Python中的未绑定本地错误

时间：2016-03-19 02:03:35

标签： python function python-3.x error-handling main

我正在尝试调整我的python程序的主文件，但它是一个＆＃34;未绑定的本地错误：本地变量＆＃39; sider＆＃39;在转让之前引用＆＃34;。我不确定这是怎么发生的，因为变量＆＃34;角度＆＃34;是在同一个位置，但它的工作原理。任何帮助表示赞赏！

from getdata import *
from trifun import *


def main():
    outfile = open("SFout.txt", "w")
    myList = getData()
    # print(myList)

    names = myList[0]
    vertices = myList[1]

    for i in range(len(names)):
        x1 = vertices[i][0][0]
        y1 = vertices[i][0][1]
        x2 = vertices[i][1][0]
        y2 = vertices[i][1][1]
        x3 = vertices[i][2][0]
        y3 = vertices[i][2][1]
        print((names[i]), (x1, y1), ",", (x2, y2), ",", (x3, y3))
        print((names[i]), (x1, y1), ",", (x2, y2), ",", (x3, y3), file = outfile)



        if duplicatePts(vertices[i]):
            print("You have duplicate points, not a triangle")
            outfile.write("You have duplicate points, not a triangle \n \n")
            print("\n")
            continue

        if collinear(vertices[i]):
            print("The points are collinear, not a triangle")
            outfile.write("\nThe points are collinear, not a triangle \n \n")
            print("\n")
            continue

        v = findAllSides(vertices[i])
        pm = str("%.1f" % (perimeter(vertices[i])))
        print("Perimeter: ", pm.ljust(25), end="")
        print("Perimeter: ", pm.ljust(25), end="", file = outfile)

        s = ("Sidelengths %0.2f, %0.2f, %0.2f" % (v[0], v[1], v[2]))
        sidelength1 = str("%.1f" % (v[0]))
        sidelength2 = str("%.1f" % (v[1]))
        sidelength3 = str("%.1f" % (v[2]))
        side = sidelength1 + "," + sidelength2 + "," + sidelength3
        print("Side lengths: ", side)
        print("Side lengths: ", side, file=outfile)
        #outfile.write("Side length:" + side + '\n')

        if acute(vertices[i]):
            angle = "Acute"
            # print("Acute")
            #           outfile.write("\nAcute")
        elif right(vertices[i]):
            angle = "Right"
        #            print("Right")
        #            outfile.write("\nRight")
        elif obtuse(vertices[i]):
            angle = "Obtuse"
        #            print("Obtuse")
        #            outfile.write("\nObtuse")
        if Scalene(vertices[i]):
            side = "Scalene"
        #            print("Scalene")
        #            outfile.write("\nScalene")
        elif Equilateral(vertices[i]):
            side = "Equilateral"
        #            print("Equilateral")
        #            outfile.write("\nEquilateral")
        elif Isosceles(vertices[i]):
            side = "Isosceles"
        #            print("Isosceles")
        #            outfile.write("\nIsosceles")
        # f.close()
        print((angle + " & " + side).ljust(37), end="")
        print((angle + " & " + side).ljust(37), end="", file=outfile)

        #outfile.write( angle + " & " + side+"\n")
        print("Area is: %0.2f" % area(vertices[i]))
        print("Area is: %0.2f" % area(vertices[i]), "\n", file=outfile)

        #outfile.write("Area is: %0.2f" % area(vertices[i])+"\n")
        print("\n")
    outfile.close()


main()

的getData 

def getData():
    names = []
    vertices = []
    myList = [names,vertices]
    with open("test.txt") as f:
        for line in f:
            x = line.split()
            names.append(x[0])
            x1 = float(x[1])
            y1 = float(x[2])
            x2 = float(x[3])
            y2 = float(x[4])
            x3 = float(x[5])
            y3 = float(x[6])

            vertices.append([[x1,y1],[x2,y2],[x3,y3]])

        return myList

Trifun 

# The text file contains vertices of three points of a triangle, separated by a space.

# Import Math for calculations.
import math


# This function checks if there are any duplicate points in te vertices of the triangle.
# If there are, it s not a triangle.
def duplicatePts(vertices):
    #vertices looks like this : [[x1,y1],[x2,y2],[x3,y3]]
    x1 = vertices[0][0]
    y1 = vertices[0][1]
    x2 = vertices[1][0]
    y2 = vertices[1][1]
    x3 = vertices[2][0]
    y3 = vertices[2][1]

    if (x1,y1)==(x2,y2) or (x2, y2) == (x3, y3) or (x1, y1) == (x3, y3):
        return True
    else:
        return False


