Extract items from array: between given values/conditions

Question

I have a number of timeseries data in arrays and wish to extract values between given dates in the simplest way possible avoiding loops. Here's an example:

from numpy import *
from datetime import *

# datetime array
date_a=array([
datetime(2000,1,1),
datetime(2000,1,2),
datetime(2000,1,3),
datetime(2000,1,4),
datetime(2000,1,5),
])

# item array, indices corresponding to datetime array
item_a=array([1,2,3,4,5])

# extract items in a certain date range
# after a certain date, works fine
item_b=item_a[date_a >= (datetime(2000,1,3))] #Out: array([3, 4, 5])

# between dates ?
item_c=item_a[date_a >= (datetime(2000,1,3)) and date_a <= (datetime(2000,1,4))]
# returns: ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Is there a one-line solution to this? I have looked at numpy any() and all(), and also where(), without being able to find a solution. I appreciate any help and point-in-direction!

Andrey Sobolev · Accepted Answer · 2011-11-30 11:48:41Z

4

If you want one-liner, then you can use

item_c=item_a[(date_a >= (datetime(2000,1,3))) * (date_a <= (datetime(2000,1,4)))]

edited Nov 30, 2011 at 11:48

answered Nov 30, 2011 at 11:43

Andrey Sobolev

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2 Comments

rhkarls Over a year ago

Brilliant, just what I was searching for! Thanks :)

Joe Kington Over a year ago

Just FYI: & is much more readable than *, here, and it does exactly the same thing.

mac · Accepted Answer · 2011-11-30 11:40:40Z

3

It's not clear to me why you are using the item_a variable. But to isolate the entries you want you can simply do:

>>> np.where(np.logical_and(date_a >= datetime(2000,1,3), date_a <= datetime(2000,1,4)))
(array([2, 3]),)

The resulting indexes are zero-based, so they correspond to the third and fourth element of your array.

EDIT: np is due to import numpy as np. Doing from numpy import * is in fact a very bad idea. You will overwrite built in functions such as sum and abs for example...

HTH!

answered Nov 30, 2011 at 11:40

mac

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1 Comment

rhkarls Over a year ago

thanks! coming from matlab there is still plenty learn about python behaviour :) this answer and the one from @Andrey Sobolev was exactly what I was looking for. item_a was just to get the values of the array, but indeed not needed as it is the indices I'm interested in

Abhijit · Accepted Answer · 2011-11-30 11:44:01Z

1

I think the following should work for you using List Comprehension

[item_a[i] for i in xrange(0,len(date_a)) if date_a[i] >= (datetime(2000,1,3)) and date_a[i] <= (datetime(2000,1,4))]

Select all items in item_a within range 0 <= i < length of date_a where datetime(2000,1,3) <= date_a[i] <= datetime(2000,1,4)

answered Nov 30, 2011 at 11:44

Abhijit

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1 Comment

rhkarls Over a year ago

Works like a charm! I'm trying to avoid loops though due to very large datasets, but I will test this implementation as well.

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