Table of Tribonacci sequence using NumPy and PANDAS

Question

What I am trying to accomplish is calculate the Tribonacci numbers and the ratio. I can do this in Excel easy like this.

So I tried do the same using Python 3 now. It works and gives such output.

But would you be so kind to review the code and see what obvious mistakes you see in it and advise on what is a better approach as I do not want to continue coding if my logic to tackle such problem is fundamentally flawed somewhere.

Priorities for me are execution time and readability I guess

What would you have done differently and why? I know it may look like a mess. From what I heard using exec and eval (code as string) are the bad ideas and also that function calls are expensive?

I don't like my code at all especially how next() is used so many times but hopefully you can see how I was thinking trying to replicate the Excel spreadsheet

To run the code you need numpy and pandas

import pandas as pd
import numpy as np

column_names = ["Number sequence", "Ratio"]
column_counter = len(column_names) -1
header_columns = ['']
while column_counter >= 0:
    column_counter = column_counter - 1
    header_columns.append(column_names[column_counter])

tribonacci_numbers = [0,1]    
tribonacci_counter = 19 #Reach
while tribonacci_counter -1 >= 0:
    tribonacci_counter = tribonacci_counter - 1
    tribonacci_numbers.append(sum(tribonacci_numbers[-3:]))
tribonaccis_list_length = len(tribonacci_numbers)-1

index_builder = 1
rows = []
while index_builder <= tribonaccis_list_length:
    try:
        index_builder = index_builder + 1
        rows.append((["Number: "+str(index_builder),tribonacci_numbers[index_builder],tribonacci_numbers[index_builder]/tribonacci_numbers[index_builder-1]]))       
    except IndexError:
        continue

def get_row():
    the_row = [x for x in rows]
    row_counter = 0
    while row_counter <= len(the_row):
        try:
            row_counter = row_counter + 1
            yield (the_row[row_counter])            
        except IndexError:
            continue

a = get_row()
def therow():
    for q in a:
        yield q

datas = np.array([
header_columns,
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow()),
next(therow())])

df = pd.DataFrame(data=datas[1:,1:],index=datas[1:,0],columns=datas[0,1:])
print(df)

If you want to provide an example that does the same and produces a Pandas dataframe like this it would be very useful for me to study your code.

Answer 1

You’re using the wrong tool for the job. Basically, you do all the computation in Python, use numpy for intermediate storage and pandas for display.

Instead, you should compute the list of tribonacci numbers and from there on use pandas for anything else as it would be much more efficient / readable. I’d keep building the tribonacci numbers in Python as I don't know enough pandas to be able to come up with an efficient way that wouldn't involve Dataframe.append and also because it can be expressed very nicely as a generator:

def tribonacci_sequence():
    a, b, c = 0, 1, 1
    yield a
    yield b
    while True:
        yield c
        a, b, c = b, c, a + b + c

You can then select the amount of numbers you like using itertools.islice and give them to a pandas.Dataframe.

From there on, you can easily compute the ratio by shifting the dataframe and dividing element-wise:

def compute_tribonacci(upper_limit, skip_first_rows=3):
    tribonacci = list(islice(tribonacci_sequence(), upper_limit + 1))
    df = pd.DataFrame({'Number sequence': tribonacci})
    df['Ratio'] = df['Number sequence'] / df.shift(1)['Number sequence']
    df.index = df.index.map('Number: {}'.format)
    return df.iloc[skip_first_rows:]

Last thing to do is to actually ask for a certain amount of rows. Full code:

from itertools import islice
import pandas as pd


def tribonacci_sequence():
    a, b, c = 0, 1, 1
    yield a
    yield b
    while True:
        yield c
        a, b, c = b, c, a + b + c


def compute_tribonacci(upper_limit, skip_first_rows=3):
    tribonacci = list(islice(tribonacci_sequence(), upper_limit + 1))
    df = pd.DataFrame({'Number sequence': tribonacci})
    df['Ratio'] = df['Number sequence'] / df.shift(1)['Number sequence']
    df.index = df.index.map('Number: {}'.format)
    return df.iloc[skip_first_rows:]


if __name__ == '__main__':
    df = compute_tribonacci(50)
    print(df)

