python - python中的简单L系统

您好，我正在寻找一种在python中将简单的L系统实现为函数的方法，该方法需要三个参数：公理，规则和交互次数（如果迭代次数= 0，则以前输入的是公理）。我想出了一些代码，它仅适用于1次迭代，而且我不知道如何实现更多代码。

我想出的代码：

#  x = axiom
#  y = rules
#  z would be iterations which I dont know how to implement

def lsystem(x,y):
    output = ''
    for i in x:
        if i in y:
            output += y[i]
        else:
            output += i
    print(output)

rules = { "A" : "ABA" , "B" : "BBB"}
# output    lsystem("AB",rules) ---> ABABBB

最佳答案

如果axioms，则需要返回给定的iterations == 0。在此函数中，您将返回已给定的参数axioms，因此，如果iterations == 0，则将返回给定的，不变的公理。

然后，稍后，在iteration的末尾，如果存在iteration，则将从iteration获得的新创建的公理转移到axioms中，以便您返回正确的值，并且在需要时是，下一个iteration将具有要迭代的新创建的公理。 :)

def lsystem(axioms, rules, iterations):
    #    We iterate through our method required numbers of time.
    for _ in range(iterations):
        #    Our newly created axioms from this iteration.
        newAxioms = ''

        #    This is your code, but with renamed variables, for clearer code.
        for axiom in axioms:
            if axiom in rules:
                newAxioms += rules[axiom]
            else:
                newAxioms += axiom
        #    You will need to iterate through your newAxioms next time, so...
        #    We transfer newAxioms, to axioms that is being iterated on, in the for loop.
        axioms = newAxioms
    return axioms

rules = { "A" : "ABA" , "B" : "BBB"}
print(lsystem('AB', rules, 0))
# outputs : 'AB'

print(lsystem('AB', rules, 1))
# outputs : 'ABABBB'

print(lsystem('AB', rules, 2))
# outputs : 'ABABBBABABBBBBBBBB'

关于python - python中的简单L系统，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/48113662/