如何从python中的字典列表中提取特定键的值?
如果我有以下类型的数据-字典列表,如何从中提取一些键值如何从python中的字典列表中提取特定键的值?,python,list,Python,List,如果我有以下类型的数据-字典列表,如何从中提取一些键值 comps = [ { "name":'Test1', "p_value":0.02, "group0_null": 0.0, "group1_null": 0.0, },{ "name":'Test2', "p_value":0.05, "group0_n
comps = [
{
"name":'Test1',
"p_value":0.02,
"group0_null": 0.0,
"group1_null": 0.0,
},{
"name":'Test2',
"p_value":0.05,
"group0_null": 0.0,
"group1_null": 0.0,
},{
"name":'Test3',
"p_value":0.03,
"group0_null": 0.0,
"group1_null": 0.0,
},{
"name":'Test4',
"p_value":0.07,
"group0_null": 0.0,
"group1_null": 0.0,
},{
"name":'Test5',
"p_value":0.03,
"group0_null": 0.0,
"group1_null": 0.0,
},{
"name":'Test6',
"p_value":0.02,
"group0_null": 0.0,
"group1_null": 0.0,
},{
"name":'Test7',
"p_value":0.01,
"group0_null": 0.0,
"group1_null": 0.0,
}]
结果
从上面的数据来看,假设我只想要name
和p_值
。我怎样才能得到这个结果
[{
"name":'Test1',
"p_value":0.02,
},{
"name":'Test2',
"p_value":0.05,
},{
"name":'Test3',
"p_value":0.03,
},{
"name":'Test4',
"p_value":0.07,
},{
"name":'Test5',
"p_value":0.03,
},{
"name":'Test6',
"p_value":0.02,
},{
"name":'Test7',
"p_value":0.01,
}]
这说明了一切
[c for c in comps]
这只显示名称
[c['name']代表公司中的c]
但如果我这样做:
[c['name','p_value'] for c in comps ]
我得到一个错误:
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
<ipython-input-94-b29459f7b089> in <module>
----> 1 [c['name','p_value'] for c in comps['continuous_explainers'] ]
2
3 # cont_comps = []
4
5 # for c in comps['continuous_explainers']:
<ipython-input-94-b29459f7b089> in <listcomp>(.0)
----> 1 [c['name','p_value'] for c in comps['continuous_explainers'] ]
2
3 # cont_comps = []
4
5 # for c in comps['continuous_explainers']:
KeyError: ('name', 'p_value')
数据
[{'name': 'Gender',
'column_index': 2,
'ks_score': 0.0022329709328575142,
'p_value': 1.0,
'quartiles': [[0.0, 0.0, 1.0, 1.0, 2.0], [0.0, 0.0, 1.0, 1.0, 2.0]],
't_test_p_value': 0.8341377317414621,
'diff_means': 0.0014959875249118681,
'primary_group_mean': 0.6312769010043023,
'secondary_group_mean': 0.6297809134793905,
'ks_sign': '+',
'group0_percent_null': 0.0,
'group1_percent_null': 0.0},
{'name': 'Gender_Missing_color',
'column_index': 3,
'ks_score': 2.220446049250313e-16,
'p_value': 1.0,
'quartiles': [[1.0, 1.0, 1.0, 1.0, 1.0], [1.0, 1.0, 1.0, 1.0, 1.0]],
't_test_p_value': 1.0,
'diff_means': 0.0,
'primary_group_mean': 1.0,
'secondary_group_mean': 1.0,
'ks_sign': '0',
'group0_percent_null': 0.9966523194643712,
'group1_percent_null': 0.9959153360564427},
{'name': 'Gender_Missing',
'column_index': 4,
'ks_score': 0.0007369834078797544,
'p_value': 1.0,
'quartiles': [[0.0, 0.0, 0.0, 0.0, 1.0], [0.0, 0.0, 0.0, 0.0, 1.0]],
't_test_p_value': 0.40301091478187256,
'diff_means': -0.0007369834079284866,
'primary_group_mean': 0.0033476805356288893,
'secondary_group_mean': 0.004084663943557376,
'ks_sign': '-',
'group0_percent_null': 0.0,
'group1_percent_null': 0.0},
{'name': 'Male',
'column_index': 5,
'ks_score': 0.0029699543407862294,
'p_value': 0.9999999999915384,
'quartiles': [[0.0, 0.0, 1.0, 1.0, 1.0], [0.0, 0.0, 1.0, 1.0, 1.0]],
't_test_p_value': 0.6740956861786738,
'diff_means': 0.0029699543407684104,
'primary_group_mean': 0.6245815399330444,
'secondary_group_mean': 0.621611585592276,
'ks_sign': '+',
'group0_percent_null': 0.0,
'group1_percent_null': 0.0}]
这是我得到的输出。如上所述,我只需要此词典列表中的一些数据。试试看
[{'name':c['name'],'p_value':c['p_value']}表示comps中的c]
您可以在比较中为每个对象创建一个新的dict,并仅使用名称
和p\u值
键对其进行初始化
ex=[{'name':d['name'],'p_value':d['p_value']}用于比较中的d]
我仍然不确定如何让上述答案对我有效。然而,我想出了另一种方法:
test = [(c['name'],c['p_value'], c['group0_percent_null']) for c in comps]
pd.DataFrame(test)
0 1 2
0 ID 5.374590e-13 0.000000
1 Gender 1.000000e+00 0.000000
2 Gender_Missing_color 1.000000e+00 0.996652
3 Gender_Missing 1.000000e+00 0.000000
4 Male 1.000000e+00 0.000000
... ... ... ...
它给了我我想要的结果。[(c['name',c['p\u value'])表示comps中的c]
?您期望的结果是什么?[{'name':项['name'],'p_值]:comps中项的项['p_值']}
@juanpa.arrivillaga他们给出了期望的结果?只是一份精简的目录?哈,太棒了。所以你根本就没有这种结构,它可能被序列化为JSON,所以一个巨大的string@roganjosh,我已经更新了问题,以显示我从服务器收到的实际数据。但问题仍然存在,即使您认为结构不一样,而且由于我在python方面没有那么丰富的经验,让我们假设它不是。我该如何解决这个问题呢?虽然这段代码可能会回答这个问题,但它如何或为什么解决这个问题会真正有助于提高你文章的质量。请记住,你是在将来回答读者的问题,而不仅仅是现在提问的人。请在回答中添加解释,并说明适用的限制和假设。
test = [(c['name'],c['p_value'], c['group0_percent_null']) for c in comps]
pd.DataFrame(test)
0 1 2
0 ID 5.374590e-13 0.000000
1 Gender 1.000000e+00 0.000000
2 Gender_Missing_color 1.000000e+00 0.996652
3 Gender_Missing 1.000000e+00 0.000000
4 Male 1.000000e+00 0.000000
... ... ... ...