通过 while 循环的用户交互/跳出 2 while 循环
User interaction via while-loops / jumping out of 2 while loops
各位程序员大家好，
我是一个 Python 整体编程新手，我不知道如何最好地处理用户输入。我的一个朋友建议 while 循环，他们工作得很好，但现在我遇到了一个严重的问题。
该程序是关于询问用户他想要 3 种计算 Pi 的方法中的哪一种。 his/her 选择后，例如使用多边形计算 Pi 的方法在每次迭代后将角数加倍，它会询问用户想要查看多少位数字或应该计算多少个循环等。我还添加了一些代码来检查用户编写的内容，例如如果要求用户键入 y 或 n 但键入 b，程序会提醒他并再次询问。
这种风格导致了 while、inside while、inside while，现在看起来更像 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
我已经评论了受影响的行...
我无法用 main-while-loop 解决这个问题，所以我想问你我该如何解决这个问题。有没有更好的方法来处理用户输入并检查它是否正确？我被困在一个我希望我能去的地方，因为我无法想象任何其他解决方案。
希望您能帮助我，感谢您阅读这么长的代码！我很感激！ :)
holofox
正如罗伯特·罗斯尼在对 another question 的回答中所说：
My first instinct would be to refactor the nested loop into a function and use return to break out.
我的第一个想法是简化事情，请看这个"pseudo-code":
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
各位程序员大家好，
我是一个 Python 整体编程新手，我不知道如何最好地处理用户输入。我的一个朋友建议 while 循环，他们工作得很好，但现在我遇到了一个严重的问题。
该程序是关于询问用户他想要 3 种计算 Pi 的方法中的哪一种。 his/her 选择后，例如使用多边形计算 Pi 的方法在每次迭代后将角数加倍，它会询问用户想要查看多少位数字或应该计算多少个循环等。我还添加了一些代码来检查用户编写的内容，例如如果要求用户键入 y 或 n 但键入 b，程序会提醒他并再次询问。
这种风格导致了 while、inside while、inside while，现在看起来更像 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
我已经评论了受影响的行...
我无法用 main-while-loop 解决这个问题，所以我想问你我该如何解决这个问题。有没有更好的方法来处理用户输入并检查它是否正确？我被困在一个我希望我能去的地方，因为我无法想象任何其他解决方案。
希望您能帮助我，感谢您阅读这么长的代码！我很感激！ :)
holofox
正如罗伯特·罗斯尼在对 another question 的回答中所说：
My first instinct would be to refactor the nested loop into a function and use return to break out.
我的第一个想法是简化事情，请看这个"pseudo-code":
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