我有一個包含微小資料的大資料框，如下所示，我在其中執行一些功能然后洗掉列。
| 約會時間 | E10 | E11 | E12 | E132 | E21 | E31 | E32 | E37 | E40 | E43 | E44 | E45 | E47 | E65 | E8 | E28 | E41 | E46 | E48 | E50 | E51 | E52 | E53 | E54 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 24-11-2021 07:30 | 22 | 122 | 62 | 232 | 55 | 874.2 | 32.8 | 351.2 | 1.4 | 0.3 | 5.1 | 4 | 1 | 24.2 | 76 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 24-11-2021 07:31 | 23 | 120 | 60 | 232 | 55 | 0 | 33.3 | 0 | 1.3 | 0.3 | 0 | 2 | 1 | 24.7 | 80 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 24-11-2021 07:32 | 22 | 123 | 61 | 208 | 54 | 0 | 32.9 | 0 | 1 | 0.3 | 0 | 3 | 1 | 24.7 | 79 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 24-11-2021 07:33 | 21 | 120 | 54 | 296 | 55 | 0 | 33 | 0 | 0.9 | 0.3 | 0 | 2 | 1 | 24.9 | 79 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 24-11-2021 07:34 | 21 | 122 | 58 | 272 | 50 | 0 | 32.4 | 0 | 0.9 | 0.3 | 0 | 3 | 1 | 24.3 | 77 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
# Arithmetic Functions
df['E08'] = df['E8'] df['E132'].multiply(0.1467)
df['E037'] = df['E37'] - df['E65']
df['E032'] = df['E32'] - df['E40']
df['E010'] = df['E10'] df['E45'] (df['E28'] - df['E48'])
df['E011'] = df['E11'] - df['E44'] - df['E43']
df['E021'] = df['E21'] - df['E50'] - df['E51'] - df['E52'] - df['E53'] - df['E54']
df['E031'] = df['E31'] - df['E41']
# Drop
df.drop([
'E8', 'E132', 'E37', 'E65', 'E32', 'E40', 'E10', 'E45',
'E48', 'E11', 'E44', 'E43', 'E21', 'E50', 'E51', 'E52',
'E53', 'E54', 'E31', 'E41'], axis=1, inplace=True)
我想知道哪個是有效的方法
問題 1：創建一個新列或保持相同的列。示例：選項 1：df['E8'] = df['E8'] df['E132'].multiply(0.1467) 或選項 2：df['E08'] = df['E8'] df[ 'E132'].multiply(0.1467)
問題 2：我應該像上面一樣單獨應用洗掉代碼，還是有一種方法可以在創建 E08 后洗掉“E8”和“E132”。我試過 df['E08'] = df['E8'] df['E132'].multiply(0.1467).drop(['E8', 'E132'], axis=1, inplace=True)-它的不作業
uj5u.com熱心網友回復：
如果要使用然后從原始使用中洗掉列DataFrame.pop，則drop沒有必要：
df['E08'] = df.pop('E8') df.pop('E132').multiply(0.1467)
df['E037'] = df.pop('E37') - df.pop('E65')
...
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