Empresas
Empleos
  • Sobre nosotros
  • Soluciones
    • Publicación de vacantes
      Publica tu vacante y recibe candidatos calificados en 48h.
    • Evaluación de candidatos
      500+ pruebas técnicas y psicológicas, más anti-fraude.
    • Headhunting
      Búsqueda ejecutiva a la medida de principio a fin.
    • Nómina + EOR
      Dispersión de nómina y EOR en más de 15 países de LATAM.
  • Precios
  • Empleos

0

5.7K
Vistas
Concatenate pandas DataFrames generated with a loop

I am creating a new DataFrame named data_day, containing new features, for each day extrapolated from the day-timestamp of a previous DataFrame df.

My new dataframes data_day are 30 independent DataFrames that I need to concatenate/append at the end in a unic dataframe (final_data_day).

The for loop for each day is defined as follow:

num_days=len(list_day)

#list_day= random.sample(list_day,num_days_to_simulate)
data_frame = pd.DataFrame()

for i, day in enumerate(list_day):

    print('*** ',day,' ***')

    data_day=df[df.day==day]
    .....................
    final_data_day = pd.concat()

Hope I was clear. Mine is basically a problem of append/concatenation of data-frames generated in a non-trivial for loop

over 4 years ago · Santiago Trujillo
3 Respuestas
Responde la pregunta

0

Pandas concat takes a list of dataframes. If you can generate a list of dataframes with your looping function, once you are finished you can concatenate the list together:

data_day_list = []
for i, day in enumerate(list_day):
    data_day = df[df.day==day]
    data_day_list.append(data_day)
final_data_day = pd.concat(data_day_list)
over 4 years ago · Santiago Trujillo Denunciar

0

Exhausting a generator is more elegant (if not more efficient) than appending to a list. For example:

def yielder(df, list_day):
    for i, day in enumerate(list_day):
        yield df[df['day'] == day]

final_data_day = pd.concat(list(yielder(df, list_day))
over 4 years ago · Santiago Trujillo Denunciar

0

Appending or concatenating pd.DataFrames is slow. You can use a list in the interim and then create the final pd.DataFrame at the end with pd.DataFrame.from_records() e.g.:

interim_list = []
for i,(k,g) in enumerate(df.groupby(['[*name of your date column here*'])):
    if i % 1000 == 0 and i != 0:
        print('iteration: {}'.format(i)) # just tells you where you are in iteration
    # add your "new features" here...
    for v in g.values:
        interim_list.append(v)

# here you want to specify the resulting df's column list...
df_final = pd.DataFrame.from_records(interim_list,columns=['a','list','of','columns'])
over 4 years ago · Santiago Trujillo Denunciar
Responde la pregunta
Encuentra empleos remotos

¡Descubre la nueva forma de encontrar empleo!

Top de empleos
Top categorías de empleo
Empresas
Publicar vacante Precios Comercial
Legal
Términos y condiciones Política de privacidad
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Recomiéndame algunas ofertas
Necesito ayuda