Trying to code a complex algorithm with `OrderedDict` data Structure

Question

I am trying to write an algorithm:

i have this input of OrderedDict data type like the following:

odict_items([(3, [(0, 1), (1, 1), (1, 1)]), (11, [(0, 0), (1, 1), (1, 1)]), (12, [(0, 0), (1, 1), (1, 1)])])

I am trying to write a function to added the number of same tuple in the each key for example the expected output like the following: if there is (1,1) same tuple then 1 and if twice the 2 and so one:

odict_items([(3, [(0, 1), (1, 1), (1, 1)],2), (11, [(0, 0), (1, 1), (1, 1)],2), (12, [(0, 0), (1, 0), (1, 1)]),1])

this is my try but how to improve it and add it to the OrderedDict ?

def foo(OrderedDict):
    listOfDic = list(makeDataStruc().items())
    tupleCounter = 0
    for i in range(len(listOfDic[1])):
        if listOfDic[1][1][i][0] == 1 and listOfDic[1][1][i][1] == 1:
            tupleCounter += 1
    return tupleCounter

Where am i making error?

Answer 1

I am making the following assumptions,

• You simply want to add the count of (1,1) to the OrderedDict value
• You don't want to create a new data structure which overrides the current OrderedDict.
• You can modify the original Dictionary

Now based on the information available in the question and the above mentioned assumptions, one possible solution could be to replace every value with a list containing two elements, i.e. `[original value, count of (1, 1)]

from collections import OrderedDict

odict_items = [(3, [(0, 1), (1, 1), (1, 1)]),
               (11, [(0, 0), (1, 1), (1, 1)]),
               (12, [(0, 0), (1, 1), (1, 1)])]

my_odict = OrderedDict()

for item in odict_items:
    k, v = item
    my_odict[k] = v

Now count the occurrence of (1,1) in every value and update the values accordingly

pattern_to_find = (1, 1)

for key, value in my_odict.items():
    tuple_count = value.count(pattern_to_find)
    new_value = [value, tuple_count]
    my_odict[key] = new_value

Now the dictionary has the following contents:

OrderedDict([(3, [[(0, 1), (1, 1), (1, 1)], 2]),
             (11, [[(0, 0), (1, 1), (1, 1)], 2]),
             (12, [[(0, 0), (1, 1), (1, 1)], 2])])

Now you can create additional function to access only the value or tuple count

# Returns the count of (1, 1) only
def get_count(my_dict, key):
    return my_dict[key][1]

# Return the original value only
def get_tuple(my_dict, key):
    return my_dict[key][0]

So you can use them like this

print(my_odict[3])
# [[(0, 1), (1, 1), (1, 1)], 2]

print(get_count(my_odict,3))
# 2

print(get_tuple(my_odict, 3))
# [(0, 1), (1, 1), (1, 1)]