Hello其他程序员，
我是一个Python和整体编程新手，我不知道如何处理用户输入bestimmt。我的一个朋友建议循环，他们工作得很好，但现在我有一个严重的问题。
该程序是关于向用户询问3种方法中的哪一种来计算他想要的Pi。在他/她的选择之后，例如使用多边形计算Pi的方法是在每次迭代后将角数加倍，它会询问用户想要看多少位数或者应该计算多少循环等等。我还添加了一些代码来检查用户所写的内容，例如如果要求用户键入y或n但键入b，则程序会提醒他并再次询问。
这种风格导致了内部时间，内部时间，内部时间，现在看起来更像是HTML代码而不是Python，并且理解起来有点复杂......
看看：
#!/usr/bin/python
# -*- coding: iso-8859-15 -*-
prgrm_ctrl_main_while_start = 0
while prgrm_ctrl_main_while_start == 0:
print "Please select the type of algorithm you want to calculate Pi."
usr_ctrl_slct_algrthm = raw_input("'p' to use polygons, 'm' to use the Monte-Carlo algorithm or 'c' for the Chudnovsky algorithm: ")
print " "
if usr_ctrl_slct_algrthm == 'p':
#import libraries
import time
from mpmath import *
#starting values
n_corners = mpf(8)
counter = 0
while_loop_check_itertions = 0
while while_loop_check_itertions == 0:
print "Pi will be calculated more precisely the more iterations you select but it'll also take longer!"
loops = int(input("Please select number of iterations (1 - 10.000.000): "))
print " "
if 1 <= loops <= 10000000:
loops = loops - 1
decimals = int(input("Please select the amount of decimals you want to see: "))
print " "
decimals = decimals + 1
starttime = time.clock()
mp.dps = decimals
while counter <= loops:
counter = counter + 1
n_corners = mpf(n_corners) * mpf(2)
circle = mpf(360)
alpha = mpf(circle) / mpf(n_corners)
beta = mpf(alpha) / mpf(2)
radiansBeta = mpf(mp.radians(beta))
opp_leg = mpf(mp.sin(radiansBeta))
edge_lngth = mpf(opp_leg) * mpf(2)
circmfrnce = mpf(n_corners) * mpf(edge_lngth)
pi = mpf(circmfrnce) / mpf(2)
print "Pi: ", mpf(pi)
print "Time to calculate: ", time.clock() - starttime, "seconds"
print " "
prgrm_ctrl_p_algrthm_while_start = 0
while prgrm_ctrl_p_algrthm_while_start == 0:
usr_ctrl_slct_p_algrthm = raw_input("Would you like to try this algorithm again? (y/n): ")
print " "
if usr_ctrl_slct_p_algrthm == 'y':
usr_ctrl_slct_p_algrthm_slction = 0
break
elif usr_ctrl_slct_p_algrthm == 'n':
usr_ctrl_slct_p_algrthm_slction = 1
break
else:
print "You must either type 'y' or 'n'!"
continue
if usr_ctrl_slct_p_algrthm_slction == 0:
continue
else:
usr_ctrl_slct_diffrt_algrthm_while_start = 0
while usr_ctrl_slct_diffrt_algrthm_while_start == 0:
usr_ctrl_slct_diffrt_algrthm = raw_input("Do you want to use another algorithm? (y/n): ")
print " "
if usr_ctrl_slct_diffrt_algrthm == 'y':
usr_ctrl_slct_diffrt_algrthm_slction = 0
break
elif usr_ctrl_slct_diffrt_algrthm == 'n':
print "See you next time!"
print " "
usr_ctrl_slct_diffrt_algrthm_slction = 1
break
else:
print "You must either type 'y' or 'n'!"
continue
if usr_ctrl_slct_diffrt_algrthm_slction == 0: #The program gets to this line and in case of the user selecting 'y' should continue in the first while-loop
continue
else: #if the user says 'n' the program should interrupt and stop, in the best case it should close the command line (in my case IDLE)
break #instead of doing all this the code gets executed again in the 2nd while loop not the 1st
"""NOTE: I also know the lines above continue or quit the 2nd while-loop"""
elif loops < 1:
print "Please select at least one iteration!"
while_loop_check_algrthm_slct = 0
while while_loop_check_algrthm_slct == 0:
usr_ctrl_slct_algrthm_p_try_agn = raw_input("Do you want to try it again? (y/n): ")
print " "
if usr_ctrl_slct_algrthm_p_try_agn == 'y':
usr_ctrl_slct_algrthm_p_try_agn_slction = 0
break
elif usr_ctrl_slct_algrthm_p_try_agn == 'n':
usr_ctrl_slct_algrthm_p_try_agn_slction = 1
break
else:
print "You must either type 'y' or 'n'!"
continue
if usr_ctrl_slct_algrthm_p_try_agn_slction == 1:
break
else:
continue
else:
print "The maximum amount of iterations is 10 million!"
while_loop_check_algrthm_slct = 0
while while_loop_check_algrthm_slct == 0:
usr_ctrl_slct_algrthm_p_try_agn = raw_input("Do you want to try it again? (y/n): ")
print " "
if usr_ctrl_slct_algrthm_p_try_agn == 'y':
usr_ctrl_slct_algrthm_p_try_agn_slction = 0
break
elif usr_ctrl_slct_algrthm_p_try_agn == 'n':
usr_ctrl_slct_algrthm_p_try_agn_slction = 1
break
else:
print "You must either type 'y' or 'n'!"
continue
if usr_ctrl_slct_algrthm_p_try_agn_slction == 1:
break
else:
continue
我评论了影响的行......
我无法解决主循环问题，因此我问你如何解决这个问题。是否有更好的方法来处理用户输入并检查它是否正确？我已经陷入了一个我希望自己已经转向的地步，因为我无法想象任何其他解决方案。
希望您能帮助我并感谢您阅读这段长代码！我很感激！ ：）
holofox
答案 0 :(得分：0)
正如罗伯特罗斯尼在回答another question时所说的那样：
我的第一直觉是将嵌套循环重构为函数并使用return来突破。
答案 1 :(得分：0)
我的第一个想法是简化事情，请看这个＆＃34;伪代码＆＃34;：
while True: #select algorithm type
#code: ask for algorithm
if answer not in "pmc":
#some message
continue #non-valid algorithm selected, ask again
#...
if answer == 'p':
while True: #select iteration range
#code: ask for iteration
if answer_not_in_range:
#some message
continue #non-valid iteration value, ask again
#...
#perform calculation upon valid selections
#...
#calculation ends
#code: ask for try again with other iteration (algorithm stays same)
if answer != "yes":
break #break out of "iteration" loop
#else loop continues asking for new iteration
#code: ask for try again with other algorithm
if answer != "yes":
break #"algorithm" loop ends, program ends