Faster method to conditionally search nested dictionary Python

Question

I have a nested dictionary that I am attempting to conditionally search through to grab some values. I am iterating through another file and grabbing val1, val2, and val3. From there I search through the dictionary to find an ID based on some conditions.

However, for 55M rows of data this is very expensive. I cannot find anywhere a faster way to do this and I am putting this on a spark job. I tried to have it so that if an ID was found we stopped searching through the dict, but im unsure if I did this correctly.

It appears I go through every key in the dictionary to find values, not sure how to optimize this. Any help is appreciated. Here is the code:

for key, val in dict[val1].items():
    if key[0]==val2 or key[1] == val2:
        if len(val3)==1:
            if val3[0]%2==0:
                for key2, val2 in val.items():
                    if key2[2]<=val3[0] and key2[3]>=val3[0]:
                        ID = val2[0]
            if val3[0]%2!=0:
                for key2, val2 in val.items():
                    if key2[0]<=val3[0] and key2[1]>=val3[0]:
                        ID = val2[0]
      if ID!=None:
          break

edit: Input values are like this

val1 = zone#
val2 = 'name'
val3 = score in tuple form like (2,)

and the nested dictionary looks something like this:

{3: defaultdict(<function __main__.<lambda>.<locals>.<lambda>()>,
                         {('jeff', 'jeff A'): defaultdict(list,
                                      {(23,
                                        41,
                                        28,
                                        40,): [61814],

Answer 1

@Gal posted the answer but here is what the set-up now looks like. Runs over 10 times faster than the for loop iteration.

if val2 in dict[val1]:
            if len(val3)==1:
                if val3[0]%2==0:
                    for key2, val5 in dict[val1][val2].items():
                        if key2[2]<=val3[0] and key2[3]>=val3[0]:
                            ID = val5[0]

to meet the condition of val2 having two possible outcomes we create two dict's and run this check twice, its much faster now.