我正在尝试调整我的python程序的主文件,但它是一个"未绑定的本地错误:本地变量' sider'在转让之前引用"。我不确定这是怎么发生的,因为变量"角度"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
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
修改强>
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。