python - 假设 x 属于 [x1,x2]，则获取 x1,x2

Question

给定:

xvalues = [0.0, 1829.0, 3658.0, 5487.0]

和

nodesF = [[1, 0, 0, 0], [2, 0.5, 0, 0], [3, 5487, 0, 0]]

我想循环nodesF并返回x1和x2每个 x 的值的nodesF位于之间，即 x 属于 [x1,x2]与 x1<x2 .

我的代码是:

     for nodeID, x, y, z in nodesF:
        x2= min(value for value in xvalues if value >= x)
        x1= max(value for value in xvalues if value <= x)
        if x1==x2:
            x1=None
            x2=None
            x2= min(value for value in xvalues if value > x)
            x1= max(value for value in xvalues if value <= x)
            if x2==None or x2<=x1:
                x2= min(value for value in xvalues if value >= x)
                x1= max(value for value in xvalues if value < x)
            elif x1==None or x2<=x1:
                print "Error"

对于 x=5487 我得到:

x2= min(value for value in xvalues if value > x) ValueError: min() arg is an empty sequence.

所以我的问题是如何传递这个错误？如果我可以设置 x2=None当min()是空的就OK了!谢谢!

Answer 1

您的方向是正确的，但您需要自己检查列表:

valid_values = [value for value in xvalues if value > x] #changed >= by >
if valid_values:
    x2 = min(valid_values)
else:
    x2 = None

或者您可以在事后发现错误:

try:
    x2 = min(value for value in xvalues if value > x) #changed >= by >
except ValueError:
    x2 = None