使用条件语句在多个for循环内中断并返回 [英] Break and return inside multiple for-loops with conditional statement
问题描述
我创建了一个长代码，由for循环内的多个列表组成.计算没有错，它确实获得了预期的结果.列表的代码和构造工作正常.问题是当它运行时，我已定义了满足特定条件时的中断.但是它不会在第一次运行时中断，而是继续运行并遍历第一次循环范围功能内的所有值.我想实现在条件匹配时返回一个真值，并停止并且不会根据第一个循环中的范围值继续增长.
I have created a long code, consist of multiple lists inside for-loops. There are nothing wrong with calculation.It does obtain results as it is expected to. The code and construction of lists work fine. The problem is when it runs, I have defined a break when a certain condition is matched. But it does not break at first run, continues and run through all values within range function of first loop. I want to achieve to return a true value, when the condition is matched, and stops and will not keep growing according to range values in first loop.
我将解释代码的工作原理！
I would explain how the code would work!
代码: 第一部分是一致的，是输入
The code: First part is consistent and are inputs
import math
import numpy as np
Ned = -500
fcd = 20
fyd = 435
E = 2e5
h = 200
cb = 35
ct = 35
ca = 35
b= 150
y = 12
d = h - cb
ds = ct
a = 25
yb = 8
ecu = 0.0035
rebarnumber = math.floor((b-(2*cb+2*yb+y))/a)
PI_Y2_4 = int(math.pi/4*(y)**2)
disc = []
dis = []
Asi = []
Asci = []
Esc = []
Esci = []
Sc = []
Sci =[]
#############################
# Calculation starts here
#############################
for n in range(0,10): # <------- First for-loop
cbb = cb + yb + y/2
ctt = ct + yb + y/2
if 0 < n <= rebarnumber:
Asi.append(PI_Y2_4)
dis.append( h - cbb)
Asci.append(PI_Y2_4)
disc.append( ctt )
if rebarnumber < n <= (2 * rebarnumber):
Asi.append(PI_Y2_4)
dis.append( h - cbb - ca)
Asci.append(PI_Y2_4)
disc.append(cbb + ca)
if (2*rebarnumber) < n <= (3 * rebarnumber):
Asi.append(PI_Y2_4)
dis.append( h - cbb - 2*ca)
Asci.append(PI_Y2_4)
disc.append(cbb + 2*ca)
if (3*rebarnumber) < n <= (4 * rebarnumber):
Asi.append(PI_Y2_4)
dis.append( h - cbb - 3*ca)
Asci.append(PI_Y2_4)
disc.append(cbb + 3*ca)
if (4*rebarnumber) < n <= (5 * rebarnumber):
Asi.append(PI_Y2_4)
dis.append( h - cbb - 4*ca)
Asci.append(PI_Y2_4)
disc.append(cbb + 4*ca)
if (5*rebarnumber) < n <= (6 * rebarnumber):
Asi.append(PI_Y2_4)
dis.append( h - cbb - 5*ca)
Asci.append(PI_Y2_4)
disc.append(cbb + 5*ca)
if (6*rebarnumber) < n <= (7 * rebarnumber):
Asi.append(PI_Y2_4)
dis.append( h - cbb - 6*ca)
Asci.append(PI_Y2_4)
disc.append(cbb + 6*ca)
if (7*rebarnumber) < n <= (8 * rebarnumber):
Asi.append(PI_Y2_4)
dis.append( h - cbb - 7*ca)
Asci.append(PI_Y2_4)
disc.append(cbb + 7*ca)
for i in range(0,len(dis)):
Esc.insert(i, dis[i])
Esci.insert(i, disc[i])
Sc.insert(i, dis[i])
Sci.insert(i, disc[i])
for x in np.linspace(1,h,10000): # <-------- Second for-loop
for k, _ in enumerate(Esc):
try:
if x < dis[k]:
Esc[k]=( ecu/x*(dis[k]-x) )
else:
Esc[k]=(- ecu/x*(x-dis[k] ) )
if x < disc[k]:
Esci[k]=( -ecu/x*(x-disc[k]) )
else:
Esci[k]=(- ecu/x*(x-disc[k]) )
except (ZeroDivisionError, RuntimeWarning):
Esc[k]=( 0 )
Esci[k]=( 0 )
for k, _ in enumerate(Sc): # <-------- Third for-loop
ss = Esc[k]*E
if ss <= -fyd:
Sc[k]= -fyd
elif ss >= -fyd and ss < 0:
Sc[k]=ss
else:
Sc[k]=min(ss,fyd)
for k, _ in enumerate(Sci):
sci = Esci[k]*E
if sci <= -fyd:
Sci[k]= -fyd
elif sci >= -fyd and sci < 0:
Sci[k]=sci
else:
Sci[k]=min(sci,fyd)
FS = 0
FSC = 0
for a, _ in enumerate(Sc):
FS += Sc[a]*Asi[a]
FSC+=Sci[a]*Asci[a]
MS = 0
MSC = 0
for m, _ in enumerate(Sc):
MS += Sc[a]*Asi[a]*(dis[m]-h/2)
MSC+= Sci[a]*Asci[a]*(h/2-disc[m])
Nrd = 0
Mrd = 0
Nrd = int((-0.8*x*b*fcd+FSC+FS)/1000)
Mrd = (0.8*x*b*fcd*(h/2-0.4*x)+MS-MSC)/1000000
if 0 <= (float(Nrd) - Ned) <= 1:
print(Nrd, x, Asi)
break
break
它如何工作?
