del a variable in python given a string containing the variable name

Question

I hacked up some code to test my hypothesis of algo complexity for set intersection in python:

s3, s4, s5, s6, t3, t4, t5, t6 are all defined and are rather large - their intersections also happen to be quite large.

import timeit

s3, t3 = set(), set()
s4, t4 = set(), set()
s5, t5 = set(), set()
s6, t6 = set(), set()

for x in xrange(int(1e3)):
    s3.add(x)
    t3.add(x*2 + x)

for x in xrange(int(1e4)):
    s4.add(x)
    t4.add(x*2 + x)

for x in xrange(int(1e5)):
    s5.add(x)
    t5.add(x*2 + x)

for x in xrange(int(1e6)):
    s6.add(x)
    t6.add(x*2 + x)


def _test():
    for i in [3, 4, 5, 6]:
        for j in [3, 4, 5, 6]:
            if i >= j:
                s, t = 's' + str(i), 't' + str(j)
                print i, j
                print timeit.timeit('{0}.intersection({1})'.format(s, t), setup="from __main__ import {0}, {1}".format(s, t))
        eval('del s' + str(i))

But the following statement 'blows up'..

eval('del s' + str(i))

Any ideas?

I would also be interested in any ideas to make my hacky code less hacky.

Thanks

what is "blows up"?

Shuo
– Shuo

2017-03-13 11:33:11 +00:00
Commented Mar 13, 2017 at 11:33 — Shuo
– Shuo, Commented Mar 13, 2017 at 11:33

Nils Werner · Accepted Answer · 2017-03-13 11:53:48Z

2

First of all, you should be aware that the slow execution time you are seeing is due to timeit. The algorithm itself is super fast.

Secondly you can create list of sets:

s = [set(), set(), ...]

and then delete the element from that list

del s[i]

in your case:

import timeit

s = [set(), set(), set(), set()]
t = [set(), set(), set(), set()]

for x in xrange(int(1e3)):
    s[0].add(x)
    t[0].add(x*2 + x)

for x in xrange(int(1e4)):
    s[1].add(x)
    t[1].add(x*2 + x)

for x in xrange(int(1e5)):
    s[2].add(x)
    t[2].add(x*2 + x)

for x in xrange(int(1e6)):
    s[3].add(x)
    t[3].add(x*2 + x)


def _test():
    for i, _ in enumerate(s):
        for j, _ in enumerate(t):
            if i >= j:
                print i, j
                print timeit.timeit('s[{0}].intersection(t[{1}])'.format(i, j), setup="from __main__ import s, t")
        del s[i]

Thirdly for larger datasets you can probably speed up the process using NumPy:

import timeit
import numpy
import itertools

s = [numpy.arange(l) for l in [1e3, 1e4, 1e5, 1e6]]
t = [ss * 3 for ss in s]  # x*2 + x == x*3

for (i, ss), (j, tt) in itertools.product(enumerate(s), enumerate(t)):
    if i >= j:
        print i, j
        numpy.intersect1d(ss, tt, assume_unique=True)

edited Mar 13, 2017 at 11:53

answered Mar 13, 2017 at 11:31

Nils Werner

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4 Comments

Jonathon Hill Over a year ago

Yep, that makes everything better. Thanks

Jonathon Hill Over a year ago

You can also create another list for xrange sizes, so there aren't four for loops anymore.

Jonathon Hill Over a year ago

and you could use 'for i in range(len(s))' instead of enumerate. Thanks alot :).

Nils Werner Over a year ago

I have added another example using NumPy, if performance is a concern.

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