[我只是创建word2vec模型，并在单词（键）和值（向量）之间创建字典。
dictionary = dict({})
for idx, key in enumerate(model.wv.vocab):
dictionary[key] = model.wv[key]
这样的字典：
{'petani': array([-1.5856586e-01, -3.7320998e-01, 5.2052790e-01, 4.2783752e-01,
7.6572634e-02, -6.9080216e-01, -6.2848133e-01, -8.8271695e-01,
-5.0581765e-01, -3.0301598e-01, 1.2088260e-01, -6.9143041e-03,
4.4222975e-01, -3.6879349e-01, 5.5616591e-02, -1.1631098e-01,
-4.1717857e-02, -1.0702840e-01, -3.8082123e-01, -3.2646924e-01,
-6.2404245e-01, 1.5188067e-01, 4.4228908e-01, -1.2028643e-02,
-1.1128881e-01, -1.2848943e-01, -2.2734607e-02, -4.8964986e-01,
8.8901877e-01, 3.4160933e-01, -2.7001941e-01, 1.9693324e-01,
2.0979826e-01, 2.9565904e-01, -1.2944296e-01, 7.1377866e-02,
-2.3514537e-02, -8.9601561e-02, 2.5837755e-01, 2.4173041e-01,
1.3690867e-01, -2.9203954e-01, -2.2040294e-01, 4.3522149e-01,
-1.3773613e-01, 1.5158685e-01, -8.4017359e-02, -4.2125726e-01,
-4.1426069e-01, -3.0153343e-01, -3.7908006e-01, 6.4364660e-01,
4.3095109e-01, 7.0161112e-02, 2.5523776e-01, 1.0336957e-01,
2.0642248e-01, -3.3600163e-01, 4.1162384e-01, 3.1530520e-01,
-7.8101611e-01, -2.8255533e-02, 3.7680027e-01, 1.4239751e-01,
-2.4524818e-01, 4.0836743e-01, 5.3868568e-01, -3.5229959e-02,
-2.4528666e-03, 6.3001931e-01, 5.2766448e-01, -4.7586882e-01,
3.9797042e-02, 6.7331716e-02, 9.5685691e-02, -1.2028452e-01,
-4.5608345e-01, 2.8427744e-01, -2.9394981e-01, -1.2994224e-01,
5.1969510e-01, 1.2906422e-01, -7.3352844e-02, 1.4258710e-01,
-5.1817909e-02, -9.9871524e-02, -7.8935099e-01, -7.4661225e-02,
7.3149313e-05, -5.7802260e-01, 9.4037224e-03, -3.1019258e-01,
-1.2252379e-01, -2.6833829e-01, 6.4406760e-02, 2.1438387e-01,
-4.6708840e-01, 4.6972111e-01, 6.5475023e-01, 2.5733718e-01],
dtype=float32),
'tewas': array([-0.5524042 , 0.06416334, 0.43303823, 0.08412471, -0.26415887,
0.81605107, -0.30849332, -0.31477442, 0.55429876, -0.34576067,
0.20641974, -0.4408319 , 0.6560267 , -0.27099684, 0.01890246,
0.19507472, 1.0473057 , -0.4251338 , 0.47794574, -0.87928706,
-0.17208456, 0.42365453, -0.1005183 , -0.4278884 , 0.5542767 ,
-0.2686175 , 0.08917291, -0.67033154, -0.5816741 , 0.6316814 ,
0.02794617, -0.36843893, 0.28338206, -0.20059416, -0.45663786,
-0.01763846, 0.0384933 , -0.6548334 , 0.30371612, -0.31790268,
-0.2999138 , -0.44880405, -0.23181435, -0.4012081 , -0.39137393,
0.40417886, -0.24790864, -0.4002588 , 0.5552438 , -0.13635041,
-0.2759215 , 0.5185368 , 1.1411811 , 0.02402515, -0.22694567,
-0.13629675, 0.00484379, 0.21743643, -0.09752762, -0.13332742,
-0.5833612 , 0.18499097, 0.16189235, -0.6939921 , -0.8281921 ,
0.22095513, 0.390632 , -0.23755077, -0.36395508, 0.8814506 ,
0.4552586 , -0.36801606, 0.06618993, 0.47150037, 0.3567098 ,
-0.9401551 , 0.53242975, -0.84471244, -0.8138361 , 0.28408146,
0.23675315, -0.28799447, -0.33678758, 0.57642925, 0.37340602,
-0.12505923, -0.02738794, -0.6040789 , -0.17035136, -0.3264485 ,
0.5719107 , -0.3058263 , -0.11770777, -0.73283607, 0.07872504,
-0.3070875 , -0.3163887 , -0.2119802 , -0.11972545, -0.27509072],
dtype=float32),
...}
我可以通过以下方式获得基于钥匙的价值：
dictionary.get('petani')
结果：
array([-1.5856586e-01, -3.7320998e-01, 5.2052790e-01, 4.2783752e-01,
7.6572634e-02, -6.9080216e-01, -6.2848133e-01, -8.8271695e-01,
