在组合中增加迭代的最有效的方法是什么？

Question

我有一个numpy数组：NxM

比如说：

input_data = np.random.rand(10,5)

我想创建一个新数组，其中新数组是输入数据列之间的所有可能差异，这将为您提供一个大小为：（10，10）的数组

到目前为止，我的代码是：

def get_data_differences(read_data):
    '''Finds every possible differences between the columns of the read_data
    read_data: NxM variable where M are the features
    returns diff_data, and NxR variables
    R is the number of every possible combination of 2 columns

    '''
    if len(read_data.shape) != 2:
        print 'The data format is not consistent'
    data_rows, data_columns = read_data.shape
    data_difference = np.zeros((data_rows, 1))
    for combination_pair in itertools.combinations(read_data.T, 2):
    #iterate over every possible pairing of columns (hence the .T)
        minuend_, substraend_ = combination_pair
        difference_ = minuend_ - substraend_
        data_difference = np.append(data_difference, difference_[:, None], axis = 1)
    data_difference = np.delete(data_difference, 0, 1)
    return data_difference

我发现删除我创建的原始零数组效率不高。你知道吗

如果你有更好的建议，那就太好了

Answer 1

为什么不同时索引多个列呢？你知道吗

np.diff(read_data[:, list(combinations(range(read_data.shape[1]), 2))])[..., 0]