背景:
我有兴趣通过在dwave的绝热量子计算机上编写模拟来研究各种材料的量子相变。为了更容易生成作为参数函数的相位图,我编写了一些实用程序来扫描参数,使用这些参数集运行仿真,并收集数据。
输入条件背景:
在DWave上,我可以设置两组参数,h偏差和J耦合输入如下:h = {qubit0: hvalue0, qubit1: hvalue1,...}J = {(qubit0, qubit1): J01, (qubit2, qubit3): J23, ...}。到目前为止,我有一个工具可以对给定的输入进行参数扫描,比如:{qubit: [hz1, hz2,..., hzn]}将量子位映射到要扫描的h值,{coupler: [J1, J2,..., Jn]}将耦合器映射到要扫描的J值在这两种情况下,输出都是[{trial1}, {trial2}, ... {trialn}]形式的列表,表示每个单独的qubit和耦合上的hJ输入的笛卡尔积。
我到底想要什么,到目前为止我写了什么:
在上面,我遇到了一个严重的问题。假设我想扫描一系列参数,其中一些量子位或耦合器在任何给定的运行中彼此之间都有固定的关系。这一点很重要,因为逻辑问题必须以非常复杂的方式映射到DWave上例如,假设我想运行一个问题,其中qubit0h中有[0, 1, 2]qubit1h中有[1, 2, 3],而qubit3h中有[5, 8],但必须保留关系qubit1_h = qubit0_h + 1;即,我希望值的乘积是[(0, 1, 5), (0, 1, 8), (1, 2, 5), (1, 2, 8), ...],而不是笛卡尔积给出的所有组合。
下面的代码将对h参数执行此操作,但对J参数不起作用,因为字典键是元组另外,如果我不想要这个功能,我必须运行我的原始代码来生成caretesian产品,因此它似乎生成了“3个案例”

def fixed_relationship_sweep(input_params, together):
    """
    Inputs
    ------
    input_params: {0:[x1, x2], 1:[x3, x4], 2:[y1, y2], 3:[y3, y4]]}
    dictionary mapping qubits to parameter lists to iterate through
    together: [[0, 1], [2, 3]]
    list of qubit lists that specify which qubit parameters to sweep with a fixed relationship

    Output
    ------
    fixed_rel_sweep: [{trial1}, {trial2}, ...{trialn}] where qubits labelled as "together" are
    swept with fixed 1-1 relationship, ie, above produces:
    [{0:x1, 1:x3, 2:y1, 3:y3}, {0:x1, 1:x3, 2:y2, 3:y4}, {0:x2, 1:x4, 2:y1, 3:y3},
    {0:x2, 1:x4, 2:y2, 3:y4}]
    """
    qsorcs = []
    params = []
    #index representation of params, as cartesian product must respect fixed positions
    #of arguments and not their values, ie [x1, x3] vary together in example
    tempidxrep = []
    for key, value in input_params.items():
        qsorcs.append(key)
        params.append(value)
        tempidxrep.append([i for i in range(len(value))])

    idxrep = []
    #remove redundancy in index representation governed by fixed relationships in together
    for fix_rel in together:
        idxrep.append(tempidxrep[fix_rel[0]])

    #sweep combinations via carteisan product
    idxcombos = itertools.product(*idxrep)

    #reconstruct actual parameter combinations
    param_combos = []
    for idxcombo in idxcombos:
        trial = {qsorcs[j]: params[j][idxcombo[i]] for i in range(len(idxcombo)) for j in together[i]}
        param_combos.append(trial)

    return param_combos

有没有一种更简单、更好的方法来使用内置工具(如itertools)来实现这一点,这些工具可以处理整数或元组形式的键,而无需编写单独的复杂函数?换句话说,我是不是从错误的方向来处理这个看似简单的问题?

最佳答案

在发布了这个问题之后,我写了一个原始代码的改进版本,它允许原始输入(格式字典{h_n: hval}和h a n iteger代表n个qubit)以及格式的附加输入{J_nm: Jval}和J_mnM一个qn和m之间耦合强度的元组(qn,qm)一些量子位/耦合器被排除在“一起”的lits之外,就像原来那样。因此,这个新代码在功能上可以满足我的需求。不过,我想有更好的出路。

def fixed_relationship_sweep(input_params, together):
"""
Inputs
------
input_params: {qorc1:[x1, x2], qorc2:[x3, x4], qorc3:[y1, y2], qorc4:[y3, y4]]}
dictionary mapping qubits or couplers to parameter lists to iterate through
together: [[qorc1, qorc2], [qorc3, qorc4]]
list of qubit lists that specify which qubit parameters to sweep with a fixed relationship

Output
------
fixed_rel_sweep: [{trial1}, {trial2}, ...{trialn}] where qubits labelled as "together" are
swept with fixed 1-1 relationship, ie, above produces:
[{qorc1:x1, qorc2:x3, qorc3:y1, qorc4:y3}, {qorc1:x1, qorc2:x3, qorc3:y2, qorc4:y4},
{qorc1:x2, qorc2:x4, qorc3:y1, qorc4:y3},{qorc1:x2, qorc2:x4, qorc3:y2, qorc4:y4}]
"""
#list of qubits or couplers
qsorcs = []
#index representation of params, as cartesian product must respect fixed positions
#of arguments and not their values, ie [x1, x3] vary together in example
idxrep = {}
for key, value in input_params.items():
    qsorcs.append(key)
    idxrep[key] = [i for i in range(len(value))]

#remove redundancy in index representation governed by fixed relationships in together
count = 0
for fix_rel in together:
    for j in range(len(fix_rel)):
        if j != 0:
            del idxrep[fix_rel[j]]

#sweep combinations via cartesian product
idxdict_combos = (list(dict(zip(idxrep, x)) for x in itertools.product(*idxrep.values())))

#reconstruct actual parameter combinations with "redundant" parameter values
dict_combos = []
for combo in idxdict_combos:
    #add back in "redundant" parameters
    for fix_rel in together:
        for qorc in fix_rel[1::]:
            combo[qorc] = combo[fix_rel[0]]

    #add back in true values corresponding to indices
    tempdict = {}
    for key, value in combo.items():
        tempdict[key] = input_params[key][value]

    dict_combos.append(tempdict)

return dict_combos

10-06 05:22