Question

我有一个DataFrame，看起来像这样：

import Data.List

keepl k n [] = []
keepl k 1 (x:xs) = x : (keepl k k xs)
keepl k n (x:xs) = keepl k (n-1) xs

main = do
 print $ keepl 3 3 [1,2,3,4,5,6,7,8,9]

我想添加一个新列，在Excel中计算如下：= IF（（B2-B1）＆gt; 0; B2-B1; 0），所以它应该如下所示：

    Date            Price
0   Jan 04, 1999    1.1827
1   Jan 05, 1999    1.1764
2   Jan 06, 1999    1.1618
3   Jan 07, 1999    1.1709
4   Jan 08, 1999    1.1575
5   Jan 11, 1999    1.1502
6   Jan 12, 1999    1.1572
7   Jan 13, 1999    1.1673
8   Jan 14, 1999    1.1698
9   Jan 15, 1999    1.1557

实际上，我有一个解决方案，但我对此并不满意：

    Date            Price   up
0   Jan 04, 1999    1.1827  0.0000
1   Jan 05, 1999    1.1764  0.0000
2   Jan 06, 1999    1.1618  0.0000
3   Jan 07, 1999    1.1709  0.0091
4   Jan 08, 1999    1.1575  0.0000
5   Jan 11, 1999    1.1502  0.0000
6   Jan 12, 1999    1.1572  0.0070
7   Jan 13, 1999    1.1673  0.0101
8   Jan 14, 1999    1.1698  0.0025
9   Jan 15, 1999    1.1557  0.0000

还有更优雅的解决方案吗？谢谢！

Answer 1

diff使用where：

dif = df['Price'].diff()
df['up'] = dif.where(dif > 0, 0)
#alternative
#df['up'] = np.where(dif > 0, dif, 0)
print (df)
           Date   Price      up
0  Jan 04, 1999  1.1827  0.0000
1  Jan 05, 1999  1.1764  0.0000
2  Jan 06, 1999  1.1618  0.0000
3  Jan 07, 1999  1.1709  0.0091
4  Jan 08, 1999  1.1575  0.0000
5  Jan 11, 1999  1.1502  0.0000
6  Jan 12, 1999  1.1572  0.0070
7  Jan 13, 1999  1.1673  0.0101
8  Jan 14, 1999  1.1698  0.0025
9  Jan 15, 1999  1.1557  0.0000

根据现有列的条件语句将新列添加到pandas数据框中

1 个答案: