试图找到一种方法来简化我的代码块

Question

我知道这不是最简洁的代码块，正在寻找简化它的方法nine = fb_posts2[fb_posts2['year']==2009].groupby('title').size()ten = fb_posts2[fb_posts2['year']==2010].groupby('title').size()eleven = fb_posts2[fb_posts2['year']==2011].groupby('title').size()twelve = fb_posts2[fb_posts2['year']==2012].groupby('title').size()thirteen = fb_posts2[fb_posts2['year']==2013].groupby('title').size()fourteen = fb_posts2[fb_posts2['year']==2014].groupby('title').size()fifteen = fb_posts2[fb_posts2['year']==2015].groupby('title').size()sixteen = fb_posts2[fb_posts2['year']==2016].groupby('title').size()seventeen = fb_posts2[fb_posts2['year']==2017].groupby('title').size()eighteen = fb_posts2[fb_posts2['year']==2018].groupby('title').size()a1 = lambda x: x/sum(nine)*100a2 = lambda x: x/sum(ten)*100a3 = lambda x: x/sum(eleven)*100a4 = lambda x: x/sum(twelve)*100a5 = lambda x: x/sum(thirteen)*100a6 = lambda x: x/sum(fourteen)*100a7 = lambda x: x/sum(fifteen)*100a8 = lambda x: x/sum(sixteen)*100a9 = lambda x: x/sum(seventeen)*100a10 = lambda x: x/sum(eighteen)*100nine = a1(nine)ten = a2(ten)eleven = a3(eleven) twelve = a4(twelve)thirteen = a5(thirteen)fourteen = a6(fourteen)fifteen = a7(fifteen)sixteen = a8(sixteen)seventeen = a9(seventeen)eighteen = a10(eighteen)my_names = [2009,2010,2011,2012,2013,2014,2015,2016,2017,2018]cols = ['link', 'post','shared','timeline','status']ser = [nine, ten, eleven, twelve, thirteen, fourteen, fifteen, sixteen, seventeen, eighteen]df = pd.concat(ser, axis=1, keys=my_names)df[2009].fillna(0, inplace=True)df[2011].fillna(0, inplace=True)df[2012].fillna(0, inplace=True)df = df.transpose()这样做的目的是返回一个数据框，以百分比形式显示给定年份中每个“标题”出现的次数。这是样本输入这是示例输出

Answer 1

一般形式将是

ser = []

for year in my_names:

    ser.append(

        x/sum(fb_posts2[fb_posts2['year']==year].groupby('title').size()) * 100

或者，作为列表理解：

ser = [x/sum(fb_posts2[fb_posts2['year']==year].groupby('title').size()) * 100

    for year in my_names]

那应该能够取代你的 3 组 10 重复行。

Answer 2

因此，我通过在 2009-2018 年的列表中运行 for 循环并应用函数将每个列表中的每个项目除以每个列表中的总计数并将其乘以 100，然后使用 pd.DataFrame 创建来简化此代码一个数据框并指定我将使用的索引名称

a = [x/sum(x)*100 for x in [nine,ten,eleven,twelve,thirteen,fourteen,fifteen,sixteen,seventeen,eighteen]]

pd.DataFrame(a, index= my_names)