Question

关于扁平化哈希的话题很多，但是我找不到关于我的情况的任何信息。

我有哈希表等的哈希表...

类似：

[:medical_address,:address_1, nil, :city, nil, :state, nil, :zip5, nil, :zip9, nil, :pharmacy_address, :address_1, nil, :city, nil, :state, nil, :zip5, nil, :zip9, nil,, :address,, :member_phone, "8000000000", :provider_phone, nil, :phone,"800-000-0000", "8000000000", "800-000-0000", "8000000000", "800-624-5060", "8006245060", "804-673-1678", "8046731678", "888-258-3432", "8882583432", "800-000-0000", "8000000000", "800-000-0000", "8000000000", "800-000-0000", "8000000000", "800-624-5060", "8006245060", "804-673-1678", "8046731678", "888-258-3432", "8882583432", :website,"www.fopblue.org", "www.fepblue.org", "www.fepblue.org", "www.fopblue.org"]

我想将其展平为单个数组。

有人方便地使用了一个很好的简单递归函数吗？我希望结果看起来像这样：

# NODE consist of value and pointer to the next node
class node():
    def __init__(self,x):
        self.val=x
        self.next=None

list1=node('m')
list1.next=node('a')
list1.next.next=node('d')
list1.next.next.next=node('a')
list1.next.next.next.next=node('m')

# head is declared as a global variable for checking purpose
# i and j will keep track of the linked list length
# for the odd number of nodes i==j
# for the even number of nodes i>j
# we will check until i>=j(EXIT CONDITION)


head=list1
i,j=0,0

def palindrome(list):
    # base condition

    if list is None:
        return 0

    # j variable will keep track of the nodes

    global j
    j+=1
    x = palindrome(list.next)

    #if we get a single FALSE then there is no need to check further
    # we will return FALSE in each case
    if x is False:
        return False
    global head,i
    i+=1
    j-=1

    #EXIT CONDITION

    if i>=j:
        return True
    #if the value is evaluated to be false then return false
    if head.val is list.val:
        head=head.next
        return True
    else:
        return False


print(palindrome(list1))

Answer 1

如果您适合使用宝石，则可以使用多功能Iteraptor宝石。

input = [your structure]
input.iteraptor.flatten
# ⇒ {"medical_address.0.address_1"=>nil,
#    "medical_address.0.city"=>nil,
#    "medical_address.0.state"=>nil,
#    "medical_address.0.zip5"=>nil,
#    "medical_address.0.zip9"=>nil,
#    "pharmacy_address.0.address_1"=>nil,
#    "pharmacy_address.0.city"=>nil,
#    "pharmacy_address.0.state"=>nil,
#    "pharmacy_address.0.zip5"=>nil,
#    "pharmacy_address.0.zip9"=>nil,
#    "member_phone"=>"8000000000",
#    "provider_phone"=>nil,
#    "phone.0"=>"800-000-0000",
#    "phone.1"=>"8000000000",
#    "phone.2"=>"800-000-0000",
#    "phone.3"=>"8000000000",
#    "phone.4"=>"800-624-5060",
#    "phone.5"=>"8006245060",
#    "phone.6"=>"804-673-1678",
#    "phone.7"=>"8046731678",
#    "phone.8"=>"888-258-3432",
#    "phone.9"=>"8882583432",
#    "phone.10"=>"800-000-0000",
#    "phone.11"=>"8000000000",
#    "phone.12"=>"800-000-0000",
#    "phone.13"=>"8000000000",
#    "phone.14"=>"800-000-0000",
#    "phone.15"=>"8000000000",
#    "phone.16"=>"800-624-5060",
#    "phone.17"=>"8006245060",
#    "phone.18"=>"804-673-1678",
#    "phone.19"=>"8046731678",
#    "phone.20"=>"888-258-3432",
#    "phone.21"=>"8882583432",
#    "website.0"=>"www.fopblue.org",
#    "website.1"=>"www.fepblue.org",
#    "website.2"=>"www.fepblue.org",
#    "website.3"=>"www.fopblue.org"}

请注意，拼合是可逆的，result.recoger将为您返回哈希值。

Ruby-展平数组和哈希的哈希

1 个答案: