Fast replacement of values in a numpy array
I have a very large numpy array (containing up to a million elements) like the one below:
[0,1,6,5,1,2,7,6,2,3,8,7,3,4,9,8,5,6,11,10,6,7,12,11,7,
8,13,12,8,9,14,13,10,11,16,15,11,12,17,16,12,13,18,17,13,
14,19,18,15,16,21,20,16,17,22,21,17,18,23,22,18,19,24,23]
and a small dictionary map for replacing some of the elements in the above array
{4: 0, 9: 5, 14: 10, 19: 15, 20: 0, 21: 1, 22: 2, 23: 3, 24: 0}
I would like to replace some of the elements according to the map above. The numpy array is really large, and only a small subset of the elements (occurring as keys in the dictionary) will be replaced with the corresponding values. What is the fastest way to do this?
Solution 1:
I believe there's even more efficient method, but for now, try
from numpy import copy
newArray = copy(theArray)
for k, v in d.iteritems(): newArray[theArray==k] = v
Microbenchmark and test for correctness:
#!/usr/bin/env python2.7
from numpy import copy, random, arange
random.seed(0)
data = random.randint(30, size=10**5)
d = {4: 0, 9: 5, 14: 10, 19: 15, 20: 0, 21: 1, 22: 2, 23: 3, 24: 0}
dk = d.keys()
dv = d.values()
def f1(a, d):
b = copy(a)
for k, v in d.iteritems():
b[a==k] = v
return b
def f2(a, d):
for i in xrange(len(a)):
a[i] = d.get(a[i], a[i])
return a
def f3(a, dk, dv):
mp = arange(0, max(a)+1)
mp[dk] = dv
return mp[a]
a = copy(data)
res = f2(a, d)
assert (f1(data, d) == res).all()
assert (f3(data, dk, dv) == res).all()
Result:
$ python2.7 -m timeit -s 'from w import f1,f3,data,d,dk,dv' 'f1(data,d)'
100 loops, best of 3: 6.15 msec per loop
$ python2.7 -m timeit -s 'from w import f1,f3,data,d,dk,dv' 'f3(data,dk,dv)'
100 loops, best of 3: 19.6 msec per loop
Solution 2:
Assuming the values are between 0 and some maximum integer, one could implement a fast replace by using the numpy-array as int->int
dict, like below
mp = numpy.arange(0,max(data)+1)
mp[replace.keys()] = replace.values()
data = mp[data]
where first
data = [ 0 1 6 5 1 2 7 6 2 3 8 7 3 4 9 8 5 6 11 10 6 7 12 11 7
8 13 12 8 9 14 13 10 11 16 15 11 12 17 16 12 13 18 17 13 14 19 18 15 16
21 20 16 17 22 21 17 18 23 22 18 19 24 23]
and replacing with
replace = {4: 0, 9: 5, 14: 10, 19: 15, 20: 0, 21: 1, 22: 2, 23: 3, 24: 0}
we obtain
data = [ 0 1 6 5 1 2 7 6 2 3 8 7 3 0 5 8 5 6 11 10 6 7 12 11 7
8 13 12 8 5 10 13 10 11 16 15 11 12 17 16 12 13 18 17 13 10 15 18 15 16
1 0 16 17 2 1 17 18 3 2 18 15 0 3]