remove key of matched number in [String:[NSNumber]] - arrays
How to remove a value from dictionary with array of values. I have response from server like below:
[
"business_proof":[
0,
0,
0,
0,
0,
0,
0,
0,
-1,// business_proof contains -1, I want to remove this key like wise any other contains
0,
0
],
"reference_proof":[
1,
2,
1
],
"vehicle_proof":[
1,
1,
2
],
"previous_loan_track":[
2,
2,
0,
0,
2,
2
],
"banking_proof":[
1,
1
],
"income_proof":[
0,
0,
2,
0,
2,
1,
2,
0,
0
],
"signature_proof":[
2,
2,
1,
2,
2,
2
],
"employment_proof":[
2,
1,
2,
2,
2,
2,
2
],
"guarantor_proof":[
1,
2,
2
],
"pdc_proof":[
1,
0
],
"address_proof":[
2,
2,
2,
2,
2,
2,
2,
2,
2,
2,
2,
2,
2,
2,
3
],
"age_proof":[
2,
2,
2,
2,
2,
2,
1,
2
],
"contact_proof":[
0,
2,
2
],
"photo_id_proof":[
2,
2,
2,
2,
2,
2,
2,
2,
2,
2
]
]
Second reponse
[
"signature_proof":[
"pan_card",
"driving_licence",
"accepted_documents",
"passport",
"cancelled_cheque",
"bank_report"
],
"guarantor_proof":[
"accepted_documents",
"co_applicant",
"guarantor"
],
"previous_loan_track":[
"housing_loan",
"vehicle_loan",
"over_draft_limit",
"accepted_documents",
"business_loan",
"personal_loan"
],
"address_proof":[
"bank_statement",
"voter_id",
"rental_agreement",
"eb_bill",
"registration_document",
"hr_letter",
"driving_licence",
"property_tax_receipt",
"telephone_bill",
"cc_statement",
"gas_bill",
"aadhaar_card",
"passport",
"ration_card",
"accepted_documents"
],
"vehicle_proof":[
"vehi_insurance",
"vehi_rc",
"accepted_documents"
],
"business_proof":[
"business_commencement_certificate",
"ssi_msme_certificate",
"business_transactions",
"mou",
"aoa",
"gst_no",
"tan_no",
"business_agreements",
"accepted_documents",
"shop_and_establishment_certificate",
"incorporation_certificate"
],
"banking_proof":[
"bank_statement",
"accepted_documents"
],
"income_proof":[
"form_16",
"profit_loss_statement",
"rental_income_proof",
"payslip",
"income_in_cash_proof",
"accepted_documents",
"brokerage_income_proof",
"it_returns",
"audited_balance_sheet"
],
"reference_proof":[
"ref2",
"accepted_documents",
"ref1"
],
"employment_proof":[
"employee_id_card",
"accepted_documents",
"payslip",
"relieving_letter",
"comp_app_letter",
"hr_letter",
"epf_no_uan_no"
],
"age_proof":[
"employee_id_card",
"ration_card",
"pan_card",
"passport",
"voter_id",
"school_certificate",
"accepted_documents",
"aadhaar_card"
],
"contact_proof":[
"accepted_documents",
"landline_bill",
"mobile_bill"
],
"photo_id_proof":[
"employee_id_card",
"nrega_card",
"ration_card",
"bank_passbook",
"pan_card",
"passport",
"voter_id",
"driving_licence",
"accepted_documents",
"aadhaar_card"
],
"pdc_proof":[
"cheque_book",
"accepted_documents"
]
]
Both are array of dictionary, Both will have same key only. I know this structure is completely wrong.
business_proof only contains - 1, so i want to remove both places.
Here i need to remove key and values if anyone key values contains -1.
I am trying like this but it shows compiler error
finalValueArray.removeAll(where: { $0.contains(-1) })
You can use filter as far as I understand
var filteredItems = object.filter { !$0.value.contains(-1)}
You can get all elements which contain -1
var minusOneItems = object.filter { $0.value.contains(-1)}
and than
for negativeItem in minusOneItems {
object.removeValue(forKey: negativeItem.key)
}
It depends what you need.
