Create multidimensional numpy array from specific keys of dictionary

Question

I have dictionary like this:

a = dict(zip( ['k1', 'k2', 'k3', 'k4'], 
          ... [ [1,2,3,4], [10,20,30,40], [100,200,300,400], [1000,2000,3000,4000]])

>>> a
{'k1': [1, 2, 3, 4], 'k2': [10, 20, 30, 40], 'k3': [100, 200, 300, 400], 'k4': [1000, 2000, 3000, 4000]}

What I want to do: get values for several keys and create multidimensional numpy array from them. Something like this:

result = numpy.array( [a[x] for x in ('k1' , 'k3')]

I tried this code:

ar = numpy.array([])
for el in ['k1', 'k3']:
     ar = numpy.r_[ar, num_dict[el]]
ar = ar.reshape(2,len(ar)/2)

But are there some built in functions or more elegant ways?

I consider your first option pretty elegant already.

akuiper
– akuiper

2017-10-30 15:37:57 +00:00
Commented Oct 30, 2017 at 15:37 — akuiper
– akuiper, Commented Oct 30, 2017 at 15:37

hpaulj · Accepted Answer · 2017-10-30 17:14:14Z

1

A list of lists is normal input to np.array, so your list comprehension makes sense.

In [382]: [a[x] for x in ['k1','k3']]
Out[382]: [[1, 2, 3, 4], [100, 200, 300, 400]]

Or for the whole dictionary

In [385]: np.array(list(a.values()))    # list required in py3
Out[385]: 
array([[1000, 2000, 3000, 4000],
       [   1,    2,    3,    4],
       [  10,   20,   30,   40],
       [ 100,  200,  300,  400]])

Normally dictionary items are selected one by one as in the comprehension. operator has a convenience class for fetching several keys with one call (I don't think it makes much difference in speed):

In [386]: import operator
In [387]: operator.itemgetter('k1','k3')(a)
Out[387]: ([1, 2, 3, 4], [100, 200, 300, 400])

I don't think the iteration with r_ is a good choice. r_ is just a cover for concatenate. If you must iterate, repeated concatante is slower. It's better to build a list, and make the array at the end (as in the list comprehension).

answered Oct 30, 2017 at 17:14

hpaulj

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1 Comment

Mikhail_Sam Over a year ago

Thank you for your answer! itemgetter is awesome function and I already want to use it!

Mikhail_Sam · Accepted Answer · 2017-10-31 08:24:00Z

0

I need exactly one numpy array from data, so I can't find the way without loops. I create function:

def loadFromDict( fieldnames, dictionary ):
    ''' fieldnames - list of needed keys, dictionary - dict for extraction
     result - numpy.array size of (number of keys, lengths of columns in dict)'''
    ar = numpy.zeros( (len(fieldnames), len(dictionary[fieldnames[0]])) )
    for c,v in enumerate(fieldnames,0):
        ar[c,:] = dictionary[v]
    return ar

In my case dictionary has the same length for all columns. Anyway it is easy to implement either they are different: use [len(v) for v in dictionary.values()] to get all lengths, or find lengths for current keys.

answered Oct 31, 2017 at 8:24

Mikhail_Sam

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