总结一致的列表元素

Question

我有一个包含元素的数组，我想总结一下准确性。我想对具有相同顺序的元素的数组求和。我宁愿不用zip编写for循环遍历每个元素并将它们相加，是否有更简单的方法来执行此操作？

这两个数组如下，目前我的代码用于计算总和。

yp = [[0, 1], [0, 1], [0, 1], [0, 1], [0, 1], [0, 1]]
y = [[0, 1], [0, 1], [0, 1], [0, 1], [0, 1], [0, 1]]

sums = np.sum(yp == y)

我的准确度为零。

Answer 1

使用您的示例：

yp = [[0, 1], [0, 1], [0, 1], [0, 1], [0, 1], [0, 1]]
y = [[0, 1], [0, 1], [0, 1], [0, 1], [0, 1], [0, 1]]

# First make the two arrays in question numpy arrays.
yp = np.array(yp)
y = np.array(y)
array_length = y.shape[1] # store length of sub arrays

equal_elements = np.array(yp) == np.array(y)  # check all equal elements
sums = np.sum(equal_elements, 1) # sum the number of equal elements in each sub array, use axis 1 as each array/sample is axis 0
equal_arrays = np.where(sums==array_length)[0] # returns a tuple, so index first element immediately

number_equal_arrays = equal_arrays.shape[0]  # What elements are equal
print('Number of equal arrays %d' % number_equal_arrays)
print('Accuracy %0.2f' % (number_equal_arrays/yp.shape[0]))

打印

  

相等数组的数量6

     

准确度1.00