Compare two lists in my python

Question

I have a python function that has as an input two variables the so-called current_level and current_affect. The function calculates some rules using training data and those rules can calculating decisions using those two variables the current_level and current_affect. The code contained two lists list_A and list_B. The lists contained the following:

list_A = [["'easy'", "'C'", "'4'", '0.714', '\n'], 
          ["'easy'", "'D'", "'5'", '0.778', '\n'], 
          ["'easy'", "'E'", "'5'", '0.500', '\n'], 
          ["'easy'", "'F'", "'6'", '0.750', '\n']]
list_B =  [["'easy'", "'B'", "'2'", '1.000', '\n'], 
           ["'easy'", "'C'", "'3'", '1.000', '\n'], 
           ["'easy'", "'D'", "'4'", '1.000', '\n'], 
           ["'easy'", "'E'", "'5'", '0.875', '\n'], 
           ["'easy'", "'F'", "'6'", '1.000', '\n']]

The first element of the lists correspond to the current_level, the second to the current_affect and the third to the variable score. Therefore what I want is by knowing the input of my method for current_affect to find which of the two lists for this specific case has greater score and return a correspondant value 1 in the case of list_A and -1 for the case of list_B or zero in the equality case. Variable current_affect takes values between A-F. Firstly, how can I add the non-existent current_affect values in my lists. For example list_A is missing levels A and B while list_B is missing level A (I want to add the sublists with zeros) and finally, how can I calculate the return value?

My function is the following:

def association_rules_adaptation(level, current_affect):

Then in the code i have zipped the two lists:

res = zip(list_A,list_B)

Withe the zip I want to link together the two lists regarding the current_affect in order to be able to compare which list has greater score for a specific current_affect. The problem is that when the lists doesn not contained the same exactly values of current_affect I link non-similar things:

(["'easy'", "'C'", "'3'", '0.714', '\n'], ["'easy'", "'B'", "'2'", '1.000', '\n'])
(["'easy'", "'D'", "'4'", '0.778', '\n'], ["'easy'", "'C'", "'3'", '1.000', '\n'])
(["'easy'", "'E'", "'5'", '0.500', '\n'], ["'easy'", "'D'", "'4'", '1.000', '\n'])
(["'easy'", "'F'", "'6'", '0.750', '\n'], ["'easy'", "'E'", "'5'", '0.875', '\n'])

So i want to add all the possible values in my lists in order to be able to link them correctly and then to perform my comparisons.

    for row in res:
    print (row)
    print ("Level", level, ": ", row[0][1])
    if (row[0][2] > row[1][2]) or (row[0][2] == row[1][2]):
        print ("1")
    else:
        print ("-1")

EDIT: I want to check which lists are missing and the to add them to lists list_A and list_B, like this:

list_A = [["'easy'", "'A'", "'0'", '0', '\n'], //added sublist
          ["'easy'", "'B'", "'0'", '0', '\n'], //added sublist
          ["'easy'", "'C'", "'4'", '0.714', '\n'], 
          ["'easy'", "'D'", "'5'", '0.778', '\n'], 
          ["'easy'", "'E'", "'5'", '0.500', '\n'], 
          ["'easy'", "'F'", "'6'", '0.750', '\n']]
list_B =  [["'easy'", "'A'", "'0'", '0', '\n'], //added subilst
           ["'easy'", "'B'", "'2'", '1.000', '\n'], 
           ["'easy'", "'C'", "'3'", '1.000', '\n'], 
           ["'easy'", "'D'", "'4'", '1.000', '\n'], 
           ["'easy'", "'E'", "'5'", '0.875', '\n'], 
           ["'easy'", "'F'", "'6'", '1.000', '\n']]

Then to zip my lists and perform the comparisons.

current_affect. Using the current_affect I want to see which of the two lists for this specific current_affect has greater score and then to return the correspondant 1, 0, -1. I have posted some code which just check my lists for all the cases. The problem is that the zip thing misses the real link between the current_affect since the lists do not contain the same current_affect so thus I want to add all of them in order to be able to compare for all cases. — konstantin
– konstantin, Commented Apr 14, 2017 at 13:26

jL4 · Accepted Answer · 2017-04-14 14:48:48Z

1

list_A = [["'easy'", "'C'", "'4'", '0.714', '\n'], 
          ["'easy'", "'D'", "'5'", '0.778', '\n'], 
          ["'easy'", "'E'", "'5'", '0.500', '\n'], 
          ["'easy'", "'F'", "'6'", '0.750', '\n']]
list_B =  [["'easy'", "'B'", "'2'", '1.000', '\n'], 
           ["'easy'", "'C'", "'3'", '1.000', '\n'], 
           ["'easy'", "'D'", "'4'", '1.000', '\n'], 
           ["'easy'", "'E'", "'5'", '0.875', '\n'], 
           ["'easy'", "'F'", "'6'", '1.000', '\n']]

A = zip(*list_A)
a = A[1]

B = zip(*list_B)
b = B[1]

def char_range(c1, c2):     //From http://stackoverflow.com/questions/7001144/range-over-character-in-python
    """Generates the characters from `c1` to `c2`, inclusive."""
    for c in xrange(ord(c1), ord(c2)+1):
        yield chr(c)

i=0
for c in char_range('A','F'):
    ch = "'{}'".format(c)
    if ch not in a:
        list_A.insert(i, ["'easy'",ch,"'0'",'0','\n'])
    if ch not in b:
        list_B.insert(i, ["'easy'",ch,"'0'",'0','\n'])
    i+=1

res = zip(list_A,list_B)

for r in res:
    if r[0][2] > r[1][2]:
        print 1
    elif r[0][2] < r[1][2]:
        print -1
    else:
        print 0

Result :

answered Apr 14, 2017 at 14:48

jL4

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5 Comments

konstantin Over a year ago

I am testing your example and I am receiving: TypeError: 'zip' object is not subscriptable

jL4 Over a year ago

I tested it in Python 2. Check this stackoverflow.com/questions/27431390/…

konstantin Over a year ago

Yes it seems that it is a dif between python 2 and 3.

konstantin Over a year ago

Is it also xrange and range the same thing? I got issues also with xrange.

jL4 Over a year ago

They are similar. range creates the entire list in memory while xrange doesn't.

Ben Quigley · Accepted Answer · 2017-04-14 13:28:42Z

1

I would try transforming those lists into a data structure more suited for looking stuff up, such as dictionaries. This is a bit messy, but:

level_lookup = {l[0]: int(l[2]) for l in list_B}
affect_lookup = {l[1]: int(l[2]) for l in list_B}

if current_level in level_lookup:
    level_score = level_lookup[current_level] 
else:
    level_score = 0
if current_affect in affect_lookup:
    affect_score = affect_lookup[current_affect]
else:
    affect_score = 0

return max(level_score, affect_score)

answered Apr 14, 2017 at 13:28

Ben Quigley

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1 Comment

konstantin Over a year ago

Probably I wasnt really clear enough. I have two lists list_A and list_B every lists contain 4 variables (level, current_affect, score, confidence). The variable for this example is constant and take the same value so I dont care for it. Therefore, by taking into account the current_affect variable I want to link the two lists list_A and list_B. Ideally I want to link them for every different current_affect {A, B, C, D, E, F} even in the case that is empty in the initial lists (so in that case I want to initialize it with score=0 and confidence =0).

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