Python循环遍历不同长度的嵌套序列

Question

我正在尝试制作一个简单的程序，通过参考关键字列表来分配代码给课程。

现在，我能够处理一个关键字列表，其中每行关键字的长度固定为2：

#The list of keyword with length fixed to 2
keyword = pd.DataFrame({
        'code':['001','002','003'], 
        'keyword': [
                ['edu|teach','primary sch|secondary sch|junior sch|preliminary sch'],  # length = 2
                ['elderly|disabled|special','care'],        # length = 2
                ['digital|social media','marketing']]       # length = 2
            })

# The list of educational programmed for which codes are to be assigned
course = pd.DataFrame({
        'course': 
            ['certificate in digital marketing',
             'certificate in elderly care',
             'diploma in primary school education',
             'bachelor in traditional chinese medicine',
             'master of law']
            })

# To generate shortlist of coded courses

courseresult = pd.DataFrame()
for i in range(0,len(keyword['keyword'])):
    courseshortlist = course[
            (course.course.str.contains(keyword['keyword'][i][0]) & course.course.str.contains(keyword['keyword'][i][1])) 
           ]
    courseshortlist['autocode'] = keyword['code'][i]
    courseresult = courseresult.append(courseshortlist)

但是，我不确定如何处理长度可变的关键字列表的循环：

keyword_variable = pd.DataFrame({
        'code':['001','002','003','004','005'], 
        'keyword': [
                ['law'],                                # length = 1
                ['edu|teach','primary sch|secondary sch|junior sch|preliminary sch'], # length = 2
                ['elderly|disabled|special','care'],  # length = 2
                ['digital|social media','marketing'], # length = 2
                ['traditional','chinese','medicine']  # length = 3
                ] 
            })

更新:我只是通过一些丑陋和笨拙的尝试和例外代码得到了我想要的：

courseresult = pd.DataFrame()
for i in range(0,len(keyword_variable['keyword'])):
    try: 
        condition0 = course.course.str.contains(keyword_variable['keyword'][i][0])
        condition1 = course.course.str.contains(keyword_variable['keyword'][i][1])
        condition2 = course.course.str.contains(keyword_variable['keyword'][i][2])
        condition = condition0 & condition1 & condition2
    except IndexError: 
        try: 
            condition0 = course.course.str.contains(keyword_variable['keyword'][i][0])
            condition1 = course.course.str.contains(keyword_variable['keyword'][i][1])
            condition = condition0 & condition1 
        except IndexError: 
            condition = course.course.str.contains(keyword_variable['keyword'][i][0])
    courseshortlist = course[(condition)]
    courseshortlist['autocode'] = keyword_variable['code'][i]
    courseresult = courseresult.append(courseshortlist)

courseresult
Out[1]: 
                                     course autocode
4                             master of law      001
2       diploma in primary school education      002
1               certificate in elderly care      003
0          certificate in digital marketing      004
3  bachelor in traditional chinese medicine      005

但我相信一定有更好的方法来做到这一点？非常感谢!

Answer 1

假设您并不真正需要将结果放在单独的DataFrame中：

for i in range(0,len(keyword_variable['keyword'])):
    condition = pd.Series([True]*len(course))
    for k in keyword_variable['keyword'][i]:
        condition = condition & course.course.str.contains(k)
    course.loc[condition, 'autocode'] = keyword_variable['code'][i]

print(course)

如果您确实需要一个新的副本，只需先创建一个副本，相同的解决方案。