CODEWITHSUNDEEP
pythonrecursiondictionary
someday
someday

Reputation: 29

python recursive in multiple dict

i have a problem with multiple dict in recursive , the origin non-recursive code is 

mylist = (["AA","BB","CC","DD"])
tempdict = dict()
n = len(mylist)
for j in range(0 , n ):
    if j == 0:
        if mylist[j] not in tempdict:
            tempdict[mylist[j]] = "1"
    if j == 1:
        if mylist[j] not in tempdict[mylist[0]]:
            tempdict[mylist[0]] = dict()
            tempdict[mylist[0]][mylist[1]] = "1"
    if j == 2:
        if mylist[j] not in tempdict[mylist[0]][mylist[1]]:
            tempdict[mylist[0]][mylist[1]] = dict() 
            tempdict[mylist[0]][mylist[1]][mylist[2]] = "1"
    if j == 3:
        .......
    if j == 4:
        .......
    if j == n:
        .......
print tempdict

Result : {'AA' {'BB': {'CC': {'DD': '1'}}}} and it is work when i need build a multiple key by dict(). however , it is impossible to list all. so i would like to refine code in a recursive function 

def rfun(tmpdict, mylist, idx, listlen):
    if idx < listlen:
        if idx == 0:
            if mylist[idx] not in tmpdict:
                tmpdict[mylist[idx]] = "1"
            rfun(tmpdict [mylist[idx]], list, idx + 1, listlen)
        else:
            if list[idx] not in tmpdict:
                tmpdict = dict()
                tmpdict [mylist[idx]] = "1"
            rfun(tmpdict [mylist[idx]], mylist, idx + 1, listlen)

newdict = dict()
mylist = (["AA","BB","CC","DD"]
print rfun(newdict, mylist, 0, len(mylist))

Result : {'AA':'1'}

but , the result is not in my expect ,
please help me to find what is wrong in my recursive code , thanks all.

Upvotes: 0

Views: 287

Answers (2)

Friendly Fire
Friendly Fire

Reputation: 55

mylist = ["AA","BB","CC","DD"]

A recursive version is concise, but in extreme cases could cause stack overflow:

def l2rd(*args):
   return  { args[0] : (len(args)>1) and l2rd(*args[1:]) or "1" }

result = l2rd(*mylist)
print(result)

Result: {'AA': {'BB': {'CC': {'DD': '1'}}}

A non-recursive version could do the same thing by looping over the list in reverse order and doing something like:

curr = "1"
for key in reversed_list:
   curr = { key : curr }
return curr

But that requires copying the list in order to reverse it (which should still be more efficient than recursion). Iteration via index would also work using range(len(args)-1,-1,-1)

Upvotes: 1

Christopher C. S. Ke
Christopher C. S. Ke

Reputation: 265

Here you go.

def rfun(tmpdict, mylist, idx, listlen):
    if idx < listlen:
        if idx == listlen - 1: # for last element in mylist
            tmpdict[mylist[idx]] = "1"
        else:
            tmpdict[mylist[idx]] = {}
            rfun(tmpdict[mylist[idx]], mylist, idx + 1, listlen)

newdict = {}
mylist = ["AA","BB","CC","DD"]
rfun(newdict, mylist, 0, len(mylist))
print newdict

The key idea is to pass the newly create dictionary to next recursive function call if the element is not the last one.

Upvotes: 1

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