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Jesus_jbs
Asked: 2020-12-10 03:18:41 +0800 CST 2020-12-10 03:18:41 +0800 CST

Error between datetime.date and NoneType operations in Python

  • 772

Taking into account that it partidosis a list with sublists that contain tuples. Ex:partidos = [[(datetime.date(2000, 9, 9), 'FC Barcelona', 2, 'Málaga CF', 1)], [(datetime.date(2000, 9, 9), 'RC Deportivo', 2, 'Athletic Club', 0)], [(datetime.date(2000, 9, 9), 'Real Sociedad', 2, 'R. Racing C.', 2)]]

I have tried to build the following function to do what you ask: 

def partidos_por_fecha(partidos, inicio=None, fin=None):
    ''' Filtra los partidos jugados en un rango de fechas

    ENTRADA: 
    - partidos: lista de partidos -> [Partido(datetime.date, str, int, str, int)]
    - inicio: fecha inicial del rango -> datetime.date
    - fin: fecha final del rango -> datetime.date
    SALIDA: 
    - lista de partidos seleccionados -> [Partido(datetime.date, str, int, str, int)] 

    Se devuelven aquellos partidos que se han jugado entre las fechas inicio
    y fin. Ambos parámetros serán objetos date del módulo datetime.
    Si inicio es None, se incluirán los partidos desde el principio de
    la serie, y si fin es None se inlcuirán los partidos hasta el último de
    la serie.
    '''
    if inicio == None:
        inicio = partidos[0][0][0]
    elif fin == None:
        fin = partidos[-1][-1][0]
    elif inicio == None and fin == None:
        inicio = partidos[0][0][0]
        fin = partidos[-1][-1][0]
    else:
        inicio == inicio
        fin == fin

    result = []
    for i in partidos:
        for fecha, local, goles_local, visitante, goles_visitante in i:
            if fecha >= inicio and fecha <= fin:
                result.append((fecha, local, goles_local, visitante, goles_visitante))
    return result

This function is executed by the following code:

def test_partidos_por_fechas(partidos):
    inicio = datetime(2007, 9, 15).date()
    fin = datetime(2008, 7, 1).date()
    print(len(partidos_por_fecha(partidos, inicio, fin)))
    print(len(partidos_por_fecha(partidos, inicio, None)))
    print(len(partidos_por_fecha(partidos, None, fin)))
    print(len(partidos_por_fecha(partidos, None, None)))

test_partidos_por_fechas(partidos)

The error that I get after executing the function is: TypeError: '<=' not supported between instances of 'datetime.date' and 'NoneType'And it tells me that it is in the ifone that is inside the loops for. I don't understand how after filtering the arguments inicioand finwhen they are Nonebefore the loops for, I still get variables Noneto compare withdatetimes

I would like to get some solution to make the function work. Sorry if there are indentation errors in the code having passed it here and thanks in advance.

python

1 Answers

  1. Best Answer

    There are some bugs in the code:

    • Comparison with Noneis recommended to be done using the operator is, ie if inicio is None, etc. instead of if inicio == None. The reason for this is quite technical (if you're interested I can explain it in a final note).

      In this particular case, this detail does not affect the operation of the code, but get used to doing it with it isas a general rule.

    • In it else:you have a ==where I think you wanted to do an assignment ( =). This doesn't matter either because it never enters through else, as we'll see right away, but it was obviously wrong.

    • The most important bug is the logic of the ones ifwith which you want to assign values ​​to inicioand finin case they don't have one. Note that if both are None, the first if inicio == None:will be fulfilled, and its body will be executed without executing any of the other cases , so in that case finit will remain with its Noneinitial value, and this case is the one that "breaks" you.

    The simplest solution would be the following:

    if inicio is None:
    inicio = partidos[0][0][0]
if fin is None:
    fin = partidos[-1][-1][0]

    That is, I use a second ifinstead of a elif. If it iniciois None, it is assigned a default value. Then, independently, the same thing is done with fin. If both were None, both will be assigned. The weird assignment inicio=inicio; fin=finyou had in else:it is obviously not necessary, as it leaves the variables as they were.

    Another possibility

    In some python code you may come across the following "trick". I personally have mixed feelings about it. On the one hand, the number of lines of code is reduced. On the other hand, it makes use of a "weird" feature that can be considered poorly readable. Although if you get used to seeing it, it ends up being quite readable...

    It is like this:

    inicio = inicio or partidos[0][0][0]
fin = fin or partidos[-1][-1][0]

    It can almost be read effortlessly as "assign to iniciowhat was in inicio, or else this other value (if what was in iniciowas None)" .

    The reason it works is because the result of a orPython expression is the value of the last term evaluated that makes the expression true, or the last term if neither is true.

    Thus, if for example inicioit is worth 20, it inicio or otra_cosais true (since 20 is considered as Truewhen it is part of a Boolean expression, and therefore the expression is true just by evaluating the first term, so the second is not even evaluated) , and the result would be 20. However if it iniciois None(either zero, or the empty string, or some other value that is considered Falseby python) then python must evaluate otra_cosato determine the result of the or. The final result of the expression will be the value of otra_cosa.

    It is a rare hack , and I don't know whether to recommend its use or advise against it, but you should know about it because you will see it written many times.

    Finally, python also has something equivalent to the C ternary operator, v = condicion?valor_cierto:valor_falso, which in python would be v = valor_cierto if condicion else valor_falso.

    Using this in your case:

    inicio = partidos[0][0][0] if inicio is None else inicio
fin = partidos[-1][-1][0] if fin is None else fin

    ifThis is basically just another way of putting the one I put in the first part of the answer in fewer lines . I almost prefer the "in more lines" version as it is clearer.

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