Question

该脚本执行SQL查询并返回如下结果：

subtract,'a','wrong'
subtract,'b','wrong'
add,a,'wrong'
add,b,'wrong'
add,c,'correct'
add,d,'wrong'
subtract,'a','wrong'
subtract,'b','wrong'

我遍历结果以逐行读取并将每个元素存储在变量中，但这是我不知道下一步该做什么的地方。

flag = 0
for rec in allRecords:
    operation = rec[0]
    value = rec[1]
    answer = rec[2]

    #if flag == 1:
       #pass
    #else:
        if operation == 'add':
            #start an inside loop to 'read ahead' and continue if operation == 'add' and stop when operation != 'add'
            #find 'c' inside this loop and get the 'correct' element which is next to it and store in a new variable.
            #break the loop to go back to main loop
            #getVar = 'correct'
            #print(getVar)
            #flag = 1 
        else:
            flag = 0

        #after breaking out of the loop above, continue to the next records
        print(rec)

期望的输出：

correct
add,a,'wrong'
add,b,'wrong'
add,c,'correct'
add,d,'wrong'

为什么我这样做？我想先显示正确的答案，然后列出其余的选项。也练习编程。

我为什么要问这里？我已经耗尽了所有的资源，我真的陷入困境，需要指导。我搜索了我从所有试验中得到的所有错误，并且找不到答案。

尽力解释。我是编程和学习python的新手。感谢您的帮助。

Answer 1

一种方法是遍历记录两次：

for operation, value, answer in allRecords:
    if answer == 'correct':
        print 'correct'

for operation, value, answer in allRecords:
    if answer != 'correct':
        print (operation, value, answer)

Answer 2

您可以先制作所有添加项的子集，然后从该子集中获取正确的答案。

# Filter all additions from the results
additions = [[o, v, a] for o, v, a in results if o == 'add']

# Filter the addition with the correct answer
correct = [v for o, v, a in additions if a == 'correct']

# Output information
print "Correct : {}".format(correct)
for addition in additions:
    print "{}, {}, {}".format(addition[0], addition[1], addition[2])

上面的代码输出，

Correct : ['c']
add, a, wrong
add, b, wrong
add, c, correct
add, d, wrong

在Python

2 个答案: