我知道這不是最簡潔的代碼塊，正在尋找簡化它的方法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()這樣做的目的是返回一個數據框，以百分比形式顯示給定年份中每個“標題”出現的次數。這是樣本輸入這是示例輸出
2 回答
MYYA
TA貢獻1868條經驗 獲得超4個贊
一般形式將是
ser = []
for year in my_names:
ser.append(
x/sum(fb_posts2[fb_posts2['year']==year].groupby('title').size()) * 100
或者，作為列表理解：
ser = [x/sum(fb_posts2[fb_posts2['year']==year].groupby('title').size()) * 100
for year in my_names]
那應該能夠取代你的 3 組 10 重復行。
ibeautiful
TA貢獻1993條經驗 獲得超6個贊
因此，我通過在 2009-2018 年的列表中運行 for 循環并應用函數將每個列表中的每個項目除以每個列表中的總計數并將其乘以 100，然后使用 pd.DataFrame 創建來簡化此代碼一個數據框并指定我將使用的索引名稱
a = [x/sum(x)*100 for x in [nine,ten,eleven,twelve,thirteen,fourteen,fifteen,sixteen,seventeen,eighteen]]
pd.DataFrame(a, index= my_names)
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