Python comparing two uneven arrays

Question

I am trying to generate two arrays, a and b, each containing ~1000 random numbers. The random number are between 1 and 5.

I want to then compare each element in a with the corresponding element of b such that if a[i] > b[i] a variable, counter, will be incremented by 1. This is considered a "success". Otherwise, if a[i] <= b[i] nothing happens (i.e counter += 0). This is considered a "fail".

However, a and b can be of variable length such that both len(a) == len(b) and len(a) != len(b) are possibilities.

In the case of the latter, if len(a) > len(b) I'd like all the "extra" elements of a to automatically be counted as "successes". If len(b) > len(a) things should proceed normally (i.e the "extra" elements of b are ignored).

For example:

If a = [1, 3, 4, 2] and b = [2, 4, 0]

Then counter = 2 because (1 < 2, 3 < 4, 4 > 0, and 2 is extra and is an automatic success`)

How would I go about doing this?

Here is some quick code, which returns the expected list index out of range... error:

import random

a = []
b = []
counter = 0

for i in range(1000):
    a += [random.randint(1, 5)]
for i in range(900):
    b += [random.randint(1, 5)]
for i in range(len(a)):
    if a[i] > b[i]:
        counter += 1

print counter

Jay Kominek · Accepted Answer · 2015-05-25 17:03:02Z

2

Just check to see if you've run off the end of b in your loop:

for i in range(len(a)):
    if (i >= len(b)) or a[i] > b[i]:
        counter += 1

You could also be a bit more clever, and do something like:

for i in range(min(len(a), len(b)):
    if a[i] > b[i]:
        counter += 1
if len(a) > len(b):
    counter += (len(a) - len(b))

(Using xrange here would be a minor improvement, as well.)

edited May 25, 2015 at 17:03

answered May 25, 2015 at 16:52

Jay Kominek

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1 Comment

Dider Over a year ago

This is a really good answer, short and easy to understand. Thanks.

Andrzej Pronobis · Accepted Answer · 2015-05-25 17:01:15Z

2

One way of doing that would be:

len([_ for (i, j) in zip(a, b) if i > j]) + max(len(a) - len(b), 0)

edited May 25, 2015 at 17:01

answered May 25, 2015 at 16:53

Andrzej Pronobis

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3 Comments

Ami Tavory Over a year ago

Why not len([_ for (i, j)...; no use to actually generate the tuples, no?

Andrzej Pronobis Over a year ago

It's less typing for sure :)

Ami Tavory Over a year ago

Great answer in any case.

Ajay · Accepted Answer · 2015-05-25 16:59:10Z

1

Using izip_longest from itertools .This works for two lists which are not of equal length.

In [47]: 2>None
Out[47]: True

Since you need to check the greatest this works fine.

In [35]: from itertools import izip_longest

In [37]: lst=list(izip_longest(a,b))

In [38]: lst
Out[38]: [(1, 2), (3, 4), (4, 0), (2, None)]

In [42]: mylst=[(i>j) for (i,j) in lst]

In [44]: c=Counter(mylst)

In [45]: c[True]
Out[45]: 2

edited May 25, 2015 at 16:59

answered May 25, 2015 at 16:52

Ajay

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1 Comment

gabhijit Over a year ago

None may not be a good idea if the start integer is 0. Better it should be a negative number. so one should use fillvalue=-1

gabhijit · Accepted Answer · 2015-05-25 17:21:37Z

1

A simple oneliner using itertools.izip_longest

import random
import itertools
l1 = [random.randint(1,20) for x in range(20)]
l2 = [random.randint(1,20) for x in range(25)]
sum([1 if y < x else 0 for (x,y) in itertools.izip_longest(l1, l2, fillvalue=random.randint(1,20)])

Here in case the lengths of the lists are not identical, izip_longest would generate random numbers again to fill in remaining values (but after rereading your question - it seems more correct way is fillvalue=-1). Basically you can use any numerical value there (or anything that can be compared with an int, if the lists are of integer types).

edited May 25, 2015 at 17:21

answered May 25, 2015 at 17:14

gabhijit

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1 Comment

gabhijit Over a year ago

I think one below is more elegant, but less explicit( It's a sum of True values - where True value is equal to 1).

Deep Shah · Accepted Answer · 2015-05-25 17:01:12Z

0

I think you just need to add one more condition to check whether i's value is less than list B size

for i in range(len(a)):
    if i < len(b) and a[i] > b[i]:
        counter += 1

answered May 25, 2015 at 17:01

Deep Shah

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jwilner · Accepted Answer · 2015-05-25 17:02:32Z

0

Use itertools.izip_longest https://docs.python.org/2/library/itertools.html#itertools.izip_longest, counting the number of situations where a is greater than b, and use a fill value for b that is less than the lowest possible a. This will result in the simplest possible control flow and most legible code. Shoot, you can ever force bools to ints with sum as here: sum(a > b for a, b in itertools.izip_longest(a_list, b_list, fillvalue=-1)).

edited May 25, 2015 at 17:02

answered May 25, 2015 at 16:55

jwilner

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1 Comment

gabhijit Over a year ago

Aah sorry didn't see this when I wrote the answer!

Karoly Horvath · Accepted Answer · 2015-05-25 16:52:40Z

-1

You can iterate over the pairs like so:

for x, y in zip(a,b):
    # if x > y...

The rest can be solved just as you described, by checking the lengths:

counter += len(a) - len(b) if len(a) > len(b) else 0

answered May 25, 2015 at 16:52

Karoly Horvath

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