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
.
Answers:
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.
your list comphresnion will, work but will return list of None because append return 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]
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)
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
.
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.
your list comphresnion will, work but will return list of None because append return 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]
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)