = np.random.rand(10, 5, 5, 4)>0.5

def single_dice_coef(y_true, y_pred_bin):
    # shape of y_true and y_pred_bin: (height, width)
    intersection = np.sum(y_true * y_pred_bin)
    if (np.sum(y_true)==0) and (np.sum(y_pred_bin)==0):
        return 1
    return (2*intersection) / (np.sum(y_true) + np.sum(y_pred_bin))

def mean_dice_coef(y_true, y_pred_bin):
    # shape of y_true and y_pred_bin: (n_samples, height, width, n_channels)
    batch_size = y_true.shape[0]
    channel_num = y_true.shape[-1]
    mean_dice_channel = 0.
    for i in range(batch_size):
        for j in range(channel_num):
            channel_dice = single_dice_coef(y_true[i, :, :, j], y_pred_bin[i, :, :, j])
            mean_dice_channel += channel_dice/(channel_num*batch_size)
    return mean_dice_channel

def dice_coef2(y_true, y_pred):
    y_true_f = y_true.flatten()
    y_pred_f = y_pred.flatten()
    union = np.sum(y_true_f) + np.sum(y_pred_f)
    if union==0: return 1
    intersection = np.sum(y_true_f * y_pred_f)
    return 2. * intersection / union

print(mean_dice_coef(true, pred))
print(dice_coef2(true, pred))

# 0.4884357140842496
# 0.499001996007984