-5.0581765e-01, -3.0301598e-01, 1.2088260e-01, -6.9143041e-03,
4.4222975e-01, -3.6879349e-01, 5.5616591e-02, -1.1631098e-01,
-4.1717857e-02, -1.0702840e-01, -3.8082123e-01, -3.2646924e-01,
-6.2404245e-01, 1.5188067e-01, 4.4228908e-01, -1.2028643e-02,
-1.1128881e-01, -1.2848943e-01, -2.2734607e-02, -4.8964986e-01,
8.8901877e-01, 3.4160933e-01, -2.7001941e-01, 1.9693324e-01,
2.0979826e-01, 2.9565904e-01, -1.2944296e-01, 7.1377866e-02,
-2.3514537e-02, -8.9601561e-02, 2.5837755e-01, 2.4173041e-01,
1.3690867e-01, -2.9203954e-01, -2.2040294e-01, 4.3522149e-01,
-1.3773613e-01, 1.5158685e-01, -8.4017359e-02, -4.2125726e-01,
-4.1426069e-01, -3.0153343e-01, -3.7908006e-01, 6.4364660e-01,
4.3095109e-01, 7.0161112e-02, 2.5523776e-01, 1.0336957e-01,
2.0642248e-01, -3.3600163e-01, 4.1162384e-01, 3.1530520e-01,
-7.8101611e-01, -2.8255533e-02, 3.7680027e-01, 1.4239751e-01,
-2.4524818e-01, 4.0836743e-01, 5.3868568e-01, -3.5229959e-02,
-2.4528666e-03, 6.3001931e-01, 5.2766448e-01, -4.7586882e-01,
3.9797042e-02, 6.7331716e-02, 9.5685691e-02, -1.2028452e-01,
-4.5608345e-01, 2.8427744e-01, -2.9394981e-01, -1.2994224e-01,
5.1969510e-01, 1.2906422e-01, -7.3352844e-02, 1.4258710e-01,
-5.1817909e-02, -9.9871524e-02, -7.8935099e-01, -7.4661225e-02,
7.3149313e-05, -5.7802260e-01, 9.4037224e-03, -3.1019258e-01,
-1.2252379e-01, -2.6833829e-01, 6.4406760e-02, 2.1438387e-01,
-4.6708840e-01, 4.6972111e-01, 6.5475023e-01, 2.5733718e-01],
dtype=float32)
而且我尝试通过以下方式获取基于值的密钥：
def get_key(val):
for key, value in dictionary.items():
if val == value:
return key
return "key doesn't exist"
contoh_value = [ 0.3296796, 0.39876923, 0.47616847, -0.85435633, 0.08756444, -0.3249461, -0.9938304, -0.94971858, 0.96680486, -0.1293482, 0.76376391, 0.56389303, 1.74900006, 1.25801649, -0.24322121, -0.37309024, -0.07400302, -0.31166066, 0.02108994, 0.18042984, 0.23819409, 0.72364472, 0.72723592, 0.47477901, -0.76086017, 1.67910596, 0.5221492, 0.07191082, -0.99006812, 0.30615227, 1.00488158, -0.73874328, 0.89766186, -0.5589205, 0.23986163, 0.59018145, -0.29154907, -0.40612788, 0.68320352, -0.32813038, -0.95184149, -0.58031032, -0.54268715, -0.68725495, -0.35027478, -0.30614226, 0.81348022, 0.40609737, 1.54977913, -0.04884687, -0.47331776, -0.22339089, 0.58311762, 0.13232404, 0.81241847, 0.72163447, 0.23984061, -0.32305742, 0.52193057, 0.506366, -0.56128758, -0.30134899, -0.53561509, -0.90930432, -1.22216257, 0.00341641, 0.41335883, -1.16226898, -0.15699308, 0.56443178, 0.1427298, -0.24072925, -1.06598538, 0.56613415, 1.16162036, 1.10104698, 0.93053441, 0.88296479, 0.28975529, -0.97775083, -1.45139772, -0.16452539, -1.02581363, 0.65278189, 0.6345808, 0.41010778, -2.14354638, 0.25915023, -0.0490873, -0.25783952, -0.40967067, -0.96420361, -0.99802028, -0.35535042, -0.09994081, 0.12680747, 1.29304774, -0.35715534, -0.98996135, -0.90942983]
print(get_key(contoh_value))
但是结果是“键不存在”。contoh_value的值是从字典中现有矢量复制的。从np.array值获取密钥的最佳解决方案是什么？
由于这些值是浮点值，因此在尝试直接比较它们是否相等时可能会遇到问题。使用numpy.isclose功能可能会获得更好的成功。此函数至少需要两个参数，分别为数组a和数组b，并返回一个新数组，每个数组的索引分别为True或False，这与[ C0]和a的值接近。然后，您可以使用b检查所有这些值是否都接近，该参数至少使用一个像数组numpy.all这样的参数，对于一维数组，请检查所有值是否均为[C0 ]并返回a a。您可以像这样修改True函数：
bool这样，如果两个数组中的所有值都接近，它将返回键，否则将继续到下一个键值对以进行比较。