list comprehension python with if statement

Question

While trying to workout list comprehension , I got stuck with desired value repeated.

I have 2 lists: L1, L2. Result required is list of items from L2 if these items are smaller than/ equal to, at least one of the items in L1.

L1=[10,20,30,40,50]
L2=[3,11,51]
L3=[d2 for d2 in L2 for d1 in l1 if d2<=d1]

L3 is returned as [3, 3, 3, 3, 3, 11, 11, 11, 11]

Answer contains valid items, but are repeated. I know using set(), we can get rid of repetitions but may be I am using list-comprehension in a wrong way. Any clarification would be appreciated.

Loops to achieve the desired result would be:

L3=[]
for d2 in L2:
    for d1 in L1:
        if d2<=d1:
            L3.append(d2)

Eli Rose · Accepted Answer · 2015-04-05 18:53:55Z

5

As you can see from your expansion into loops, the inner loop runs over every element in L1, so the d2 element will get appended once for every time there's a bigger element in L2.

You could use a set() (equivalently, { } brackets as HepaKKes suggests) to get rid of repetitions, but this is inefficient -- you're still creating the unneeded intermediate results. It will be O(n^2) time.

You're selecting items based on whether they're <= than any item in L1. This is the same as asking whether they're <= max(L1). So the following:

L3 = [e for e in L2 if e < max(L1)]

will achieve the same result. If you save the value of L1 ahead of time

L1_max = max(L1)
L3 = [e for e in L2 if e < L1_max]

then this solution is O(n).

answered Apr 5, 2015 at 18:53

Eli Rose

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Greg Edelston · Accepted Answer · 2015-04-05 18:51:42Z

2

What you wrote goes through two nested for-loops, and whenever the condition d2<=d1 is true, that value is appended. So we're seeing both 3 and 11 added for each value in L2 that they're less than.

Instead, you'll want a stricter condition for adding d2 to L3, so that your comprehension can be of the form L3 = [d2 for d2 in L2 if (x)], where x is equivalent to "d2 is less than some value in L1". There are a few ways to do that, but I found the following to work:

L3 = [d2 for d2 in L2 if (filter(lambda d1: d2<=d1, L1))]

answered Apr 5, 2015 at 18:51

Greg Edelston

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3 Comments

Eli Rose Over a year ago

filter returns a list. I think you want to use any, as in: if any(d2 <= d1 for d1 in L1). Map would also work.

Greg Edelston Over a year ago

ah, yup. filter works here because any non-empty list evaluates to True, but any would be cleaner.

Eli Rose Over a year ago

oh, it does work. ha! (yeah, I would still use any).

Mike Bessonov · Accepted Answer · 2015-04-05 19:14:37Z

1

If the task is to write a single list comprehension, I would go for something like this:

In [22]: [x for x in L2 if any([y >= x for y in L1])]
Out[22]: [3, 11]

answered Apr 5, 2015 at 19:14

Mike Bessonov

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