了解Python

Question

I am reading the book Programming Collective Intelligence, What exactly the following piece of python code do?

  # Add up the squares of all the differences 
  sum_of_squares=sum([pow(prefs[person1][item]-prefs[person2][item],2) 
                      for item in prefs[person1] if item in prefs[person2]]) 

I am trying to play with the examples in Java.

Prefs is a map of person to movie ratings, movie ratings is another map of names to ratings.

Answer 1

First it constructs a list containing the results from:

for each item in prefs for person1:
    if that is also an item in the prefs for person2:
        find the difference between the number of prefs for that item for the two people
        and square it (Math.pow(x,2) is "x squared")

Then it adds those up.

Answer 2

This might be a little more readable if the call to pow were replaced with an explicit use of '**' exponentiation operator:

sum_of_squares=sum([(prefs[person1][item]-prefs[person2][item])**2
                   for item in prefs[person1] if item in prefs[person2]])

Lifting out some invariants also helps readability:

p1_prefs = prefs[person1]
p2_prefs = prefs[person2]

sum_of_squares=sum([(p1_prefs[item]-p2_prefs[item])**2
                      for item in p1_prefs if item in p2_prefs])

Finally, in recent versions of Python, there is no need for the list comprehension notation, sum will accept a generator expression, so the []'s can also be removed:

sum_of_squares=sum((p1_prefs[item]-p2_prefs[item])**2
                      for item in p1_prefs if item in p2_prefs)

Seems a bit more straightforward now.

Ironically, in pursuit of readability, we have also done some performance optimization (two endeavors that are usually mutually exclusive):

• lifted invariants out of the loop
• replaced the function call pow with inline evaluation of '**' operator
• removed unnecessary construction of a list

Is this a great language or what?!

Answer 3

01 sum_of_squares =
02 sum(
03  [
04      pow(
05         prefs[person1][item]-prefs[person2][item],
06         2
07      ) 
08    for
09       item
10    in
11       prefs[person1]
12    if
13       item in prefs[person2]
14  ]
15 )

Sum (line 2) a list, that consists of the values computed in lines 4-7 for each 'item' defined in the list specified on line 11 which the condition on line 13 holds true for.

Answer 4

It computes the sum of the squares of the difference between prefs[person1][item] and prefs[person2][item] , for every item in the prefs dictionary for person1 that is also in the prefs dictionary for person2 .

In other words, say both person1 and person2 have a rating for the film Ratatouille , with person1 rating it 5 stars, and person2 rating it 2 stars.

prefs[person1]['Ratatouille'] = 5
prefs[person2]['Ratatouille'] = 2

The square of the difference between person1 's rating and person2 's rating is 3^2 = 9 .

It's probably computing some kind of Variance .