假設(shè)我有兩個(gè) pandas DataFrame 即 df1, df2df1 = {name : [tom, jerry, jennifer, hafiz, kitty]}df2 = {name : [tom, jerry, alex, hafiz, samdin, unnar]}從這兩個(gè)數(shù)據(jù)集中，我想生成good_boy = [tom, jerry] # present in both the datasetsbad_boy = [jenifer, hafiz, kitty] # present in df1 but not in df2new_boy = [alex, samdin, unnar] # in df2 but not in df1實(shí)際數(shù)據(jù)集非常大，有數(shù)百萬(wàn)行，我嘗試進(jìn)行迭代檢查，但速度太慢了。Pandas 中是否已經(jīng)存在任何 tric（并行處理）。請(qǐng)幫我解決這個(gè)問(wèn)題，我的注意力是時(shí)間。謝謝
2 回答
慕標(biāo)琳琳
TA貢獻(xiàn)1830條經(jīng)驗(yàn) 獲得超9個(gè)贊
正如@QuangHoang 在評(píng)論中所說(shuō)，這里的關(guān)鍵是merge. 該indicator=True選項(xiàng)需要一個(gè)額外的_merge列來(lái)指示該行是否存在于其中一個(gè)數(shù)據(jù)幀（以及哪個(gè)數(shù)據(jù)幀）或兩者中：
df1 = pd.DataFrame({'name' : ['tom', 'jerry', 'jennifer', 'hafiz', 'kitty']})
df2 = pd.DataFrame({'name' : ['tom', 'jerry', 'alex', 'hafiz', 'samdin', 'unnar']})
tmp = pd.merge(df1, df2, how='outer', on='name', indicator=True)
good_boy = tmp.loc[tmp['_merge']=='both', 'name'].to_list()
bad_boy = tmp.loc[tmp['_merge']=='left_only', 'name'].to_list()
new_boy = tmp.loc[tmp['_merge']=='right_only', 'name'].to_list()
慕桂英3389331
TA貢獻(xiàn)2036條經(jīng)驗(yàn) 獲得超8個(gè)贊
您可以使用DataFrame.join
https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.join.html
good_boy = df1.join(df2, on = 'name', how = 'inner')[['name_left']].rename(columns = {'name_left' : 'name'})
bad_boy = df1[~df1['name'].isin(df2['name'].tolist())]
new_boy = df2[~df2['name'].isin(df1['name'].tolist())]
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