Sort lists in Python 3 depending on another list

define a classEntrada

class Entrada:
    def __init__(codigo, fecha):
        self.codigo = codigo
        self.fecha = fecha

Then if you have two lists with entries you can mix and order them as follows

entradas = [
    Entrada('code 1', 234567890),
    Entrada('code 2', 726578236),
    Entrada('code 3', 328957359)
]

entradas2 = [
    Entrada('code 4', 789275883),
    Entrada('code 5', 349092090),
    Entrada('code 6', 952893850)
]

for i in range(len(entradas2)):
    entradas.append(entradas2[i])

entradas.sort(key=lambda entrada: entrada.fecha)

If the positions of both lists match, that is, the code in CODIGOS[x]corresponds to the date in FECHAS[x], what you can do is use zip:

>>> codigos
[10, 20, 30, 40]
>>> fechas
[datetime.date(2017, 7, 15), datetime.date(2017, 7, 20), datetime.date(2017, 7, 1)]
>>> zip(codigos, fechas)
[(10, datetime.date(2017, 7, 15)), (20, datetime.date(2017, 7, 20)), (30, datetime.date(2017, 7, 1))]

Jorge, the following code explains my comment and gives you a solution to your question. I agree with @FjSevilla, if you are already using pandas, why transform the data into lists and then do this ordering? pandasoffers a lot of functionality to operate with dataframes, you may have to investigate on that side, if not this can surely help you

codigo = [1, 2, 4]
fechas = ["2017-01-01", "2017-01-02", "2017-01-04"]

codigo2 = [3]
fechas2 = ["2017-01-03"]


print("Las dos listas con el orden natural y sincronizadas")
print("\n".join([str(e) for e in zip(codigo, fechas)]))

print("Agrego las listas nuevas")
print("\n".join([str(e) for e in zip(codigo2, fechas2)]))

fechas = fechas + fechas2
codigo = codigo + codigo2

print("El resultado sigue manteniendo el orden natural, sincronizadas, pero ahora la fecha esta desordenada")
print("\n".join([str(e) for e in zip(codigo, fechas)]))


print("Si ordeno la lista de fechas pierdo la sincronía con la de códigos")
fechas.sort()
print("\n".join([str(e) for e in zip(codigo, fechas)]))

print("Si armo tuplas (codigo,fecha) puedo ordenar sin problema por una de las columnas")
tuplelist = [e for e in zip(codigo, fechas)]
tuplelist.sort(key=lambda tuplelist: tuplelist[1])
print("\n".join([str(e) for e in tuplelist]))

This is the output:

Las dos listas con el orden natural y sincronizadas
(1, '2017-01-01')
(2, '2017-01-02')
(4, '2017-01-04')
Agrego las listas nuevas
(3, '2017-01-03')
El resultado sigue manteniendo el orden natural, sincronizadas, pero ahora la fecha esta desordenada
(1, '2017-01-01')
(2, '2017-01-02')
(4, '2017-01-04')
(3, '2017-01-03')
Si ordeno la lista de fechas pierdo la sincronía con la de códigos
(1, '2017-01-01')
(2, '2017-01-02')
(4, '2017-01-03')
(3, '2017-01-04')
Si armo tuplas (codigo,fecha) puedo ordenar sin problema por una de las columnas
(1, '2017-01-01')
(2, '2017-01-02')
(4, '2017-01-03')
(3, '2017-01-04')

Take note that each "code" corresponds to the day in the "date". By sorting the "date" we lost the "synchrony", it's like when you sort by one column in Excel and forget to indicate the others. The solution is to maintain a data structure that allows "code" and "date" to be related. One possibility is to use "tuples", with this: you tuplelist = [e for e in zip(codigo, fechas)]transform the two lists into another list of tuples, something like [(codigo, fecha), (codigo, fecha),...]. With this structure you can already do: tuplelist.sort(key=lambda tuplelist: tuplelist[1])to sort everything by date (column 2). Finally, if you want to leave everything as the original lists, you just have to do an "unzip":

codigo, fecha = [list(e) for e in zip(*tuplelist)]

As I have told you, it is possible using the Schwartzian transform which, although the idea comes from Perl in Python, can be implemented using zip.

The idea is simple, it is used zipto obtain the pairs of elements and is applied to them sorted. This gives us a list with the pairs ordered by the first iterable passed to zip(if this element is equal, the next one is used).

This done, we unpack the list returned by sortedand pass all its elements as arguments to zipreturn to obtain two lists using the elements of the tuples. In Python you can unpack an iterable using *in front of it.

The only complicated thing is to understand the operation of zip, I think that with a couple of examples it can be explained better than with words:

>>> a = (1, 2, 3, 4)
>>> b = ('a', 'b', 'c', 'd')
>>> c = list(zip(a, b))
>>> c
[(1, 'a'), (2, 'b'), (3, 'c'), (4, 'd')]
>>> d, e = zip(*c)
>>> d
(1, 2, 3, 4)
>>> e
('a', 'b', 'c', 'd')

zip(*c)It is equivalent to:

zip((1, 'a'), (2, 'b'), (3, 'c'), (4, 'd'))

An example of how to sort using the idea above based on your code:

from datetime import date

# Las dos listas originales
CODIGOS = [1, 2, 3, 4, 5]
FECHAS = [date(2017, 7, 24), date(2017, 7, 25), date(2017, 7, 26), date(2017, 7, 27), date(2017, 7, 28)]

# Concatenas nuevos valores
CODIGOS.extend([7, 10, 24, 14])
FECHAS.extend([date(2017, 7, 30), date(2017, 7, 20), date(2017, 8, 1), date(2017, 7, 21)])

# Ordenamos aplicando lo explicado antes
FECHAS, CODIGOS = map(list, zip(*sorted(zip(FECHAS, CODIGOS))))

Departure:

>>> DATES
[datetime.date(2017, 7, 20), datetime.date(2017, 7, 21), datetime.date(2017, 7, 24), datetime.date(2017, 7, 25), datetime .date(2017, 7, 26), datetime.date(2017, 7, 27), datetime.date(2017, 7, 28), datetime.date(2017, 7, 30), datetime.date(2017, 8 , 1)]
>>> CODES
[10, 14, 1, 2, 3, 4, 5, 7, 24]

If you notice use list.extendto concatenate the lists, consider using this in your code instead of the fory -loop appendyou use now.