Create a list of int & DateTime in Python

Question

How can I create a list (or a numpy array if possible) in python that takes datetime objects in the first column and other data types in other columns?

For example, the list would be something like this:

list = [[<datetime object>, 0, 0.]
        [<datetime object>, 0, 0.]
        [<datetime object>, 0, 0.]]

What is the best way to create and initialize a list like this?

So far, I have tried using np.empty, np.zeros, and a list comprehension, similar to this:

list = [[None for x in xrange(3)] for x in xrange(3)]

But if I do this, I would need a for loop to populate the first column and there doesn't seem to be a way to assign it in a simpler way like the following:

list[0][:] = another_list_same_length

Padraic Cunningham · Accepted Answer · 2014-08-07 23:03:20Z

2

You mean something like:

In [17]: import numpy as np

In [18]: np.array([[[datetime.now(),np.zeros(2)] for x in range(10)]])
Out[18]: 
array([[[datetime.datetime(2014, 8, 7, 23, 45, 12, 151489),
         array([ 0.,  0.])],
        [datetime.datetime(2014, 8, 7, 23, 45, 12, 151560),
         array([ 0.,  0.])],
        [datetime.datetime(2014, 8, 7, 23, 45, 12, 151595),
         array([ 0.,  0.])],
        [datetime.datetime(2014, 8, 7, 23, 45, 12, 151619),
         array([ 0.,  0.])],
        [datetime.datetime(2014, 8, 7, 23, 45, 12, 151634),
         array([ 0.,  0.])],
        [datetime.datetime(2014, 8, 7, 23, 45, 12, 151648),
         array([ 0.,  0.])],
        [datetime.datetime(2014, 8, 7, 23, 45, 12, 151662),
         array([ 0.,  0.])],
        [datetime.datetime(2014, 8, 7, 23, 45, 12, 151677),
         array([ 0.,  0.])],
        [datetime.datetime(2014, 8, 7, 23, 45, 12, 151691),
         array([ 0.,  0.])],
        [datetime.datetime(2014, 8, 7, 23, 45, 12, 151706),
         array([ 0.,  0.])]]], dtype=object)

edited Aug 7, 2014 at 23:03

answered Aug 7, 2014 at 22:37

Padraic Cunningham

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4 Comments

oxtay Over a year ago

Thank you for the answer! The first one is similar to what I wrote in the question. The use of numpy in the second one might be what I am looking for, but it might have the same issue I pointed out at the end of my question. I have a separate list of timestamps (datetime objects) and I want to populate the first column using that list. Using this method, I will have to use a for loop to achieve that goal. I was curious to know if there is a better way to do it, to avoid a for loop and take advantage of capabilities of lists or arrays in python.

Padraic Cunningham Over a year ago

what is your datetime data in, a list or array?

oxtay Over a year ago

It is a list. What I did for now is to use for loops and your answer definitely is helpful. But do you think there is a way to assign to the first column of this new array a list of datetime objects without the use of a for loop? something like this: array[0,:] = list_of_time

oxtay Over a year ago

I just tried doing something like this with the array that you had suggested: my_array[:,0]=datetime_list and it worked without the need for a for loop. Thanks!

Sumukh Barve · Accepted Answer · 2014-08-07 22:52:46Z

1

Use zip

>>> column1 = [1, 1, 1]
>>> column2 = [2, 2, 2]
>>> column3 = [3, 3, 3]
>>> zip(column1, column2, column3)
[(1, 2, 3), (1, 2, 3), (1, 2, 3)]
>>> # Or, if you'd like a list of lists:
... 
>>> [list(tup) for tup in zip(column1, column2, column3)]
[[1, 2, 3], [1, 2, 3], [1, 2, 3]]
>>>

This would allow you to build-up the columns separately and then combine them. column1 could be dates (or anything else.)

Hope this helps.

answered Aug 7, 2014 at 22:52

Sumukh Barve

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