CODEWITHSUNDEEP
python
M.E.
M.E.

Reputation: 5496

Initialize an empty multidimensional list

I would like to initialize a multidimensional list capable of storing an object for every minute since year 1950-2050.

Something like:

minute = [None]*61
hour = [minute]*25
day = [hour]
month = [day]
data = [month]*100

So you can do:

data[89][1][29][23][55] = 'It was a good minute the one from January 29th in 1989 when it was 23:55'

How would be such a multidimensional list be initialized in Python? Would it be an actual different object than the one created with the above code?

Initially the multidimensional list would contain objects None.

Python 2.7

Following answer I tried:

# Data structure
minute = 60
hour = 24
day = 31
month = 12
year = 100
test = [[[[[None for _minute in range(minute)] for _hour in range(hour)] for _day in range(day)] for _month in range(month)] for _year in range(year)]

But it seems too much for multidimensional lists, as I get "Killed" when trying to execute this.

Upvotes: 2

Views: 5980

Answers (2)

sacuL
sacuL

Reputation: 51335

I also don't recommend this, but you could use a numpy.chararray for this:

import numpy as np
arr = np.chararray((100, 12, 31, 24, 60, 60), itemsize=100)

arr[52, 7, 12, 12, 44, 54] = 'year 1950+52, 7th month, 12th day, 12th hour, 44th minute, 54th second'

I'm not exactly sure what your desired structure is, but the string I inserted into the array should explain the structure I proposed, and you can change it however you need. Note that itemsize limits how many characters you can put in at any index.

Again, with a caveat that this is not necessarily the most efficient thing in the world to do, but if you wish to store lists of ints and/or floats in that array (as per your comment), one way to do it would be to convert that list to strings, and then when retrieving it, re-transform back to a list:

data_to_insert = [1,2,3,4.5]
# store as string
arr[52, 7, 12, 12, 44, 54] = ','.join(map(str, data_to_insert))
# retrieve
arr[52, 7, 12, 12, 44, 54].decode('utf-8').split(',')

This should be pretty fast

Upvotes: 2

Ubdus Samad
Ubdus Samad

Reputation: 1215

Though i won't recommend it, A multi dimentional empty list can be created by using list comprehension:

> >>> a = 4 #Width of elements
> >>> b = 6 #Width of main list container
>>>>> c = 4
>>>>> d = 3
> >>> l = [[[[0 for k in range(d) ] for z in range(c)] for x in range(a)] for y in range(b)]
> >>> [[[[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]]], [[[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]]], [[[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]]], [[[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]]], [[[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]]], [[[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]]]]

Keep replacing 0 with list comprehensions to add more dimentions.

Upvotes: 2

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