这里是代码：
import itertools
# import list from file
a = [line.strip() for line in open("real.txt", 'r')]
b = [line.strip() for line in open("imag.txt", 'r')]
# create file contain restore informations
f = open("restore.txt","w+")
loop = 0
try:
for r in itertools.product(a,b):
loop +=1
print "processing : "+ r[1] + ": "+ r[0] + " ("+str(loop)+")"
except Exception, e:
print str(e)
f.write("iterazions seccession: " +str(loop)+"\n")
f.write("real number : " +r[1]+"\n")
f.write("imaginary unit: " + r[0]+ "\n")
示例输出
processing: 1 : 1i (1)
processing: 1 : 2i (2)
processing: 1 : 3i (3)
processing: 1 : 4i (4)
...
processing: 2000 : 174i (348000)
processing: 2000 : 175i (348001)
...and so forth
(it does all combinations of two lists)
问题是如果错误停止了迭代，那么有一种方法可以从上次迭代重启脚本而无需从头开始？ 我尝试保存文件中的最后一个值，但我不知道如何使用它。
P.S。我知道复数的cmath函数，但我更感兴趣的是恢复问题
更具体
如果错误停止迭代： 处理：2000：175i（348001） 有一种方法可以从 2000：175i（348001）迭代重启脚本吗？
答案 0 :(得分：3)
将try / except INSIDE置于for循环中 - 如果except不会引发，则for循环将继续。
for r in itertools.product(a,b):
try:
loop +=1
print "processing : "+ r[1] + ": "+ r[0] + " ("+str(loop)+")"
#enable for testing error
#if loop == 15:
#loop = loop/0
except Exception, e:
print str(e)
f.write("iterazions seccession: " +str(loop)+"\n")
f.write("real number : " +r[1]+"\n")
f.write("imaginary unit: " + r[0]+ "\n")
答案 1 :(得分：1)
通常，您会看到使用try ... except块，这些块在Python中相对便宜。
这不会停止执行你的循环，如本例所示（可以应用于你的情况）：
numbers = [1, 3, 2, 5, 0, 4, 5]
for number in numbers:
try:
print(10/number)
except ZeroDivisionError:
print("Please don't divide by 0...")
您将在Python 2.7.11中看到输出结果为：
10
3
5
2
Please don't divide by 0...
2
2
如果您的代码可以抛出多种类型的异常，您可以将它们链接起来：
numbers = [1, 3, 2, 5, 0, 4, 5]
for number in numbers:
try:
print(10/number)
except ZeroDivisionError:
print("Please don't divide by 0...")
except ValueError:
print("Something something ValueError")
except KeyError:
print("Something something KeyError")
如果你知道你的代码可以引发的所有类型的异常，你可以适当地处理它们，这通常是一个比异常的一般解决方案更好的解决方案。
关于您的代码的一般说明：
a和b在实数和虚数方面没有多大意义 - 为什么不使用real_nums和imag_nums？没有更多的打字和更具描述性。open(file) as f enumerate(iterable) []表示法访问 - 您可以直接使用元组中的元素。把所有这些放在一起：
import itertools
# import list from file
reals = [line.strip() for line in open("real.txt", 'r')]
imags = [line.strip() for line in open("imag.txt", 'r')]
# create file contain restore informations
with open("restore.txt", "w+") as f:
for i, (real, imag) in enumerate(itertools.product(reals, imags)):
try:
print "processing: {0}, {1}, ({2})".format(real, imag, i)
# Process data here
#enable for testing error
if i == 15:
i = i/0
except ZeroDivisionError:
# Handle exception here
continue
except Exception as e:
print str(e)
f.write("iterazions seccession: {0}\n".format(i))
f.write("real number: {0}\n".format(real))
f.write("imaginary unit: {0}\n".format(imag))
答案 2 :(得分：0)
所以，我的问题是“如何从头开始重新启动脚本而不从头开始？”。 首先我尝试hard way，即使它工作正常，我也会发现一种比这更简单的方法（亲吻吧？）。代码如下：
import itertools
import pickle #for restoring last data session
fname = "restore.pck"
reals = [line.strip() for line in open("real.txt", 'r')]
imags = [line.strip() for line in open("imag.txt", 'r')]
seed_count = 0 #number last processing cycle
count_ses = 0 #number current session cycle
count_tot = 0 #number total cycle processing so far
it_tot=len(reals)*len(imgs)
ret = raw_input("recover last session?: y/n")
if ret =="y":
with open(fname,"rb") as f:
rec= pickle.load(f)
print "last session data {0} {1} {2}".format(rec[0],rec[1],rec[2])
seed_count=rec[2]
seed_img=rec[1]
seed_rea=rec[0]
try:
print "total iterations", it_tot
for r in itertools.product(reals,imgs):
# do nothing until you get last interrupted cycle
count_tot+=1
if count_tot >= seed_count:
# and now start processing
count_ses +=1
print "processing: {0}, {1}, (total : {2}), (this session: {3}) {4}% )".format(r[0], r[1], count_tot, count_ses,(count_tot/len_tot*100))
except Exception as e:
print str(e)
#save data if an error occurred
store = r[0], r[1], count_tot
with open(fname,"wb") as f:
pickle.dump(store,f,pickle.HIGHEST_PROTOCOL)
f.close()
它只是让迭代运行而不处理任何内容，直到它到达最后一个处理周期然后重新开始计算。 当然一些“wiseguy”会说这不是pythonic方式或这段代码它是一团糟，不像我的，但这是我的第一个python程序，它的工作原理而且我很高兴：）