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Create groups from an unstructured data array using Ramda or similar


Create groups from an unstructured data array using Ramda or similar

By : Christine Burnette
Date : November 22 2020, 03:01 PM
this will help I like to think about questions like this in terms of a few steps that keep moving me toward my desired output. Sometimes this means that I miss a more elegant solution, but it usually makes it easier to come up with something that works.
So let's look at your problem this way using Ramda.
code :
[
  {option: "option1", set: "set1"},
  {option: "option2", set: "set1"},
  {option: "option3", set: "set1"},
  {option: "optionA", set: "set2"}
  {option: "optionB", set: "set2"}
]
{
  set1: [
    {option: "option1", set: "set1"},
    {option: "option2", set: "set1"},
    {option: "option3", set: "set1"},
  ],
  set2: [
    {option: "optionA", set: "set2"}
    {option: "optionB", set: "set2"}
  ]
}
{
  set1: ["option1", "option2", "option3"], 
  set2: ["optionA", "optionB"]
}
[
  ["set1", ["option1", "option2", "option3"]], 
  ["set2", ["optionA", "optionB"]]
]
[
  {
    set: "set1",
    options: ["option1", "option2", "option3"]
  },
  {
    set: "set2",
    options: ["optionA", "optionB"]
  }
]
const transform = pipe(
  pluck('data'),
  groupBy(prop('set')),
  map(pluck('option')),
  toPairs,
  map(zipObj(['set', 'options']))
)
transform(myObj)


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create array with form data which groups and sorts data

create array with form data which groups and sorts data


By : LunaSa
Date : March 29 2020, 07:55 AM
it should still fix some issue I'm assuming you can rely on the hidden input fields always appearing in groups of four with an id[], name[], sales[] & price[] for each group, otherwise (obviously) you can't tell which fields are related. So rather than use .serializeArray(), which returns a single array with all the values, I'd put the ids in their own array, the names in their own, and so forth. Perhaps something like this:
code :
function showValues() {
    function getVal(el, i) {
        return el.value;
    }
    var ids = $.map($('input[name="id[]"]'), getVal),
        names = $.map($('input[name="name[]"]'), getVal),
        sales = $.map($('input[name="sales[]"]'), getVal),
        prices = $.map($('input[name="price[]"]'), getVal),
        data = {},
        i,
        $results = $("#results");

    for (i = 0; i < ids.length; i++) {
        if (!data[ids[i]]) {
            // if current id is new add a record for it:
            data[ids[i]] = {
                "id":ids[i],"name":names[i],"sales":+sales[i],"price":+prices[i]
            };
        } else {
            // otherwise add to existing record's totals
            data[ids[i]].sales += +sales[i];
            data[ids[i]].price += +prices[i];
        }
    }
    // data object now contains the details for each salesman,
    // so turn it into an array to allow sorting:
    data = $.map(data, function(val, key) { return val; });
    data.sort(function(a,b) { return a.price - b.price; });

    // now output table - assume there's already a table element with headings
    $.each(data, function(i, val) {
        var $tr = $("<tr></tr>").attr("data-id", val.id).appendTo($results);
        $("<td></td>").text(val.name).appendTo($tr);
        $("<td></td>").text(val.sales).appendTo($tr);
        $("<td></td>").text(val.price).appendTo($tr);
        $("<td></td>").text(val.price / 10).appendTo($tr);
    });
}
ids = $.map($('input[name="id[]"]'), getVal)
How to create structured array out of unstructured HTML using python

How to create structured array out of unstructured HTML using python


By : user3530646
Date : March 29 2020, 07:55 AM
To fix this issue First, I need to restate your question. The example shows a div tag which contains inside it an A tag. The A tag has an ID which you want to use as the a key for looking up the following table. The div tag is followed by a table. Each row of the table contains a name-value pair associated with the object identified in the previous A.
You have a page filled with multiple div tags, each of which is described by my previous paragraph.
code :
from bs4 import BeautifulSoup
from pprint import pprint

result = {}
soup = BeautifulSoup(page)
divs = soup.find_all("div", {"class":"info"})
for div in divs:
    name = div.find("a")["id"]
    table = div.find_next("table", {"class":"properties"})
    rows = table.find_all("tr", {"class":None})
    rowd = {}
    for row in rows:
        cells = row.find_all("td")
        rowd[cells[0].text] = cells[1].text
    result[name] = rowd
pprint (result)
result = {
    div.find("a")["id"]: {
        cells[0].text : cells[1].text
        for row in table.find_all("tr", {"class":None})
        for cells in [row.find_all("td")]
    }
    for div in soup.find_all("div", {"class":"info"})
    for table in [div.find_next("table", {"class":"properties"})]
}

pprint(result)
{'bc968f9fa2db71455f50e0c13ce50e871fS7f0e': {u'AppServer': u'WebLogic 10',
                                             u'SSL_Port': u'28443',
                                             u'configureBPMUIStaticContent': u'true',
                                             u'context': u'workspace',
                                             u'instance_home': u'/essdev1/app/oracle/Middleware/user_projects/epmsystem1',
                                             u'isCompact': u'false',
                                             u'maintVersion': u'11.1.2.2.0.66',
                                             u'port': u'28080',
                                             u'serverName': u'FoundationServices0',
                                             u'validationContext': u'workspace/status',
                                             u'version': u'11.1.2.0'}}
print result["bc968f9fa2db71455f50e0c13ce50e871fS7f0e"]["serverName"]
How can I split a dictionary into groups based on similar keys then create an array of those groups with updated keys?

How can I split a dictionary into groups based on similar keys then create an array of those groups with updated keys?


By : user2333471
Date : March 29 2020, 07:55 AM
will help you I have a dictionary with keys that are appended with _0, _1, _2.
code :
func transformDict(dict: [String: String]) -> [[String: String]] {
    var temp: [String: [String: String]] = [:]
    for (key, value) in dict {
        let components = key.components(separatedBy: "_")
        if components.count == 2 {
            let groupKey = components[1]
            let dictKey = components[0]
            if temp[groupKey] != nil {
                temp[groupKey]?[dictKey] = value
            } else {
                temp[groupKey] = [dictKey: value]
            }
        }
    }

    let res = Array(temp.values)
    return res
}

let res = transformDict(dict: ["a_0":"0",
                               "b_1":"1",
                               "c_0":"2",
                               "d_2":"3"])
print(res)
How to create unstructured data in terraform

How to create unstructured data in terraform


By : user3011466
Date : March 29 2020, 07:55 AM
it should still fix some issue There is no such thing as "unstructured data" in Terraform: every value has an associated type. However, in Terraform 0.12 introduced two structural types that allow for different element/attribute types to be mixed together inside a single value, which is not possible for the collection types.
You can use Local Values if you need to factor out the expressions for these structural values for use in multiple locations:
code :
locals {
  your_data = {
    settings = {
      foo = "bar"
      baz = []
    }
  }
}
object({
  settings = object({
    foo = string
    baz = tuple([])
  })
})
variable "example" {
  type = object({
    settings = object({
      foo = string
      baz = list(string)
    })
  })
}
module "example" {
  source = "./modules/example"

  example = local.your_data
}
How to create an array of property/value from an object using Ramda?

How to create an array of property/value from an object using Ramda?


By : MilZen22
Date : March 29 2020, 07:55 AM
Hope that helps I have an object like , A ramda solution:
code :
R.pipe(
  R.toPairs,
  R.map(R.zipObj(['name', 'value']))
)(obj)
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