Lists within repeating structure

Thiago Ramos 0
Viewed 70

I need to write the repeating structure data into named lists. The name of the list needs to be determined in the variable i. I have the following input data:

Nome Local Hora
Thiago Cantina 08:01
Ana Cantina 08:05
Thiago Cantina 08:10
Thiago Sala01 08:50
Jose Sala01 09:03
Thiago Sala01 09:30
Jose Sala01 09:45
Jose Cantina 09:49
Ana Biblioteca 11:07

Code:

#Input
list_entry=[['Thiago', 'Cantina', '08:01:00'], ['Ana', 'Cantina', '08:05:00'], ['Thiago', 'Cantina', '08:10:00'], ['Thiago', 'Sala01', '08:50:00'], ['Jose', 'Sala01','09:03:00'], ['Thiago', 'Sala01', '09:30:00'], ['Jose', 'Sala01', '09:45:00'], ['Jose', 'Cantina', '09:49:00'], ['Ana', 'Biblioteca', '11:07:00']]


i=1
lista=[]
    while i <= 4:#repeticacao1
        y=0
        inicio, fim = intervalo[i] #interval is a dictionary with time period every 1 hour
        #The data of the block below must be written in a list with the name list1. After exiting While(#repeticacao2) and the i is incremented, the data must be written to a list named list2 and so on.
        while y < 6: #repeticacao2 - Stop condition is <6 due to the number of input registers
            registro2 = df_registro['Hora'].dt.time[y]
            if inicio <= registro2 < fim: # checks if time is between the start and end of the range
                print(f'O registro esta dentro do intervalo {inicio:%H:%M} às {fim:%H:%M}')
                #Create a list with Name and Location datal
                lista.append([df_registro['Nome'][y],df_registro['Local'][y]])
            y += 1
        i += 1

Result:

[['Thiago', 'Cantina'], ['Ana', 'Cantina'], ['Thiago', 'Cantina'], ['Thiago', 'Sala01'], ['Jose', 'Sala01'], ['Thiago', 'Sala01'], ['Jose', 'Sala01'], ['Jose', 'Cantina'], ['Ana', 'Biblioteca']]

The result is a list of all records. However, in this way, the time intervals are being mixed.

The expected result would be for separate time intervals. Thus:

list1 = [['Thiago', 'Cantina'], ['Ana', 'Cantina'], ['Thiago', 'Cantina'], ['Thiago', 'Sala01']]

list2 = [['Jose', 'Sala01'], ['Thiago', 'Sala01'], ['Jose', 'Sala01'], ['Jose', 'Cantina']]

list3 = []

list4 = ['Ana', 'Biblioteca']

I tried to create the list with list[i] and then with list[i].append but it doesn't work. How could it be done?

while-loop list python
3 Answers

Do you have to use a while loop here? It seems like a good case for comprehensions.

Something along the lines of

hourly_list = [entry for entry in entry_list if interval_start <= entry.time < interval_end]

would give you the list of entries within a given interval (I haven't quite used the data structure you're working with but the method should work).

PhlebasJon0

Creating a separate variable for each location is counter-productive. Instead, consider using a dictionary keyed by location:

{k: list(g['Nome']) for k, g in df.groupby('Local')}
Mad Physicist0

Code:

df["Hora"] = pd.to_datetime(df["Hora"])
d = df.set_index('Hora').resample('1H', label='right').sum()  #Inorder to get the Hourly index including 10 o'clock

#Here I am separating the sum string by capital latter and converting them
#into list by using .split()

import re
d['Nome'] = d['Nome'].apply(lambda x: re.sub(r"\B([A-Z])", r" \1", str(x)))
d['Local'] = d['Local'].apply(lambda x: re.sub(r"\B([A-Z])", r" \1", str(x)))

#lastly zipping both column list values and define a new column
d['List'] = d.apply(lambda x: [list(a) for a in zip(x.Nome.split(' '), x.Local.split(' '))], axis=1)

Print:

d['List'][0]    #Lets say the List1

Output:

[['Thiago', 'Cantina'],
 ['Ana', 'Cantina'],
 ['Thiago', 'Cantina'],
 ['Thiago', 'Sala01']]

And to get all list into one:

[[y for y in x if '0' not in y] for x in list(d.List)] #Here Looping just to avoid '0' at 10 o'clock

Output:

[[['Thiago', 'Cantina'],
  ['Ana', 'Cantina'],
  ['Thiago', 'Cantina'],
  ['Thiago', 'Sala01']],
 [['Jose', 'Sala01'],
  ['Thiago', 'Sala01'],
  ['Jose', 'Sala01'],
  ['Jose', 'Cantina']],
 [],
 [['Ana', 'Biblioteca']]]

UPDATE:

In order to lowercase the all list element:

Option 1:By add Lambada function

[[list((map(lambda x: x.lower(), y))) for y in x if '0' not in y] for x in list(d.List)]

Option 2: By for loop

[[[x.lower() for x in y] for y in x if '0' not in y] for x in list(d.List)]

Output:

[[['thiago', 'cantina'],
  ['ana', 'cantina'],
  ['thiago', 'cantina'],
  ['thiago', 'sala01']],
 [['jose', 'sala01'],
  ['thiago', 'sala01'],
  ['jose', 'sala01'],
  ['jose', 'cantina']],
 [],
 [['ana', 'biblioteca']]]
Stackpy0
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