Python complex formula using the items of a dictionary (or a list of dictionaries)
1
2
Entering edit mode
3 months ago
Bianca ▴ 20

I have a list with 20 dictionaries that are similar to these:

dic1 = {"aa":1, "ba":1, "ca":1, "dd":2, "ee":2, "fa":2, "ga":4, "hh":4, "ia":1}

dic2 = {"aa":1, "bc":1, "cd":2, "dd":2, "ea":2, "fg":2, "gk":4, "hh":4, "mb":5}

dic3 = {"ab":1, "bd":2, "cm":2, "dj":2, "ej":2, "fa":2, "gg":2, "ha":4}


I have to apply the same formula to each dictionary and I would like to output a list with the results of the formula per dictionary (thus, a list with 20 integers).

The formula will vary according to the "value" amount:

In summary, per dictionary I need to calculate final_formula:

If value == 1, then my partial formula is: formula1 = (number of unique keys that contain value 1) [1 - x + x ((x-10/x) ^1]

If the value is anything above 1 (else), then my intermediary formulas are formulan = (number of unique keys with value i) [(value i) - x + x ((x-10/x)^(value i)]

final_formula = formula 1 + formula2 + formula3 (n formulas, where n is the number of distinct values =! 1). X is a constant that I will define. The output should be an integer.


A practical example: For my dic1, I have 4 keys with value ==1, 3 keys with value ==2, 2 keys with value ==4 The formula for dic1 thus is:

final_formula_dic1 = 4 [1 - x + x ((x-10/x)^1] + 3 [2 - x + x ((x-10/x)^2] + 2 * [4 - x + x ((x-10/x)^4]

For my dic2, I have 2 keys with value ==1, 4 keys with value == 2, 2 keys with value == 4, 1 key with value == 5 The formula for dic2 thus is:

final_formula_dic2 = 2 [1 - x + x ((x-10/x)^1] + 4 [2 - x + x ((x-10/x)^2] + 2 * [4 - x + x ((x-10/x)^4]

For my dic3, I have 1 key1 with value ==1, 6 keys with value == 2, 1 key with value == 4 The formula for dic3 thus is:

final_formula_dic3 = 1 [1 - x + x ((x-10/x)^1] + 6 [2 - x + x ((x-10/x)^2] + 1 * [4 - x + x ((x-10/x)^4]

I am applying each formula manually to each dictionary, but I have too many dictionaries in my list and this is error prone.

I would really like to iterate over all dictionaries and apply the formula (since it is the same criteria to all dictionaries) and create a list with results, for example, list = ['final_formula_dic1', 'final_formula_dic2', 'final_formula_dic3'...] which would be integers, for instance: list = ['3000', '3200', '1300'...].

I hope this makes sense. Thank you very much in advance!

python dictionary • 483 views
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1
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This has no connection to bioinformatics, and appears to be an ordinary programming problem. Given your level of explanation and understanding, I think you should be able to figure it out with some research.

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5
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3 months ago

Hope bioinformatics will benefit.

X = 2     # Define X here
DICTS = [
    {"aa":1, "ba":1, "ca":1, "dd":2, "ee":2, "fa":2, "ga":4, "hh":4, "ia":1},
    {"aa":1, "bc":1, "cd":2, "dd":2, "ea":2, "fg":2, "gk":4, "hh":4, "mb":5},
    {"ab":1, "bd":2, "cm":2, "dj":2, "ej":2, "fa":2, "gg":2, "ha":4},
]

for d in DICTS:
    dt = {}
    for key, value in d.items():
        if value not in dt:
            dt[value] = set()
        dt[value].add(key)

    final_result = 0
    final_formula = []
    for value in sorted(dt):
        n = len(dt[value])
        formula = f'{n} * ({value} - X + X * (X - 10 / X^{value})'
        final_formula.append(formula)
        result = n * (value - X + X * (X - 10 / X) ** value)
        final_result += result

    print(" + ".join(final_formula))
    print(final_result)

Output:

4 * (1 - X + X * (X - 10 / X^1) + 3 * (2 - X + X * (X - 10 / X^2) + 2 * (4 - X + X * (X - 10 / X^4)
354.0
2 * (1 - X + X * (X - 10 / X^1) + 4 * (2 - X + X * (X - 10 / X^2) + 2 * (4 - X + X * (X - 10 / X^4) + 1 * (5 - X + X * (X - 10 / X^5)
-97.0
1 * (1 - X + X * (X - 10 / X^1) + 6 * (2 - X + X * (X - 10 / X^2) + 1 * (4 - X + X * (X - 10 / X^4)
256.0
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1
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I am always reminded that there are star programmers, and more importantly kind people, who will answer even questions that are not meant for this forum.

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0
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Thank you for the kind words. I am delighted to hear them from you, Mensur Dlakic .

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0
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Thank you very much for your help

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