Pandas Dataframe，列表列，創建累積列表集合列，並按記錄差異記錄

Question

我有一個簡單的數據框df，列表爲lists。我想根據lists生成3個額外的列。

df樣子：

import pandas as pd 
lists={1:[[1]],2:[[1,2,3]],3:[[2,9,7,9]],4:[[2,7,3,5]]} 
#create test dataframe 
df=pd.DataFrame.from_dict(lists,orient='index') 
df=df.rename(columns={0:'lists'}) 
df 

      lists 
1   [1] 
2  [1, 2, 3] 
3 [2, 9, 7, 9] 
4 [2, 7, 3, 5] 

我想df看起來像這樣：

lists  cumset  adds drops 
1 [1]  {1}   {1}  {} 
2 [1,2,3] {1,2,3}  {2,3} {} 
3 [2,9,7,9] {1,2,3,7,9} {7,9} {3} 
4 [2,7,3,5] {1,2,3,5,7,9} {3,5} {9} 

基本上我需要弄清楚如何創建cumset（某些類型的應用？（是那麼對於增加和刪除，基本上我們想要比較df.lists和df.lists.shift（），並確定哪些項目是新的，哪些項目丟失了，可能類似於：

df['adds']=df[['lists',df.lists.shift()]].apply(lambda x: {i for i in x.lists if i not in x.lists.shift()}, axis=1) 

祝你玩得開心，謝謝。

Answer 1

您可以使用pandas.DataFrame.cumsum，使累積的柱，並與臺而不是列表列，並使用pandas.DataFrame.shift使「add」和「降」列：

import pandas as pd 
import numpy as np 


df['cumset'] = df['lists'].cumsum().apply(lambda x: np.unique(x)) 
df['sets'] = df['lists'].apply(lambda x: set(x)) 

shifted = df['sets'].shift(1).apply(lambda x: x if not pd.isnull(x) else set()) 

df['add'] = df['sets'] - shifted 
df['drop'] = shifted - df['sets'] 
df = df.drop('sets', axis=1) 

print(df) 
#-->Output: 
      lists    cumset  add drop 
1   [1]     [1]  {1}  {} 
2  [1, 2, 3]   [1, 2, 3] {2, 3}  {} 
3 [2, 9, 7, 9]  [1, 2, 3, 7, 9] {9, 7} {1, 3} 
4 [2, 7, 3, 5] [1, 2, 3, 5, 7, 9] {3, 5}  {9} 

Answer 2

我認爲你可以使用Series.cumsum + Series.shift + Series.iat，主要爲Series.applyset S：

df['cumset'] = df['lists'].cumsum().apply(set) 
lists_sets = df['lists'].apply(set) 
lists_shifted = lists_sets.shift() 
#replace first value - NaN to set 
lists_shifted.iat[0] = set() 
lists_shifted = lists_shifted.apply(set) 
df['add'] = lists_sets - lists_shifted 
df['drop'] = lists_shifted - lists_sets 
print (df) 
      lists    cumset  add drop 
1   [1]     {1}  {1}  {} 
2  [1, 2, 3]   {1, 2, 3} {2, 3}  {} 
3 [2, 9, 7, 9]  {1, 2, 3, 9, 7} {9, 7} {1, 3} 
4 [2, 7, 3, 5] {1, 2, 3, 5, 7, 9} {3, 5}  {9}