Inline for loop

Question

I'm trying to learn neat pythonic ways of doing things, and was wondering why my for loop cannot be refactored this way:

q  = [1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5]
vm = [-1, -1, -1, -1]

for v in vm:
    if v in q:
        p.append(q.index(v))
    else:
        p.append(99999)

vm[p.index(max(p))] = i

I tried replacing the for loop with:

[p.append(q.index(v)) if v in q else p.append(99999) for v in vm]

But it doesn't work. The for v in vm: loop evicts numbers from vm based on when they come next in q.

You may want to use -1 instead of 99999 for a flagging an inexistent value (so it would work even for a list with 99999+ elements) — Paulo Scardine, Dec 10 '14 at 22:04
you list comprehsnsion will work but it will return lists of None too — Hackaholic, Dec 10 '14 at 22:04
I used 99999 instead of -1 because later I ran a max(p) and nonexistent values need to be selected. — Will, Dec 10 '14 at 22:06

score 112 · Accepted Answer · 2014-12-10T22:05:48.607

What you are using is called a list comprehension in Python, not an inline for-loop (even though it is similar to one). You would write your loop as a list comprehension like so:

p = [q.index(v) if v in q else 99999 for v in vm]

When using a list comprehension, you do not call list.append because the list is being constructed from the comprehension itself. Each item in the list will be what is returned by the expression on the left of the for keyword, which in this case is q.index(v) if v in q else 99999. Incidentially, if you do use list.append inside a comprehension, then you will get a list of None values because that is what the append method always returns.

Thanks man, awesome! +internets for answering in less than a minute. — Will, Dec 10 '14 at 22:03

score 13 · Answer 2 · edited Dec 13 '22 at 14:21

13

your list comprehension will work but will return list of None, because append returns None:

demo:

>>> a = []
>>> [a.append(x) for x in range(10)]
[None, None, None, None, None, None, None, None, None, None]
>>> a
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

better way to use it like this:

>>> a = [x for x in range(10)]
>>> a
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

edited Dec 13 '22 at 14:21

andreas

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answered Dec 10 '14 at 22:07

Hackaholic

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Padraic Cunningham · Answer 3 · 2014-12-10T22:12:35.447

5

you can use enumerate keeping the ind/index of the elements is in vm, if you make vm a set you will also have 0(1) lookups:

vm = {-1, -1, -1, -1}

print([ind if q in vm else 9999 for ind,ele in enumerate(vm) ])

edited Dec 10 '14 at 22:12

answered Dec 10 '14 at 22:06

Padraic Cunningham

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Cool fancy syntax, thanks for sharing this looks very useful! – Will Dec 11 '14 at 19:05

score 3 · Answer 4 · answered Dec 10 '14 at 22:27

q  = [1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5]
vm = [-1, -1, -1, -1,1,2,3,1]

p = []
for v in vm:
    if v in q:
        p.append(q.index(v))
    else:
        p.append(99999)

print p
p = [q.index(v) if v in q else 99999 for v in vm]
print p

Output:

[99999, 99999, 99999, 99999, 0, 1, 2, 0]
[99999, 99999, 99999, 99999, 0, 1, 2, 0]

Instead of using append() in the list comprehension you can reference the p as direct output, and use q.index(v) and 99999 in the LC.

Not sure if this is intentional but note that q.index(v) will find just the first occurrence of v, even tho you have several in q. If you want to get the index of all v in q, consider using a enumerator and a list of already visited indexes

Something in those lines(pseudo-code):

visited = []
for i, v in enumerator(vm):
   if i not in visited:
       p.append(q.index(v))
   else:
       p.append(q.index(v,max(visited))) # this line should only check for v in q after the index of max(visited)
   visited.append(i)

Inline for loop

4 Answers4

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