python - For loop, how can i use loop variable outside loop -

lr = int(raw_input())  rpl = [] c1 = [] c2 = [] l1a = [8, 9, 14, 13, 12] l2a =[9, 12, 14, 10, 8] om = [9, 10] l3a = [26] l1b = [27, 32, 26] l2b = [30, 27, 32, 28, 31] l3b = [31, 30, 26]  def i():     in l1b:         j in l2b:             k in l3b:                 n = * j * k                 c1.append(n)     in om:         j in l1a:             k in l2a:                 n = * j * k                 c2.append(n) def ii():     in c1:         j in c2:             x = (i / j) * lr             rpl.append(x) 

in program need loop variables i,j,k 'i' function print them in 'ii' function show combination used create x. tried 2 dimensional arrays didnt work well. there easy option work out?

the following script think trying do:

import operator, itertools  lr = int(raw_input())  l1a = [8, 9, 14, 13, 12] l2a = [9, 12, 14, 10, 8] l3a = [26] l1b = [27, 32, 26] l2b = [30, 27, 32, 28, 31] l3b = [31, 30, 26] om = [9, 10]  c1 = [(reduce(operator.mul, p), p) p in itertools.product(l1b, l2b, l3b)] c2 = [(reduce(operator.mul, p), p) p in itertools.product(om, l1a, l2a)]  rpl = [(((p[0][0]) / p[1][0]) * lr, p[0][1], p[1][1])  p in itertools.product(c1, c2)]  print rpl 

this displays following type of results of lr of 10:

[(380, (27, 30, 31), (9, 8, 9)), (290, (27, 30, 31), (9, 8, 12)), (240, (27, 30, 31), (9, 8, 14)), ... etc 

each permutation stored tuple result of multiplication. used when calculating rpl value.

you format rpl follows, show permutations made each result:

for rpl, p1, p2 in rpl:     print "%8d  %15s  %15s" % (rpl, str(p1), str(p2)) 

giving output in form:

 380     (27, 30, 31)        (9, 8, 9)  290     (27, 30, 31)       (9, 8, 12)  240     (27, 30, 31)       (9, 8, 14)  340     (27, 30, 31)       (9, 8, 10)  430     (27, 30, 31)        (9, 8, 8)  340     (27, 30, 31)        (9, 9, 9) 

the script uses python itertools module. provides product function has same effect of having multiple nested for loops. result of each iteration gives values i, j , k, tuple, e.g. (27, 30, 31).

the reduce command can used multiply of numbers in list returned, applying same function each entry. can't write reduce( * , p), can use python's operator module provide function name version *, i.e. operator.mul.

the result of gets wrapped in () make tuple 2 parts, first result of multiplications , second part permutation produced it. e.g. (25110, (27, 30, 31)).

c1 list holding of these value. called list comprehension. equivalent for loop c1.append() inside.

once c1 , c2 have been created (i suggest try , print values see like), script uses similar method calculate of rpl values. each iteration gives p like:

((25110, (27, 30, 31)), (648, (9, 8, 9))) 

this tuple 2 entries (25110, (27, 30, 31)) , (648, (9, 8, 9)). python can access each value using indexes. 25110 use p[0][0] first tuple, first part. or p[0][1] (27, 30, 31).

the solution script converted not use list comprehensions follows:

c1 = []  p in itertools.product(l1b, l2b, l3b):     multiply_all = reduce(operator.mul, p)     c1.append((multiply_all, p))  c2 = []  p in itertools.product(om, l1a, l2a):     multiply_all = reduce(operator.mul, p)     c2.append((multiply_all, p))  rpl = []  p in itertools.product(c1, c2):     calculation = (p[0][0] / p[1][0]) * lr     rpl.append((calculation, p[0][1], p[1][1]))