# This function checks if the points of the triangle are collinear.
# Triangle points cannot be in the same line. They have to be in different positions on the lane.
def collinear(vertices):


    x1 = vertices[0][0]
    y1 = vertices[0][1]
    x2 = vertices[1][0]
    y2 = vertices[1][1]
    x3 = vertices[2][0]
    y3 = vertices[2][1]

    if (x2 - x1) == 0: 
        # As discussed in class, set it to a high number, to assume it is vertical
        slope1 = 9999
    else:
        slope1 = (y2 - y1) / (x2 - x1)
    if (x3 - x2) == 0:
        # As discussed in class, set it to a high number, to assume it is vertical
        slope2 = 9999
    else:
        slope2 = (y3 - y2) / (x3 - x2)

    if slope1 == slope2:
        return True
    else:
        return False


# To find the perimeter, we need to find the sides first.
# Perimeter = Sum of all three sides
def perimeter(vertices):

    x1 = vertices[0][0]
    y1 = vertices[0][1]
    x2 = vertices[1][0]
    y2 = vertices[1][1]
    x3 = vertices[2][0]
    y3 = vertices[2][1]

    sideAB = sideLength(x1, y1, x2, y2)
    sideBC = sideLength(x2, y2, x3, y3)
    sideAC = sideLength(x3, y3, x1, y1)
    perimeter = sideAB + sideAC + sideBC
    return perimeter

# Finding the length of the side using the distance formula.
def sideLength(x1, y1, x2, y2):

    length = math.sqrt((x2 - x1) ** 2 + (y2 - y1) ** 2)
    return length


# Using the min function to find out the shortest side.
def findAllSides(vertices):

    x1 = vertices[0][0]
    y1 = vertices[0][1]
    x2 = vertices[1][0]
    y2 = vertices[1][1]
    x3 = vertices[2][0]
    y3 = vertices[2][1]

    sideAB = sideLength(x1, y1, x2, y2)
    sideBC = sideLength(x2, y2, x3, y3)
    sideAC = sideLength(x3, y3, x1, y1)
    lst=[sideAB, sideAC, sideBC]

    return lst




# We know how to find the sides now.

# We determine if the triangle is a right triangle by using Pythagoras theorem.
# Pythagoras theorem states that a triangle is a right triangle when
# Square of Hypotenus is equal to the sum of other two sides.
def typeObtuseRightAcute(vertices):
    #no idea if this is a good value but works for example
    #and should be low enough to give right answers for all but crazy close triangles
    epsilon=10**-8
    x1 = vertices[0][0]
    y1 = vertices[0][1]
    x2 = vertices[1][0]
    y2 = vertices[1][1]
    x3 = vertices[2][0]
    y3 = vertices[2][1]
    # Using Pythagoras theorem
    sideAB = sideLength(x1, y1, x2, y2)
    sideBC = sideLength(x2, y2, x3, y3)
    sideAC = sideLength(x3, y3, x1, y1)

    #use this instead
    [var1,var2,largest] = sorted([sideAB, sideBC, sideAC])



    if abs((largest) ** 2-((var1 ** 2 + (var2) ** 2)))<epsilon:
        return "right"
    elif (largest) ** 2 > ((var1 ** 2 + (var2) ** 2)):
        return "obtuse"
    else:
        return "acute"

def acute(vertices):
    return typeObtuseRightAcute(vertices)=="acute"

def right(vertices):
    return typeObtuseRightAcute(vertices)=="right"

def obtuse(vertices):
    return typeObtuseRightAcute(vertices)=="obtuse"

# A triangle is a scalene triangle, when all three sides are not equal to each other.
def Scalene(vertices):
    EPSILON = 0.000001
    x1 = vertices[0][0]
    y1 = vertices[0][1]
    x2 = vertices[1][0]
    y2 = vertices[1][1]
    x3 = vertices[2][0]
    y3 = vertices[2][1]

    sideAB = sideLength(x1, y1, x2, y2)
    sideBC = sideLength(x2, y2, x3, y3)
    sideAC = sideLength(x3, y3, x1, y1)
    if (abs(sideAB-sideBC)>=EPSILON and abs(sideBC-sideAC)>=EPSILON and abs(sideAB-sideAC)>=EPSILON):
        return True


