Select an array from a multidimensional array

Question

In the following array, I want select a sub-array from array a when an id is known

a=[['id123','ddf',1],['id456','ddf',1],['id789','ddf',1]]

I know the id i.e, id456, based on this how can I select the values ['id456','ddf',1] from a without using any loop.

What do you mean by no loop? As in no normal for loop? Are generator expressions like mine allowed? — jamylak
– jamylak, Commented Jun 4, 2012 at 12:59
this is a simple one ;) , just call it like you would any other list. a[0] should do the trick :) — IT Ninja
– IT Ninja, Commented Jun 4, 2012 at 13:00
If the list is sorted it should be possible to fetch a value in O(log n) time, but a loop of some sort (explicitly with a for loop or implicitly via more nuanced syntax) would still be required. You need a dictionary, or some other kind of mapping data structure, to do better. — Andrew Gorcester
– Andrew Gorcester, Commented Jun 4, 2012 at 13:05

jamylak · Accepted Answer · 2012-06-04 13:06:44Z

2

>>> a = [['id123','ddf',1],['id456','ddf',1],['id789','ddf',1]]
>>> next(x for x in a if x[0] == 'id456')
['id456', 'ddf', 1]

I would however, recommend the use of a dictionary instead.

edited Jun 4, 2012 at 13:06

answered Jun 4, 2012 at 12:58

jamylak

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Comments

mgilson · Accepted Answer · 2012-06-04 12:57:57Z

1

I think this should work...

filter(lambda x:x[0]=='id456',a)[0]

But in this case, wouldn't a dictionary be a better data structure?

answered Jun 4, 2012 at 12:57

mgilson

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2 Comments

Christian Witts Over a year ago

That would still go through the entire list before it completes. And imo, a dictionary as you stated would be a better data structure to use.

jamylak Over a year ago

@ChristianWitts This could work efficiently in Python 3 using next instead of [0] but as it is I agree that is inefficient and the use of lambda makes it ugly when compared to a generator expression.

Benjamin · Accepted Answer · 2012-06-04 13:21:46Z

0

A dictionary structure would work a lot better.

b = {'id123': ['ddf', 1], 'id456': ['dff', 1], 'id789': ['ddf', 1]}
print b['id123']

answered Jun 4, 2012 at 13:21

Benjamin

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Comments

Janne Karila · Accepted Answer · 2012-06-04 13:26:22Z

0

If the list is sorted, you can use the bisect module:

>>> i = bisect.bisect_left(a, ['id456'])
>>> if i < len(a) and a[i][0]=='id456':
...     print a[i]
...
['id456', 'ddf', 1]

answered Jun 4, 2012 at 13:26

Janne Karila

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Comments

heltonbiker · Accepted Answer · 2012-06-04 13:39:32Z

0

You can use numpy.where() to do that:

a = numpy.array(a)

row = numpy.where(a == 'id456')[0]

sub_array = a[row,:]

This will check for elements with the desired id and give back their indexes. You can use these indexes to take slices from the original array, as the example shows.

This code will only work if there is a single row for the given ID, but it can be adapted.

Hope this helps.

answered Jun 4, 2012 at 13:39

heltonbiker

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