You can simply use a combination of forEach and contains on the dictionary like so,
var dictionary = ["business_proof": [0, 0, 1, -1, 2, -1], "reference_proof": [1, 2, 1], "vehicle_proof": [-1, 0, 0, 2]]
dictionary.forEach { (key,value) in
dictionary[key] = value.contains(-1) ? nil : value
}
print(dictionary) //["reference_proof": [1, 2, 1]]
Or you can simply apply filter on dictionary like,
dictionary = dictionary.filter({ !$0.value.contains(-1) })
print(dictionary) //["reference_proof": [1, 2, 1]]
You can filter your dictionary to remove the entries that contains -1 in their value.
let filteredArrayOnDict = dataDict.filter { value.contains{ $0 != -1 } }
The filteredArrayOnDict is the array of tuples. Now if you want to create a dictionary from it. You can do this way:
let filteredDictionary = filteredArrayOnDict.reduce(into: [:]) { $0[$1.0] = $1.1}
Now you have the only the entries in filteredDictionary that doesn't have -1 in their value.
You can use while loop to remove all key-value pairs which contain -1 from both dictionaries.
var dict1 = ["business_proof":[0,0,0,0,0,0,0,0,-1,0,0],"reference_proof":[1,2,1],"vehicle_proof":[1,1,2],"previous_loan_track":[2,2,0,0,2,2],"banking_proof":[1,1],"income_proof":[0,0,2,0,2,1,2,0,0],"signature_proof":[2,2,1,2,2,2],"employment_proof":[2,1,2,2,2,2,2],"guarantor_proof":[1,2,2],"pdc_proof":[1,0],"address_proof":[2,2,2,2,2,2,2,2,2,2,2,2,2,2,3],"age_proof":[2,2,2,2,2,2,1,2],"contact_proof":[0,2,2],"photo_id_proof":[2,2,2,2,2,2,2,2,2,2]]
var dict2 = ["signature_proof":["pan_card","driving_licence","accepted_documents","passport","cancelled_cheque","bank_report"],"guarantor_proof":["accepted_documents","co_applicant","guarantor"],"previous_loan_track":["housing_loan","vehicle_loan","over_draft_limit","accepted_documents","business_loan","personal_loan"],"address_proof":["bank_statement","voter_id","rental_agreement","eb_bill","registration_document","hr_letter","driving_licence","property_tax_receipt","telephone_bill","cc_statement","gas_bill","aadhaar_card","passport","ration_card","accepted_documents"],"vehicle_proof":["vehi_insurance","vehi_rc","accepted_documents"],"business_proof":["business_commencement_certificate","ssi_msme_certificate","business_transactions","mou","aoa","gst_no","tan_no","business_agreements","accepted_documents","shop_and_establishment_certificate","incorporation_certificate"],"banking_proof":["bank_statement","accepted_documents"],"income_proof":["form_16","profit_loss_statement","rental_income_proof","payslip","income_in_cash_proof","accepted_documents","brokerage_income_proof","it_returns","audited_balance_sheet"],"reference_proof":["ref2","accepted_documents","ref1"],"employment_proof":["employee_id_card","accepted_documents","payslip","relieving_letter","comp_app_letter","hr_letter","epf_no_uan_no"],"age_proof":["employee_id_card","ration_card","pan_card","passport","voter_id","school_certificate","accepted_documents","aadhaar_card"],"contact_proof":["accepted_documents","landline_bill","mobile_bill"],"photo_id_proof":["employee_id_card","nrega_card","ration_card","bank_passbook","pan_card","passport","voter_id","driving_licence","accepted_documents","aadhaar_card"],"pdc_proof":["cheque_book","accepted_documents"]]
while let invalid = dict1.first(where: { $0.value.contains(-1) }) {
dict1.removeValue(forKey: invalid.key)
dict2.removeValue(forKey: invalid.key)
}
