我有一個像這樣的多列的df（還有更多的列和行）：
df = pd.DataFrame([
{'ID': 1,'date': '2022-01-01', 'fruit_code':'[100,99,300]', 'vegetable_code':'[1000,2000,3000]','supermarket':'xy',},
{'ID': 2,'date': '2022-01-01', 'fruit_code':'[67,200,87]', 'vegetable_code':'[5000]','supermarket':'z, m'},
{'ID': 3,'date': '2021-01-01', 'fruit_code':'[100,5,300,78]', 'vegetable_code':'[7000,2000,3000]','supermarket':'wf, z'},
{'ID': 4,'date': '2020-01-01', 'fruit_code':'[77]', 'vegetable_code':'[1000]','supermarkt':'wf'},
{'ID': 5,'date': '2022-15-01', 'fruit_code':'[100,200,546,33]', 'vegetable_code':'[4000,2000,3000]','supermarket':'t, wf'},
{'ID': 6,'date': '2002-12-01', 'fruit_code':'[64,2]', 'vegetable_code':'[6000,8000,1000]','supermarket':'k' },
{'ID': 7,'date': '2018-12-01', 'fruit_code':'[5]', 'vegetable_code':'[6000,8000,1000]','supermarket':'p' }
])
我預期的df最終應該是這樣的：
df = pd.DataFrame([
{'ID': 1,'date': '2022-01-01', 'fruit_code':'[100,99,300]', 'vegetable_code':'[1000,2000,3000]','supermarket':'xy','new_col_1':'all'},
{'ID': 2,'date': '2022-01-01', 'fruit_code':'[67,200,87]', 'vegetable_code':'[5000]','supermarket':'z, m','new_col_1':'[5000]'},
{'ID': 3,'date': '2021-01-01', 'fruit_code':'[100,5,300,78]', 'vegetable_code':'[7000,2000,3000]','supermarket':'wf, z','new_col_1':'all'},
{'ID': 4,'date': '2020-01-01', 'fruit_code':'[77]', 'vegetable_code':'[1000]','supermarket':'wf','new_col_1':'[77]'},
{'ID': 5,'date': '2022-15-01', 'fruit_code':'[100,200,546,33]', 'vegetable_code':'[4000,2000,3000]','supermarket':'t, wf','new_col_1':'all'},
{'ID': 6,'date': '2002-12-01', 'fruit_code':'[64,2]', 'vegetable_code':'[6000,8000,1000]','supermarket':'k', 'new_col_1':'[64]', 'new_col_2':'[2]'},
{'ID': 7,'date': '2018-12-01', 'fruit_code':'[5]', 'vegetable_code':'[6000,8000,1000]','supermarket':'p' ,'new_col_1':'all'}
])
這里有多個條件我想在 cols fruit_code 和vegetable_code 上應用以獲得兩個新列：
更新
def fruits_vegetable(row):
if len(str(row['fruit_code'])) == 1: # fruit_code in new_col_1
row['new_col_1'] = row['fruit_code']
elif len(str(row['fruit_code'])) == 1 and len(str(row['vegetable_code'])) > 1: # write "all" in new_col_1
row['new_col_1'] = 'all'
elif len(str(row['fruit_code'])) > 2 and len(str(row['vegetable_code'])) == 1: # vegetable_code in new_col_1
row['new_col_1'] = row['vegetable_code']
elif len(str(row['fruit_code'])) > 3 and len(str(row['vegetable_code'])) > 1: # write "all" in new_col_1
row['new_col_1'] = 'all'
elif len(str(row['fruit_code'])) == 2 and len(str(row['vegetable_code'])) >= 0: # fruit 1 new_col_1 & fruit 2 new_col_2
row['new_col_1'] = row['fruit_code'][0]
row['new_col_2'] = row['fruit_code'][1]
return row
df = df.apply(fruits_vegetable, axis=1)
我仍然卡住了，現在我在第一列的某些行中得到“全部”，但第二列沒有改變。
如果有人有一些見解，那就太好了。
謝謝，非常感謝
uj5u.com熱心網友回復：
首先必須將串列的字串 repr 轉換ast.literal_eval為串列，然后為檢查長度洗掉轉換為字串。如果需要一個元素串列而不是標量使用[]infruit[0]和fruit[1]最后更改條件的順序len(fruit) == 1，也更改len(fruit) > 3為len(fruit) > 2匹配第一行：
def fruits_vegetable(row):
fruit = ast.literal_eval(row['fruit_code'])
vege = ast.literal_eval(row['vegetable_code'])
if len(fruit) == 1 and len(vege) > 1: # write "all" in new_col_1
row['new_col_1'] = 'all'
elif len(fruit) > 2 and len(vege) == 1: # vegetable_code in new_col_1
row['new_col_1'] = vege
elif len(fruit) > 2 and len(vege) > 1: # write "all" in new_col_1
row['new_col_1'] = 'all'
elif len(fruit) == 2 and len(vege) >= 0:# fruit 1 new_col_1 & fruit 2 new_col_2
row['new_col_1'] = [fruit[0]]
row['new_col_2'] = [fruit[1]]
elif len(fruit) == 1: # fruit_code in new_col_1
row['new_col_1'] = fruit
return row
df = df.apply(fruits_vegetable, axis=1)
print (df)
ID date fruit_code new_col_1 new_col_2 supermarket \
0 1 2022-01-01 [100,99,300] all NaN xy
1 2 2022-01-01 [67,200,87] [5000] NaN z, m
2 3 2021-01-01 [100,5,300,78] all NaN wf, z
3 4 2020-01-01 [77] [77] NaN NaN
4 5 2022-15-01 [100,200,546,33] all NaN t, wf
5 6 2002-12-01 [64,2] [64] [2] k
6 7 2018-12-01 [5] all NaN p
supermarkt vegetable_code
0 NaN [1000,2000,3000]
1 NaN [5000]
2 NaN [7000,2000,3000]
3 wf [1000]
4 NaN [4000,2000,3000]
5 NaN [6000,8000,1000]
6 NaN [6000,8000,1000]
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