python - 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并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)*100
a2 = lambda x: x/sum(ten)*100
a3 = lambda x: x/sum(eleven)*100
a4 = lambda x: x/sum(twelve)*100
a5 = lambda x: x/sum(thirteen)*100
a6 = lambda x: x/sum(fourteen)*100
a7 = lambda x: x/sum(fifteen)*100
a8 = lambda x: x/sum(sixteen)*100
a9 = lambda x: x/sum(seventeen)*100
a10 = lambda x: x/sum(eighteen)*100
nine = 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，以百分比形式显示每个“标题”在给定年份出现的次数。

这是示例输入

这是示例输出

标签： pythonpandasdata-science

Answer 1

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)