print(dict1)//["reference_proof":[1,2,1],"vehicle_proof":[1,1,2],"previous_loan_track":[2,2,0,0,2,2],"banking_proof":[1,1],"income_proof":[0,0,2,0,2,1,2,0,0],"signature_proof":[2,2,1,2,2,2],"employment_proof":[2,1,2,2,2,2,2],"guarantor_proof":[1,2,2],"pdc_proof":[1,0],"address_proof":[2,2,2,2,2,2,2,2,2,2,2,2,2,2,3],"age_proof":[2,2,2,2,2,2,1,2],"contact_proof":[0,2,2],"photo_id_proof":[2,2,2,2,2,2,2,2,2,2]]
print(dict2)//["signature_proof":["pan_card","driving_licence","accepted_documents","passport","cancelled_cheque","bank_report"],"guarantor_proof":["accepted_documents","co_applicant","guarantor"],"previous_loan_track":["housing_loan","vehicle_loan","over_draft_limit","accepted_documents","business_loan","personal_loan"],"address_proof":["bank_statement","voter_id","rental_agreement","eb_bill","registration_document","hr_letter","driving_licence","property_tax_receipt","telephone_bill","cc_statement","gas_bill","aadhaar_card","passport","ration_card","accepted_documents"],"vehicle_proof":["vehi_insurance","vehi_rc","accepted_documents"],"banking_proof":["bank_statement","accepted_documents"],"income_proof":["form_16","profit_loss_statement","rental_income_proof","payslip","income_in_cash_proof","accepted_documents","brokerage_income_proof","it_returns","audited_balance_sheet"],"reference_proof":["ref2","accepted_documents","ref1"],"employment_proof":["employee_id_card","accepted_documents","payslip","relieving_letter","comp_app_letter","hr_letter","epf_no_uan_no"],"age_proof":["employee_id_card","ration_card","pan_card","passport","voter_id","school_certificate","accepted_documents","aadhaar_card"],"contact_proof":["accepted_documents","landline_bill","mobile_bill"],"photo_id_proof":["employee_id_card","nrega_card","ration_card","bank_passbook","pan_card","passport","voter_id","driving_licence","accepted_documents","aadhaar_card"],"pdc_proof":["cheque_book","accepted_documents"]]
Create a struct and initialize the struct objects with both dictionary values. Then store the struct objects in an array. Now you can filter the array by its values
struct Proof {
var title: String
var arr: [Int]
var documents: [String]
}
var proofs = [Proof]()
dict1.forEach {
if let docs = dict2[$0.key] {
proofs.append(Proof(title: $0.key, arr: $0.value, documents: docs))
}
}
print(proofs)
let validProofs = proofs.filter { !$0.arr.contains(-1) }
print(validProofs)
Enumerate the index dictionary, look for occurrences of -1 and filter the indices. Then reverse the found indices and remove the items in both arrays. The code considers that both dictionaries are value types
var indexDict = ["business_proof":[0,0,0,0,0,0,0,0,-1,0,0] ...
var valueDict = ["signature_proof":["pan_card","driving_licence","accepted_documents","passport","cancelled_cheque","bank_report"] ...
for (key, value) in indexDict {
let foundIndices = value.indices.filter({value[$0] == -1})
for index in foundIndices.reversed() {
indexDict[key]!.remove(at: index)
valueDict[key]!.remove(at: index)
}
}
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I have an array: arr2 = [6, '(7,0)', '(15,0)', '(5,0)', 3, '(15,2)', 17] I want to parse each element of the array and get the value as follows: arr2 = [6, 0, 0, 0, 3, 2, 17] means if the array element is like (15,2) then only the second element i.e 2 should get printed in the response, and if the array element is not in the format like (7,0) then it should be printed as it is.
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This seems fine to me: arr2.map{|x| x.to_s.scan(/\d+/).last.to_i } And if you want something without regexp: arr2.map do |item| item.is_a?(Integer) ? item : item[1+item.rindex(',')..-2].to_i end