How does it work?
第一个for循环创建一个索引为0的列表，即[value1]，第二个for循环创建一个x值递增(范围)，第三个for循环创建依赖于第一个创建的列表[value1]的列表.然后Nrd值基于x值的增量确定.如果满足条件0 <= (float(Nrd) - Ned) <= 1:，则计算将停止并返回Nrd值.如果不匹配，则返回并从第一个for循环获取索引1，创建[value1, value2]，如果满足条件，则再次到达Nrd，否则将中断，否则将继续直到匹配.
First for-loop creates a list with index 0 i.e. [value1], second for-loop makes a x value increment ( range), third for-loops creates lists depend on first created list [value1]. then Nrd value determines based on increment of x values. if condition 0 <= (float(Nrd) - Ned) <= 1: is met, then the calculation would stop and return Nrd value. if not matched, it goes back and take index 1 from first for-loop, [value1, value2] is created, again reach to Nrd if condition is met, would break otherwise continues until it is matched.
我的问题是代码运行时，我得到了这些输出.
My problem is when the code runs, I get those outputs.
Nrd x Asi
---------------------------------------------------
-499 181.84938493849384 [113]
-499 162.36533653365336 [113, 113]
-499 147.3990399039904 [113, 113, 113]
-499 137.48784878487848 [113, 113, 113, 113]
-499 130.72117211721172 [113, 113, 113, 113, 113]
-499 126.10391039103911 [113, 113, 113, 113, 113, 113]
-499 122.7006700670067 [113, 113, 113, 113, 113, 113, 113]
-499 120.01390139013901 [113, 113, 113, 113, 113, 113, 113, 113]
-499 119.71537153715371 [113, 113, 113, 113, 113, 113, 113, 113, 113]
以上输出均为真正的多解.但是我想在第一个比赛之前停下来(休息)，而不是给出所有解决方案.
Above outputs are all true mutiple solutions. But I want to stop ( break) by first match, instead of giving all solutions.
我曾经使用过return True，但是当它抱怨功能失常时，它在这里并没有真正的作用.
I have used return True, but it is not really working here, when it complains of being out of function.
第二个问题是第二个for循环for x in np.linspace(1,h,10000):我真的想用很多小数来运行它以获得最佳结果，但是它会变慢并且需要很长时间来计算.有没有办法加快速度?
Second issue here is second for-loop for x in np.linspace(1,h,10000): I really want to run it with many decimals to get a best result, but it slows down and takes long time to calculate. Is there a way to speed up?
也许将上述代码行定义为函数会更有效.
Perhaps defining above codelines into functions would work more efficient.
推荐答案
最简单，最直接的解决方案是将所有代码移到一个函数中
The easiest and most straight-forward solution is to move all of your code inside a function
def calculate_results(...args...): # Bad name, use a more suitable one
...
for n in range(0, 10): # <------- First for-loop
...
for x in np.linspace(1,h,10000): # <-------- Second for-loop
...
if 0 <= (float(Nrd) - Ned) <= 1:
return Nrd, x, Asi
然后再调用该函数
Nrd, x, Asi = calculate_results(...)
print(Nrd, x, Asi)
使用一个真正描述该功能本质的功能名称，并且仅在该功能内执行单个操作.然后return语句清楚地表明此任务已完成.
Use a function name that really describes the essence of what the functions is doing and do only that single thing within the function. Then the return statement makes it clear that this task is now done.
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