python - 二维数组代表一个巨大的python字典，像COOrdinate这样的解决方案可以节省内存

我尝试用数组中的数据用元组键更新dict：

myarray = np.array([[0, 0],  # 0, 1
                    [0, 1],
                    [1, 1],  # 1, 2
                    [1, 2],  # 1, 3
                    [2, 2],
                    [1, 3]]
) # a lot of this with shape~(10e6, 2)

dict_with_tuples_key = {(0, 1): 1,
                        (3, 7): 1} # ~10e6 keys

使用数组存储dict值（感谢@MSeifert），我们得到：

def convert_dict_to_darray(dict_with_tuples_key, myarray):
    idx_max_array = np.max(myarray, axis=0)
    idx_max_dict  = np.max(dict_with_tuples_key.keys(), axis=0)
    lens = np.max([list(idx_max_array), list(idx_max_dict)], axis=0)
    xlen, ylen = lens[0] + 1, lens[1] + 1
    darray = np.zeros((xlen, ylen)) # Empty array to hold all indexes in myarray
    for key, value in dict_with_tuples_key.items():
        darray[key] = value
    return darray

@njit
def update_darray(darray, myarray):
    elements = myarray.shape[0]
    for i in range(elements):
        darray[myarray[i][0]][myarray[i][1]] += 1
    return darray

def darray_to_dict(darray):
    updated_dict = {}
    keys = zip(*map(list, np.nonzero(darray)))
    for x, y in keys:
        updated_dict[(x, y)] = darray[x, y]
    return updated_dict

darray = convert_dict_to_darray(dict_with_tuples_key, myarray)
darray = update_darray(darray, myarray)

我得到了所需的确切结果：

# print darray_to_dict(darray)
# {(0, 1): 2.0,
#  (0, 0): 1.0,
#  (1, 1): 1.0,
#  (2, 2): 1.0,
#  (1, 2): 1.0,
#  (1, 3): 1.0,
#  (3, 7): 1.0, }

对于小矩阵，它工作得很好，@njit工作得很快，
但是…
创建巨大的空darray = np.zeros((xlen, ylen))不适合内存。我们如何避免分配一个非常稀疏的数组，并且只以坐标格式存储非空值，比如稀疏矩阵？

最佳答案

Usedok_matrixfromscipy；adock_matrix是基于键的稀疏矩阵字典它们允许您以增量方式构建稀疏矩阵，并且不会分配不适合您计算机内存的巨大空darray = np.zeros((xlen, ylen))。
唯一要做的更改是从scipy导入正确的模块，并更改函数darray中convert_dict_to_darray的定义。
它将如下所示：

from scipy.sparse import dok_matrix

def convert_dict_to_darray(dict_with_tuples_key, myarray):
    idx_max_array = np.max(myarray, axis=0)
    idx_max_dict  = np.max(dict_with_tuples_key.keys(), axis=0)
    lens = np.max([list(idx_max_array), list(idx_max_dict)], axis=0)
    xlen, ylen = lens[0] + 1, lens[1] + 1
    darray = dok_matrix( (xlen, ylen) )
    for key, value in dict_with_tuples_key.items():
        darray[key[0], key[1]] = value
    return darray