# A triangle is an isosceles triangle when 2 of the sides are equal.
def Isosceles(vertices):
    EPSILON = 0.000001
    x1 = vertices[0][0]
    y1 = vertices[0][1]
    x2 = vertices[1][0]
    y2 = vertices[1][1]
    x3 = vertices[2][0]
    y3 = vertices[2][1]

    sideAB = sideLength(x1, y1, x2, y2)
    sideBC = sideLength(x2, y2, x3, y3)
    sideAC = sideLength(x3, y3, x1, y1)
    if (abs(sideAB-sideBC)<EPSILON or abs(sideBC-sideAC)<EPSILON or abs(sideAB-sideAC)<EPSILON) and not (abs(sideAB-sideBC)<EPSILON and abs(sideBC-sideAC)<EPSILON and abs(sideAB-sideAC)<EPSILON):
        return True  # or True
    else:
        return False


# A triangle is an equilateral triangle when all three sides are equal to each other.
def Equilateral(vertices):
    EPSILON = 0.000001
    x1 = vertices[0][0]
    y1 = vertices[0][1]
    x2 = vertices[1][0]
    y2 = vertices[1][1]
    x3 = vertices[2][0]
    y3 = vertices[2][1]

    sideAB = sideLength(x1, y1, x2, y2)
    sideBC = sideLength(x2, y2, x3, y3)
    sideAC = sideLength(x3, y3, x1, y1)

    if (abs(sideAB-sideBC)<EPSILON and abs(sideBC-sideAC)<EPSILON and abs(sideAB-sideAC)<EPSILON):
        return True
    else:
        return False

# To find the area, we need to find the semi perimeter first.
# We find the area of triangle using Heron's formula.
def area(vertices):

    x1 = vertices[0][0]
    y1 = vertices[0][1]
    x2 = vertices[1][0]
    y2 = vertices[1][1]
    x3 = vertices[2][0]
    y3 = vertices[2][1]

    sideAB = sideLength(x1, y1, x2, y2)
    sideBC = sideLength(x2, y2, x3, y3)
    sideAC = sideLength(x3, y3, x1, y1)
    semiPerimeter = (sideAB + sideBC + sideAC) / 2

    area = (semiPerimeter * (semiPerimeter - sideAB) * (semiPerimeter - sideBC) * (semiPerimeter - sideAC)) ** 0.5

    return area

文字档案

Alpha -4 0 4 0 0 6.9282
Bravo -2.161 -3.366 2.161 3.366 -5.83 3.743
Charlie 3.54 5.46 -4.54 5.557 3.1 -2.1
Delta 0 4 0 2 0 2
Echo -4 3 1 3 6 3
Foxtrot 0 0 5 1 10 0
Golf 0 4 0 0 4 0

2 个答案:

答案 0 :(得分：0)

我不确定这是否可以让您满意但是这可以防止获得Unbound Local Error：

sider = ("Scalene"
         if scalene(vertices[i])
         else ("Equilateral"
               if equilateral(vertices[i])
               else ("Isosceles"
                     if isosceles(vertices[i])
                     else "None of the above")
               )
         )

或者如果您想维护格式并且只是为了安全起见，只需初始化局部变量sider：

def main():
    outfile = open("SFout.txt", "w")
    myList = getData()

    names = myList[0]
    vertices = myList[1]
    sider = '' # declare this one
    angle = '' # and this one

    for i in range(len(names)):
        x1 = vertices[i][0][0]
        y1 = vertices[i][0][1]
        x2 = vertices[i][1][0]
        y2 = vertices[i][1][1]
        x3 = vertices[i][2][0]
        y3 = vertices[i][2][1]
        print((names[i]), (x1, y1), ",", (x2, y2), ",", (x3, y3))
        print((names[i]), (x1, y1), ",", (x2, y2), ",", (x3, y3), file=outfile)

....

答案 1 :(得分：0)

这将有助于为您的问题提供追溯。我认为错误发生在var c = 0; var m = 0; var t = 0; var x = 0; var y = 0; window.setInterval(function() { var e; var s; if (c === 0) { m = Date.now(); } s = window.performance.now(); x += Math.random(); y += Math.random(); c++; e = window.performance.now(); t += e - s; if (c !== 1000) { return; } console.log(t.toFixed(0).toString() + " milliseconds"); console.log((Date.now() - m).toFixed(0).toString() + " milliseconds"); c = 0; m = 0; t = 0; x = 0; y = 0; }, 1); 语句中。所有函数print((angle ...)，scalene()或equilateral()必须返回False。要找到失败的三角形，请添加这样的isosceles()块以打印顶点。

else

修改 问题在于：typeEquilateralIsoscelesScalene（顶点）返回“ E quilateral”，但在if scalene(vertices[i]): sider = "Scalene" elif equilateral(vertices[i]): sider = "Equilateral" elif isosceles(vertices[i]): sider = "Isosceles" else: # this should never happen print("Something's wrong!") print(vertices[i]) 中，您将其与“ e quilateral”进行比较。与scalene（）和isosceles（）相同。

相关问题
最新问题