我目前正在嘗試執行網格搜索以優化功能。不幸的是，計算時間太長了，這主要是由于我的網格搜索上的無用計算。到目前為止，我只嘗試做一個笛卡爾積，它給了我比我需要的更多的引數集。
我有以下輸入：
dict_params = {"params_a": [0, 1],
"params_b": ["x1", "x2", "x3"],
"params_c": {"x1": ["a1", "a2"],
"x2": ["b1", "b2"],
"x3": ["c1", "c2"]
}
}
我期望以下輸出：
output_expected = [{"params_a" : 0, "params_b" : "x1", "params_c": "a1"},
{"params_a" : 0, "params_b" : "x1", "params_c": "a2"},
{"params_a" : 0, "params_b" : "x2", "params_c": "b1"},
{"params_a" : 0, "params_b" : "x2", "params_c": "b2"},
{"params_a" : 0, "params_b" : "x3", "params_c": "c1"},
{"params_a" : 0, "params_b" : "x3", "params_c": "c2"},
{"params_a" : 1, "params_b" : "x1", "params_c": "a1"},
{"params_a" : 1, "params_b" : "x1", "params_c": "a2"},
{"params_a" : 1, "params_b" : "x2", "params_c": "b1"},
{"params_a" : 1, "params_b" : "x2", "params_c": "b2"},
{"params_a" : 1, "params_b" : "x3", "params_c": "c1"},
{"params_a" : 1, "params_b" : "x3", "params_c": "c2"}
]
回傳的任何格式都很好。我試圖用完整的笛卡爾積填充資料框，然后在“params_b”=“params_c”上合并，不幸的是我無法讓它作業。
謝謝閱讀
uj5u.com熱心網友回復：
我看到使用熊貓的一種方式：
(pd.DataFrame(product(dict_params['params_a'], dict_params['params_b']),
columns=['params_a', 'params_b'])
.assign(params_c=lambda d: d['params_b'].map(pd.Series(dict_params['params_c'])))
.explode('params_c')
)
輸出：
params_a params_b params_c
0 0 x1 a1
0 0 x1 a2
1 0 x2 b1
1 0 x2 b2
2 0 x3 c1
2 0 x3 c2
3 1 x1 a1
3 1 x1 a2
4 1 x2 b1
4 1 x2 b2
5 1 x3 c1
5 1 x3 c2
uj5u.com熱心網友回復：
使用本機 python，您可以這樣做：
dict_params = {"a": [0, 1],
"b": ["x1", "x2", "x3"],
"c": {"x1": ["a1", "a2"],
"x2": ["b1", "b2"],
"x3": ["c1", "c2"]
}
}
res = []
for param_a in dict_params['a']:
for param_b in dict_params['b']:
for param_c in dict_params['c'][param_b]:
res.append({'a': param_a, 'b': param_b, 'c': param_c})
print(res)
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