Python初学者,运行2.7
我找不到 Pythonic 的方式来编写以下公式。我的目标是拥有一个可以输入值的函数(在以前的尝试中,我只发现了一个高度重复的蛮力解决方案)。
必要信息:
如果您不知道 sigma 符号的含义:看这里
# 'v' is the formula
# 'i' 'j' 'k' are actors with numerical attributes 'c' 's' and 'x'
# 'n' = length of the following lists (length is identical for all four lists)
values_1 = [Bob, Sue, Mary, Jo...] # i, j, and k are any 3 values drawn from this list
# under the condition that i != j != k
values_2 = [5,6,7,8...] # c is drawn from this list.
# if i = values_1[0], then c(i) = values_2[0].
# same for j and k
values_3 = [9,10,11,12...] # s is drawn from this list.
# if i = values_1[0], then s(i) = values_3[0]
values_4 = [13,14,15, 16..] # x is drawn from this list.
# if i = values_1[0], then x(i) = values_4[0].
def V (j,k):
#this is where I need help!
v_Bob_Sue = # this is where V(Bob, Sue) should go.
# this is what the math would look like in English:
# SUM of (c_Mary * s_Mary * ((abs(x_Mary - x_Sue) - abs(x_Mary - x_Bob)) / x's range))
# and (c_Jo * s_Jo * ((abs(x_Jo - x_Sue) - abs(x_Jo - x_Bob)) / x's range))
# Bob and Sue are the j and k value
# we do the formula with all the other actors (from values_1) who AREN'T Bob and Sue
v_Bob_Mary = # etc
v_Bob_Jo =
v_Sue_Bob =
v_Sue_Mary =
v_Sue_Jo =
v_Mary_Bob =
v_Mary_Sue =
v_Mary_Jo =
v_Jo_Bob =
v_Jo_Sue =
v_Jo_Mary =