Notes for Session 07¶
A collection of notes to go over in class, to keep things organized.
Lightning Talks¶
Marlon M Estrada (if you are prepared)
Brian Nagata
Rajaramesh V Yaramati
Zandra Eng
Issues that came up during the week.¶
Mutating vs. re-assigning¶
I’ve seen code like this in a few trigram solutions:
output = output + [follower]
(output is a list of strings, follower is a single string)
What it does is add a new item to a list.
But is that an efficient way to do that?
If you are adding one element to a list – append() is the way to go. This works fine, but it’s creating a whole new list just to throw it away again:
output_list.append (random_trigram_followers)
and if you are adding another list of objects, you want to use extend(). The way it is now, you are actually doing:
- create a list with random_trigram_followers in it.
- create a new list with the contents of output_list the new list.
- re-assign the name output_list to that new list.
- throw away the original output_list and the temporary list you created for random_trigram_followers
That’s a LOT of overhead!
Be cognizant of when you are mutating (changing) an object vs creating a new one and assigning it to the same name. When you do assignment (=) you are probably creating a new object.
+= is different – it is the “in_place” operator, so:
a_list += another_list
does not create an new lists – it adds to the original list “in place” – it is identical to:
a_list.extend(another_list)
And it is an efficient operation.