Modify existing value in dictionary (or defaultdictionary) by incrementing that value

Question

I would like to keep a counter for every time a particular value comes up in the list I am scanning.

For example: list:

[(a, 0.2), (b, 1), (a, 0.2), (a, 1)]

I would like a dictionary that can show the following:

mydict = {"a": (# val below 1, # val equal to 1), ...}

Therefore: mydict = {"a": (2, 1), "b" :(0, 1)}

Is there a way to do this with a default dictionary or normal dictionary?

Should I do something like: mydict[mydict["a"]+1] for every value I see either below or equal to 1?

Answer 1

OK, assuming the type of input is an array of arrays and you can store the results as an array in your dictionary, here is how it could be done.

# Define list of numbers
lettersNumbersList = [["a", 0.2], ["b", 1], ["a", 0.2], ["a", 1]]

# Here is the dictionary you will populate.
numberOccurences = {}

# This function is used to increment the numbers depending on if they are less
# than or greater than one.
def incrementNumber(letter, number):

    countingArray = numberOccurences[letter]

    if number < 1:
        countingArray[0] = countingArray[0] + 1
    elif number >= 1:
        countingArray[1] = countingArray[1] + 1

    return(countingArray)

# Loops through all of the list, gets the number and letter from it. If the letter
# is already in the dictionary then increments the counters. Otherwise starts 
# both from zero.
for item in lettersNumbersList:

    letter = item[0]
    number = item[1]

    if letter in numberOccurences:
        numberOccurences[letter] = incrementNumber(letter, number)

    else:
        numberOccurences[letter] = [0, 0]
        numberOccurences[letter] = incrementNumber(letter, number)

print(numberOccurences)