Printing tabular form lists with generators

Question

I have a list that represents 5 one-month periods with 31 days, and each period contains daily target values for a retail seller. Only for elucidation purposes would the empty structure be like this:

[[], [], [], [], []]

ITERATING OVER THE LIST (WITH VALUES):

for periodo in plan:
    print(periodo)

OUTPUT:

[2000.0, 2000.0, 2000.0, 2000.0, 2000.0, 2000.0]
[2000.0, 2000.0, 2000.0, 2000.0, 2000.0, 2000.0, 2000.0]
[2000.0, 2000.0, 1250.0, 1250.0, 1250.0, 1250.0, 1250.0]
[1250.0, 1250.0, 1250.0, 1250.0, 1250.0, 1250.0, 1250.0]
[1250.0, 1250.0, 1250.0]

The periods did not become equal because there is a cut on Sundays.

I want to plot these values in a modular way, with a horizontal header that discriminates each period like this:

print(f'\033[34m{"1º PERÍODO":<15}{"2º PERÍODO":<15}{"3º PERÍODO":<15}{"4º PERÍODO":<15}{"5º PERÍODO":<15}\033[m')

OUTPUT:

1º PERÍODO     2º PERÍODO     3º PERÍODO     4º PERÍODO     5º PERÍODO 

My difficulty is to have each element of the planning list (periods) organized horizontally according to the header. I tried a solution with geradores, among others, but it just doesn’t work.

My first attempt with generators was this:

gerador_periodo = (periodo for periodo in plan)
print(f"{next(gerador_periodo)}, {next(gerador_periodo)}, {next(gerador_periodo)}, {next(gerador_periodo)}, {next(gerador_periodo)}")

Obviously, it failed, because it printed the periods in a linear way, not respecting the organization of the header above.

My second attempt was using a list comprehension looped for to print breaking lines:

print(f"{[print(num) for num in next(gerador_periodo)]}, {[print(num) for num in next(gerador_periodo)]}, ...")

Here, I received on the screen all values below each other followed by the 5 periods with values None(Why? ). I tried using the parameter end in various ways also.

Is it possible to build this structure without using Pandas, Dataframe, etc? Some other idea?

Answer 1

His second attempt generated several None because the comprehensilist on is using the return of print(num), and print always returns None.

Anyway, from what I understand, you want to have the first element of each of the periods in the first line, in the second line, the second element of each of the periods, etc.

That is, deep down you want to go through all the lists at the same time. It is usually used zip for that, but zip interrupts the loop when the smallest of the lists ends. Only in this case you want to go to the end of the largest of the lists, then you should use itertools.zip_longest:

plan = [
    [2000.0, 2000.0, 2000.0, 2000.0, 2000.0, 2000.0],
    [2000.0, 2000.0, 2000.0, 2000.0, 2000.0, 2000.0, 2000.0],
    [2000.0, 2000.0, 1250.0, 1250.0, 1250.0, 1250.0, 1250.0],
    [1250.0, 1250.0, 1250.0, 1250.0, 1250.0, 1250.0, 1250.0],
    [1250.0, 1250.0, 1250.0]
]

from itertools import zip_longest

size = 15
print(f'{"1º PERÍODO":<{size}}{"2º PERÍODO":<{size}}{"3º PERÍODO":<{size}}{"4º PERÍODO":<{size}}{"5º PERÍODO":<{size}}')

for metas in zip_longest(*plan, fillvalue=''):
    print(('{:<{size}}' * len(metas)).format(*metas, size=size))

The asterisk in *plan and *metas serves to make the unpacking. That is to say, zip_longest(*plan) is the same as doing zip_longest(plan[0], plan[1], etc...). But so I don’t need to know the exact amount of elements in plan.

At each iteration, the variable metas will be a tuple containing the elements of each of the lists contained in plan. In the first iteration, it has the first element of each, in the second iteration, the second element and so on.

The fillvalue indicates the value to be used when a list is smaller than the others and has no more elements to iterate. In this case, I used the empty string, but you can put whatever you want there.

To format, I generated a string that has several times {:<{size}} (the number of times is the number of elements in metas, ie, all values of it will be formatted equally). Then I step the values, also using unpacking, and indicate the size to be used in the alignment.

The exit is:

1º PERÍODO     2º PERÍODO     3º PERÍODO     4º PERÍODO     5º PERÍODO     
2000.0         2000.0         2000.0         1250.0         1250.0         
2000.0         2000.0         2000.0         1250.0         1250.0         
2000.0         2000.0         1250.0         1250.0         1250.0         
2000.0         2000.0         1250.0         1250.0                        
2000.0         2000.0         1250.0         1250.0                        
2000.0         2000.0         1250.0         1250.0                        
               2000.0         1250.0         1250.0                        

---

Another example, changing the fillvalue:

for metas in zip_longest(*plan, fillvalue='-'):
    print(('{:<{size}}' * len(metas)).format(*metas, size=size))

Exit:

1º PERÍODO     2º PERÍODO     3º PERÍODO     4º PERÍODO     5º PERÍODO     
2000.0         2000.0         2000.0         1250.0         1250.0         
2000.0         2000.0         2000.0         1250.0         1250.0         
2000.0         2000.0         1250.0         1250.0         1250.0         
2000.0         2000.0         1250.0         1250.0         -              
2000.0         2000.0         1250.0         1250.0         -              
2000.0         2000.0         1250.0         1250.0         -              
-              2000.0         1250.0         1250.0         -              

---

Note: for the header, you can also use:

print(('{:<{size}}' * len(plan)).format(*map(lambda x: f'{x + 1}º PERÍODO', range(len(plan))), size=size))

Or:

print(*map(lambda x: f'{str(x + 1) + "º PERÍODO":<{size}}', range(len(plan))), sep='')

So it also becomes dynamic, according to the amount of elements in plan.