我的想法是为您的标识符设置字典数据结构，如下所示:

My idea is to set up a dictionary data structure for your identifiers like this:

datemap = { 'January' :  {'day' : 1, 'month' : 1, 'quarter' : 1}, 
            'February' : {'day' : 1, 'month' : 2, 'quarter' : 1}, 
            'March' :    {'day' : 1, 'month' : 3, 'quarter' : 1}, 
            # and so on ...
            'Spring' : {'day' : 1, 'month' : 1, 'quarter' : 1}, 
            'Summer' : {'day' : 1, 'month' : 4, 'quarter' : 2}, 
            'Fall' :   {'day' : 1, 'month' : 7, 'quarter' : 3}, 
            'Winter' : {'day' : 1, 'month' : 10, 'quarter' : 4}, 
            'Q1' : {'day' : 1, 'month' : 1, 'quarter' : 1}, 
            'Q2' : {'day' : 1, 'month' : 4, 'quarter' : 2}, 
            'Q3' : {'day' : 1, 'month' : 7, 'quarter' : 3}, 
            'Q4' : {'day' : 1, 'month' : 10, 'quarter' : 4}, 
            'Year' : {'day' : 1, 'month' : 1, 'quarter' : 1} }

然后，您可以通过查看第一个单词r['period'].split()[0](或年份的第二个单词)来转换给定值r['period']，如下所示:

Then you can transform a given value r['period'] by looking at the first word r['period'].split()[0] (or second word for the year) like this:

df['day'] = df.apply (lambda r: datemap[r['period'].split()[0]]['day'], axis=1)
df['month'] = df.apply (lambda r: datemap[r['period'].split()[0]]['month'], axis=1)
df['quarter'] = df.apply (lambda r: datemap[r['period'].split()[0]]['quarter'], axis=1)
df['year'] = df.apply (lambda r: "20" + r['period'].split()[1][-2:], axis=1)