Question

我将在这里放置一个带有两个结尾的代码，一个有用，另一个没有。

我不会强调代码的目的是什么，因为这不是问题......

我的两个结局之间的区别是在第二个解决方案（崩溃）中，我想要应用一些数据修改。

这是代码的开头，它是两个.py文件的共同点：

import tetgen, geometry
from pprint import pprint
import random, csv
import numpy as np
from pprint import pprint

all_colors = [(name, float(X), float(Y), float(Z))
              for name, X, Y, Z in csv.reader(open('colors.csv'))]

priority_list = {name: int(i)
                 for i, name in csv.reader(open('priority.csv'))}

# background is marked SUPPORT
support_i = [i for i, color in enumerate(all_colors) if color[0] == 'SUPPORT']
if len(support_i)>0:
    support = np.array(all_colors[support_i[0]][1:])
    del all_colors[support_i[0]]
else:
    support = None

tg, hull_i = geometry.tetgen_of_hull([(X,Y,Z) for name, X, Y, Z in all_colors])
colors = [all_colors[i] for i in hull_i]

print ("thrown out: "
       + ", ".join(set(zip(*all_colors)[0]).difference(zip(*colors)[0])))

targets = [(name, float(X), float(Y), float(Z), float(BG))
           for name, X, Y, Z, BG in csv.reader(open('targets.csv'))]

for target in targets:
    name, X, Y, Z, BG = target
    target_point = support + (np.array([X,Y,Z]) - support)/(1-BG)
    tet_i, bcoords = geometry.containing_tet(tg, target_point)
    AT = (1-BG)

在此之后，解决方案（让我们称之为（1））起作用：

    output = open('solution_AT.txt','a')

    if tet_i == None:
        output.write(str(target[0]))
        output.write('\n')

    else:
        names = [colors[i][0] for i in tg.tets[tet_i]]
        sorted_indices = sorted(enumerate(names), key=lambda (i, name): priority_list[name])
        output.write(target[0])
        counting = 0

        for i, name in sorted(enumerate(names), key=lambda (i, name): priority_list[name]):
            output.write(',%s,%s' % (name, bcoords[i]*AT))
            counting = counting + 1

            if counting > 3:
                output.write('\n')
                counting = 0

output.close()

但不是解决方案（2）：

    output = open('solution_AT.txt','a')

    if tet_i == None:
        output.write(str(target[0]))
        output.write('\n')

    else:
        names = [colors[i][0] for i in tg.tets[tet_i]]
        sorted_indices = sorted(enumerate(names), key=lambda (i, name): priority_list[name])
        output.write(target[0])
        counting = 0

        for i, name in sorted(enumerate(names), key=lambda (i, name): priority_list[name]):
            counting = counting + 1
            top = bcoords[i]*AT
            output.write(',%s,%s' % (name, top))

            if counting > 0:
                counting = counting + 1
                cheese = bcoords[i]*AT
                output.write(',%s,%s' % (name, cheese/(1-top)))

                if counting > 1:
                    counting = counting + 1
                    meat = bcoords[i]*AT
                    output.write(',%s,%s' % (name, meat/(1-top-cheese)))    

                    if counting > 2:
                        counting = counting + 1
                        bread = bcoords[i]*AT
                        output.write(',%s,%s' % (name, bread/(1-top-cheese-meat))

                        if counting > 3:
                            output.write('\n')
                            counting = 0

output.close()

我收到错误：Failed to run script - syntax error - invalid syntax并且指针显示它在那里（我放“|”）：if counting > |3:

你知道为什么吗？

正如您所看到的，我要做的是将这些“top / cheese / meat”公式应用于bcoords[i]，见下文：

伪代码：

if counting = 0   // this is the initial value
I want: 
- top = bcoords[i]*AT
- counting = 1

if counting = 1   // the next value...
- cheese = bcoords[i]*AT
- output.write(',%s,%s' % (name, cheese/(1-top))
- counting = 2

if counting = 2
- meat = bcoords[i]*AT
- output.write(',%s,%s' % (name, meat/(1-top-cheese))    
- counting = 3

if counting > 2:
- counting = counting + 1
- bread = bcoords[i]*AT
- output.write(',%s,%s' % (name, bread/(1-top-cheese-meat))

但它并不适用于所有人！

有什么想法吗？

由于

Answer 1

您似乎有缩进问题。我在代码中观察到的是，以下代码行（if块）尝试访问在name循环中定义的变量top和for，即使它们实际上是在for循环之外。

缩进这些语句，使它们属于for循环。如果有效，请告诉我们

if counting > 0:
    counting = counting + 1
    cheese = bcoords[i]*AT
    output.write(',%s,%s' % (name, cheese/(1-top))

if counting > 1:
    counting = counting + 1
    meat = bcoords[i]*AT
    output.write(',%s,%s' % (name, meat/(1-top-cheese))    

if counting > 2:
    counting = counting + 1
    bread = bcoords[i]*AT
    output.write(',%s,%s' % (name, bread/(1-top-cheese-meat))

if counting > 3:
    output.write('\n')
    counting = 0

Answer 2

之前，您有3 (，只有2 )

因此Python正在有效地解析它

output.write(',%s,%s' % (name, cheese/(1-top)) if counting > 1:...

和if显然是语法错误

在output.write之前对列表内容进行python数学运算

2 个答案: