Question

我有一个哈希数组，如下所示。我想将一些字段的值与自定义分隔符合并。在这里，我只在阵列中显示两个哈希，它可能有更多。但是，它们总是与此处所示的顺序相同。

import tensorflow.contrib.learn as learn
classifier = learn.DNNClassifier(hidden_units=[10, 20, 10], n_classes=2)

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-30-0273d6847e2b> in <module>()
----> 1 classifier = learn.DNNClassifier(hidden_units=[10, 20, 10], n_classes=2)

TypeError: __init__() missing 1 required positional argument: 'feature_columns'

我想修改详细信息。我想构建新的领域。

{
"details": [
{
  "place": "abc",
  "group": 3,
  "year": 2006,
  "id": 1304,
  "street": "xyz 14",
  "lf_number": "0118",
  "code": 4433,
  "name": "abc coorperation",
  "group2": 3817,
  "group1": 32,
  "postal_code": "22926",
  "status": 2
},
{
  "place": "cbc",
  "group": 2,
  "year": 2007,
  "id": 4983,
  "street": "mnc 14",
  "lf_number": "0145",
  "code": 4433,
  "name": "abc coorperation",
  "group2": 3817,
  "group1": 32,
  "postalcode": "22926",
  "status": 2
}
],
"@timestamp": "2017-09-04",
"parent": {
  "child": [
  {
    "w_2": 0.5,
    "w_1": 0.1,
    "id": 14226,
    "name": "air"
  },
  {
    "w_2": null,
    "w_1": 91,
    "id": 25002,
    "name": "Water"
  }]
},
"p_name": "anacin",
"@version": "1",
 "id": 28841
}

如何为详细信息中的所有值实现此目的。以下是最终结果的样子。

Field 1)  coorperations: (details.name | details.postal_code details.street ; details.name | details.postal_code details.street)

Output:
Coorperations: (abc coorperation |22926 xyz 14; abc coorperation | 22926 mnc 14)

Field 2) access_code: (details.status-details.id-details.group1-details.group2-details.group(always two digit)/details.year(only last two digits); details.status-details.id-details.group1-details.group2-details.group(always two digit)/details.year(only last two digits))

Output: access_code (2-32-3817-03-06; 2-32-3817-02-07)

Answer 1

您可以尝试在rails控制台中运行此代码，哈希是您的json：

new_hash = hash.except(:details)
coorperations = ""
access_code = ""
elements = hash[:details]
elements.each do |element|
    coorperations = "#{coorperations}#{if coorperations.present? then '; ' else '' end}#{element[:name]} | #{element[:postal_code]} #{element[:street]}"
    access_code = "#{access_code}#{if access_code.present? then '; ' else '' end}#{element[:status]}-#{element[:id]}-#{element[:group1]}-#{element[:group2]}-#{element[:group1]}-#{element[:group]}"
end
new_hash.merge!(Coorperations: coorperations)
new_hash.merge!(access_code: access_code)
new_hash

使用相同的键合并哈希数组

1 个答案: