Random sample of paired lists in Python

Question

I have two lists x and y, both of length n, with x_i and y_i forming a pair. How could I take a random sample of m values from these two lists while preserving the pairing information (e.g. x[10] and y[10] would be together in the resulting sample)

My initial idea is this.

use zip to create a list of tuples
shuffle the list of tuples
select the first m tuples from the list
break up the tuples into new paired lists

And the code would look something like this.

templist = list()
for tup in zip(x, y):
    templist.append(tup)
random.shuffle(templist)
x_sub = [a for a, b in templist[0:m]]
y_sub = [b for a, b in templist[0:m]]

This seems rather kludgy to me. Is there any way I could make this more clear, concise, or Pythonic?

Have you looked at [`random.sample`](https://docs.python.org/2/library/random.html#random.sample)? — metatoaster, Sep 15 '15 at 02:50
@metatoaster As a replacement for the shuffle command? The whole solution would still be a bit kludgy. Unless `random.sample` could take two paired lists as input. — Daniel Standage, Sep 15 '15 at 02:53

score 14 · Accepted Answer · edited Aug 17 '21 at 01:25

14

You can sample m pairs and split those into two lists with the following code:

import random

x = list(range(1, 10))
y = list("abcdefghij")
m = 3

x_sub, y_sub = zip(*random.sample(list(zip(x, y)), m))

edited Aug 17 '21 at 01:25

iron9

397
2
12

answered Sep 15 '15 at 03:01

Jerry Day

366
1
5

1

I like that this solution creates two new lists, rather than a single list of tuples. Kind of dense otherwise, but conciseness and clarity always seem to be at odds! – Daniel Standage Sep 15 '15 at 03:04
I get an error trying to execute this (assuming m=3) – Alexander Sep 15 '15 at 03:07
@Alexander `m` is the sample size, the OP didn't provide it just the variable `m`, what error - works fine for me. – AChampion Sep 15 '15 at 03:09
@Alexander sampling m elements in the two lists – Jerry Day Sep 15 '15 at 03:09
With the following inputs, I get the following error (Python 2.7.10): x = [1,2,3,4,5] y = [6,7,8,9,10] m = 3 TypeError: random_sample() takes at most 1 positional argument (2 given) – Alexander Sep 15 '15 at 03:35
@Alexander what is `random_sample()`? please `import random` then do `random.sample()` – Jerry Day Sep 15 '15 at 05:28
@JerryDay Yes, that works. I was using `np.random.sample` Thx for clarifying the module. – Alexander Sep 15 '15 at 07:19

score 1 · Answer 2 · answered Sep 15 '15 at 02:58

If you have two lists with elements that are direct pairs of each other and simply zip them (and in python 3, cast that object into a list), then use random.sample to take a sample.

>>> m = 4
>>> x = list(range(0, 3000, 3))
>>> y = list(range(0, 2000, 2))
>>> random.sample(list(zip(x, y)), m)
[(2145, 1430), (2961, 1974), (9, 6), (1767, 1178)]

Alexander · Answer 3 · 2020-10-11T17:15:07.550

1

If you have two lists of the same dimensions, you just want to sample a subset of these elements and pair the results.

x = [1,2,3,4,5] 
y = [6,7,8,9,10]
sample_size = 3
idx = np.random.choice(len(x), size=sample_size, replace=False)
pairs = [(x[n], y[n]) for n in idx]
>>> pairs
[(5, 10), (2, 7), (1, 6)]

edited Oct 11 '20 at 17:15

answered Sep 15 '15 at 03:00

Alexander

105,104
32
201
196

when I ran the code, it shows "SyntaxError: closing parenthesis ')' does not match opening parenthesis '['. I've tried to adjust it but not successful. Can you have a look, please? – Mark K Oct 11 '20 at 15:07
1

@MarkK Corrected above. – Alexander Oct 11 '20 at 17:15

pylang · Answer 4 · 2017-08-31T02:21:11.007

You can implement the random_product itertools recipe. I will use a third-party library, more_itertools, that implements this recipe for us. Install this library via pip install more_itertools.

Code

import more_itertool as mit


x, y, m = "abcdefgh", range(10), 2

iterable = mit.random_product(x, y, repeat=m)

Results

iterable
# ('e', 9, 'f', 3)

It is not clear in what form the OP wants the results, but you can group x and y together, e.g. [(x[0], y[0]), (x[1], y[1]), ...]:

paired_xy = list(zip(*[iter(iterable)]*2))
paired_xy
# [('e', 9), ('f', 3)]

See also more_itertools.sliced and more_itertools.grouper for grouping consecutive items.

Alternatively, you may zip further to group along x and y, e.g. [(x[0], x[1], ...), (y[0], y[1], ...)]:

paired_xx = list(zip(*paired_xy))
paired_xx
# [('e', 'f'), (9, 3)]

Note, this approach accepts any number of iterables, x, y, z, etc.

# Select m random items from multiples iterables, REF 101
x, y, m = "abcdefgh", range(10), 2
a, b, c = "ABCDE", range(10, 100, 10), [False, True]
iterable = mit.random_product(x, y, a, b, c, repeat=m) 
iterable
# ('d', 6, 'E', 80, True, 'a', 1, 'D', 50, False)

Details

From the itertools recipes:

def random_product(*args, repeat=1):
    "Random selection from itertools.product(*args, **kwds)"
    pools = [tuple(pool) for pool in args] * repeat
    return tuple(random.choice(pool) for pool in pools)

We can see the function indeed accepts multiple arguments which each become a collection of pools. The size of the pool scales by the value of repeat keyword. A random selection is made from each pool and tupled together as the final result.

See also more_itertools docs for more tools.

Random sample of paired lists in Python

4 Answers4

Linked