Answer 2

Some general tips

Put code in functions

That way you can test each part individually. Here the generation of the sequence, calculation of the ratio and exporting to pandas are clear divisions in the functionality of your code

Generators

The best algorithms implement fibonacci sequences as generators

with limit

def fib(n):
    a, b = 0, 1
    for _ in range(n):
        yield a
        a, b = b, a + b

or endless

def fib():
    a, b = 0, 1
    while True:
        yield a
        a, b = b, a + b

Adapting this to tribonacci should be trivial

def trib():
    a, b, c = 0, 1, 1
    while True:
        yield a
        a, b, c = b, c, a + b + c

or more generalized:

def generalised_fibonacci(max_len = 2, begin_data = None):
    from collections import deque
    if begin_data is None:
        begin_data = [0] + [1] * (max_len-1)
    data = deque(begin_data, max_len)
    while True:
        # print(data)
        a = data.popleft()
        yield a
        data.append(a + sum(data))

optimised to @Accumulation's remark

def generalised_fibonacci(max_len = 2, begin_data = None):
    from collections import deque
    if begin_data is None:
        begin_data = [0] + [1] * (max_len - 1) + [max_len - 1]
    data = deque(begin_data, max_len + 1)
    while True:
        a = data.popleft()
        yield a
        # print(a, data)
        data.append(2 * data[-1] - a)

Get it into pandas

here list and itertools.islice are your friends

def get_trib(num_items)
    trib_nums = list(itertools.islice(trib(3), num_items + 1))
    return pd.DataFrame(data=trib_nums, index = range(num_items))

calculate the ratios

pandas has a simple DataFrame.shift method, but it is also possible to calculate it before the introduction into the DataFrame

def trib_ratio(num_items):

    trib_generator = itertools.islice(generalised_fibonacci(3), num_items + 1)
    trib_gen, trib_gen2 = itertools.tee(trib_generator, 2)
    yield next(trib_gen), None

    for trib, trib2 in zip(trib_gen, trib_gen2):
        ratio = trib / trib2 if trib2 else None  # prevent zerodivision on first element
        yield trib, ratio

pd.DataFrame(list(trib_ratio(20)), columns=['trib', 'ratio'])

    trib    ratio
0   0   
1   1   
2   1   1.0
3   2   2.0
4   4   2.0
5   7   1.75
6   13  1.8571428571428572
7   24  1.8461538461538463
8   44  1.8333333333333333
9   81  1.8409090909090908
10  149 1.8395061728395061
11  274 1.8389261744966443
12  504 1.8394160583941606
13  927 1.8392857142857142
14  1705    1.8392664509169363
15  3136    1.8392961876832845
16  5768    1.8392857142857142
17  10609   1.8392857142857142
18  19513   1.8392873974926949
19  35890   1.8392866294265362
20  66012   1.8392867093898022

This can be generalized for more and further ratios, by adapting the tee and keeping the different generators in a data structure

Answer 3

Create a dataframe with the given column names:

df = pd.DataFrame(columns=['Number sequence', 'Ratio'])

Build each row:

for row_number in range(4,number_of_rows):
     row_name = 'Number: {}'.format(row_number)
     dropped_number = df.loc['Number: {}'.format(row_number-4),'Number Sequence']
     current_number = 2*previous_number-dropped_number
     df.loc[row_name,'Number Sequence'] = current_number
     df.loc[row_name,'Ratio'] = current_number/previous_number
     previous_number = current_number

if you're having trouble understanding current_number = 2*previous_number-last_number, consider the case of generating the 7th number. This is going to be 4+7+13. But 13 was calculated by 13 = 2+4+7. So the 7th number is 4+7+(2+4+7) = 2*(2+4+7)-2, which is 2*(6th number) - 3rd number. So the nth number is 2*(n-1)th number - (n-4)th number.

This does require handling the first few rows separately. One way is to manually create them, and then if you don't want them, you can delete them afterwards. Note that if you're deleting them, then you don't have to fill in the ratio column for those rows.

You will also have to initialize previous_number to the proper value.